Using curly brackets in multidimensional array












0















I have an array of ID's used to execute mysqli queries in a foreach loop (the below code is just an example - the array will contain more than 50+ elements/ID's). For each array element/ID I query an array which I then include in a compData array. However, I am unsure of how to dynamically define the arrays in the foreach loop. I would very much appreciate if you can check whether I am using ${$val} correctly in the below or if something else is wrong. Thanks.



  $idArray  = array(
"List1",
"List2",
);

$compData = array();
foreach($idArray as $val) {
$sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = {$val}");
${$val} = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
${$val}['data'] = $r[$n];
}
}
$compData[{$val}] = ${$val}
}

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}


The above code does not work and does not echo any data from the $compData array. Below is an example that works just fine where I execute the code using the ID's directly.



  $sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = 'List1'");
$List1 = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
$List1['data'] = $r[$n];
}
}

$companyData = array();
$companyData['List1'] = $List1;

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}









share|improve this question

























  • There is rarely any use for variable variables that can't be solved with an associative array.

    – miken32
    Nov 15 '18 at 21:32











  • What exactly is the problem you're having?

    – miken32
    Nov 15 '18 at 21:33











  • My problem is that the data is not being added to the $compData array. The echo does not return anything. I am fairly sure that there is an error in the $compData[{$val}] = ${$val . "Array"} line.

    – ChartProblems
    Nov 15 '18 at 21:37











  • And I'm pretty sure your echo doesn't return anything because you're burying your data many levels deep. Your loops at the end there don't account for the fact that you've put things under the data element of the array, for example.

    – miken32
    Nov 15 '18 at 21:44











  • I have tried to add more explanation and included a piece of code that works as intended. Hopefully this will clarify what the issue is?

    – ChartProblems
    Nov 15 '18 at 21:46
















0















I have an array of ID's used to execute mysqli queries in a foreach loop (the below code is just an example - the array will contain more than 50+ elements/ID's). For each array element/ID I query an array which I then include in a compData array. However, I am unsure of how to dynamically define the arrays in the foreach loop. I would very much appreciate if you can check whether I am using ${$val} correctly in the below or if something else is wrong. Thanks.



  $idArray  = array(
"List1",
"List2",
);

$compData = array();
foreach($idArray as $val) {
$sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = {$val}");
${$val} = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
${$val}['data'] = $r[$n];
}
}
$compData[{$val}] = ${$val}
}

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}


The above code does not work and does not echo any data from the $compData array. Below is an example that works just fine where I execute the code using the ID's directly.



  $sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = 'List1'");
$List1 = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
$List1['data'] = $r[$n];
}
}

$companyData = array();
$companyData['List1'] = $List1;

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}









share|improve this question

























  • There is rarely any use for variable variables that can't be solved with an associative array.

    – miken32
    Nov 15 '18 at 21:32











  • What exactly is the problem you're having?

    – miken32
    Nov 15 '18 at 21:33











  • My problem is that the data is not being added to the $compData array. The echo does not return anything. I am fairly sure that there is an error in the $compData[{$val}] = ${$val . "Array"} line.

    – ChartProblems
    Nov 15 '18 at 21:37











  • And I'm pretty sure your echo doesn't return anything because you're burying your data many levels deep. Your loops at the end there don't account for the fact that you've put things under the data element of the array, for example.

    – miken32
    Nov 15 '18 at 21:44











  • I have tried to add more explanation and included a piece of code that works as intended. Hopefully this will clarify what the issue is?

    – ChartProblems
    Nov 15 '18 at 21:46














0












0








0








I have an array of ID's used to execute mysqli queries in a foreach loop (the below code is just an example - the array will contain more than 50+ elements/ID's). For each array element/ID I query an array which I then include in a compData array. However, I am unsure of how to dynamically define the arrays in the foreach loop. I would very much appreciate if you can check whether I am using ${$val} correctly in the below or if something else is wrong. Thanks.



  $idArray  = array(
"List1",
"List2",
);

$compData = array();
foreach($idArray as $val) {
$sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = {$val}");
${$val} = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
${$val}['data'] = $r[$n];
}
}
$compData[{$val}] = ${$val}
}

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}


The above code does not work and does not echo any data from the $compData array. Below is an example that works just fine where I execute the code using the ID's directly.



  $sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = 'List1'");
$List1 = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
$List1['data'] = $r[$n];
}
}

$companyData = array();
$companyData['List1'] = $List1;

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}









share|improve this question
















I have an array of ID's used to execute mysqli queries in a foreach loop (the below code is just an example - the array will contain more than 50+ elements/ID's). For each array element/ID I query an array which I then include in a compData array. However, I am unsure of how to dynamically define the arrays in the foreach loop. I would very much appreciate if you can check whether I am using ${$val} correctly in the below or if something else is wrong. Thanks.



  $idArray  = array(
"List1",
"List2",
);

$compData = array();
foreach($idArray as $val) {
$sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = {$val}");
${$val} = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
${$val}['data'] = $r[$n];
}
}
$compData[{$val}] = ${$val}
}

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}


The above code does not work and does not echo any data from the $compData array. Below is an example that works just fine where I execute the code using the ID's directly.



  $sth = mysqli_query($conn, "SELECT * FROM {$tableName} WHERE ID = 'List1'");
$List1 = array();
while($r = mysqli_fetch_array($sth)) {
For ($n = 1; $n <= $CI_NOYEARS; $n++){
$List1['data'] = $r[$n];
}
}

$companyData = array();
$companyData['List1'] = $List1;

foreach($compData['List1'] as $result) {
For ($n = 0; $n <= $CI_NOYEARS; $n++){
echo $result[$n];
}
}






php arrays mysqli






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 15 '18 at 22:06







ChartProblems

















asked Nov 15 '18 at 20:50









ChartProblemsChartProblems

10510




10510













  • There is rarely any use for variable variables that can't be solved with an associative array.

    – miken32
    Nov 15 '18 at 21:32











  • What exactly is the problem you're having?

    – miken32
    Nov 15 '18 at 21:33











  • My problem is that the data is not being added to the $compData array. The echo does not return anything. I am fairly sure that there is an error in the $compData[{$val}] = ${$val . "Array"} line.

    – ChartProblems
    Nov 15 '18 at 21:37











  • And I'm pretty sure your echo doesn't return anything because you're burying your data many levels deep. Your loops at the end there don't account for the fact that you've put things under the data element of the array, for example.

    – miken32
    Nov 15 '18 at 21:44











  • I have tried to add more explanation and included a piece of code that works as intended. Hopefully this will clarify what the issue is?

    – ChartProblems
    Nov 15 '18 at 21:46



















  • There is rarely any use for variable variables that can't be solved with an associative array.

    – miken32
    Nov 15 '18 at 21:32











  • What exactly is the problem you're having?

    – miken32
    Nov 15 '18 at 21:33











  • My problem is that the data is not being added to the $compData array. The echo does not return anything. I am fairly sure that there is an error in the $compData[{$val}] = ${$val . "Array"} line.

    – ChartProblems
    Nov 15 '18 at 21:37











  • And I'm pretty sure your echo doesn't return anything because you're burying your data many levels deep. Your loops at the end there don't account for the fact that you've put things under the data element of the array, for example.

    – miken32
    Nov 15 '18 at 21:44











  • I have tried to add more explanation and included a piece of code that works as intended. Hopefully this will clarify what the issue is?

    – ChartProblems
    Nov 15 '18 at 21:46

















There is rarely any use for variable variables that can't be solved with an associative array.

– miken32
Nov 15 '18 at 21:32





There is rarely any use for variable variables that can't be solved with an associative array.

– miken32
Nov 15 '18 at 21:32













What exactly is the problem you're having?

– miken32
Nov 15 '18 at 21:33





What exactly is the problem you're having?

– miken32
Nov 15 '18 at 21:33













My problem is that the data is not being added to the $compData array. The echo does not return anything. I am fairly sure that there is an error in the $compData[{$val}] = ${$val . "Array"} line.

– ChartProblems
Nov 15 '18 at 21:37





My problem is that the data is not being added to the $compData array. The echo does not return anything. I am fairly sure that there is an error in the $compData[{$val}] = ${$val . "Array"} line.

– ChartProblems
Nov 15 '18 at 21:37













And I'm pretty sure your echo doesn't return anything because you're burying your data many levels deep. Your loops at the end there don't account for the fact that you've put things under the data element of the array, for example.

– miken32
Nov 15 '18 at 21:44





And I'm pretty sure your echo doesn't return anything because you're burying your data many levels deep. Your loops at the end there don't account for the fact that you've put things under the data element of the array, for example.

– miken32
Nov 15 '18 at 21:44













I have tried to add more explanation and included a piece of code that works as intended. Hopefully this will clarify what the issue is?

– ChartProblems
Nov 15 '18 at 21:46





I have tried to add more explanation and included a piece of code that works as intended. Hopefully this will clarify what the issue is?

– ChartProblems
Nov 15 '18 at 21:46












2 Answers
2






active

oldest

votes


















1














It's a bit tough figuring out what your goal is based on the code you provided, but it seems far more complicated than it needs to be. I think what I've got here will replicate your desired output.



You should also be using prepared statements; they are safer, and take a lot of the overhead out of database queries when you're going to be repeating the same query multiple times. I've implemented them here.



$idArray = ["List1", "List2"];
$sth = $conn->prepare("SELECT * FROM `$tableName` WHERE ID = ?");

foreach($idArray as $val) {
$data = ;
$sth->bind_param("s", $val);
$sth->execute();
$result = $sth->get_result();
while ($r = $result->fetch_array()) {
$data = array_merge($data, array_splice($r, 1, $CI_NOYEARS));
}
$compData[$val]["data"] = $data;
}

foreach($compData["List1"]["data"] as $result) {
echo $result;
}





share|improve this answer


























  • This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

    – ChartProblems
    Nov 15 '18 at 22:26











  • Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

    – miken32
    Nov 15 '18 at 22:28



















0














I don't understand why you're using that curly-brace notation at all when you're just copying it over into a multidimensional compData array anyway? Why not just directly put the data into the array:



for ($n = 1; $n <= $CI_NOYEARS; $n++) {
$compdata[$val]['data'] = $r[$n];
}


(You'll also need to declare it as an array at the top level of the foreach of course.)



Side note: I would suggest not interpolating data directly into your query string like that, and instead using bound parameters: http://php.net/manual/en/mysqli-stmt.bind-param.php It's much more secure.






share|improve this answer
























  • Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

    – ChartProblems
    Nov 15 '18 at 21:06











  • Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

    – IceMetalPunk
    Nov 16 '18 at 18:35











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














It's a bit tough figuring out what your goal is based on the code you provided, but it seems far more complicated than it needs to be. I think what I've got here will replicate your desired output.



You should also be using prepared statements; they are safer, and take a lot of the overhead out of database queries when you're going to be repeating the same query multiple times. I've implemented them here.



$idArray = ["List1", "List2"];
$sth = $conn->prepare("SELECT * FROM `$tableName` WHERE ID = ?");

foreach($idArray as $val) {
$data = ;
$sth->bind_param("s", $val);
$sth->execute();
$result = $sth->get_result();
while ($r = $result->fetch_array()) {
$data = array_merge($data, array_splice($r, 1, $CI_NOYEARS));
}
$compData[$val]["data"] = $data;
}

foreach($compData["List1"]["data"] as $result) {
echo $result;
}





share|improve this answer


























  • This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

    – ChartProblems
    Nov 15 '18 at 22:26











  • Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

    – miken32
    Nov 15 '18 at 22:28
















1














It's a bit tough figuring out what your goal is based on the code you provided, but it seems far more complicated than it needs to be. I think what I've got here will replicate your desired output.



You should also be using prepared statements; they are safer, and take a lot of the overhead out of database queries when you're going to be repeating the same query multiple times. I've implemented them here.



$idArray = ["List1", "List2"];
$sth = $conn->prepare("SELECT * FROM `$tableName` WHERE ID = ?");

foreach($idArray as $val) {
$data = ;
$sth->bind_param("s", $val);
$sth->execute();
$result = $sth->get_result();
while ($r = $result->fetch_array()) {
$data = array_merge($data, array_splice($r, 1, $CI_NOYEARS));
}
$compData[$val]["data"] = $data;
}

foreach($compData["List1"]["data"] as $result) {
echo $result;
}





share|improve this answer


























  • This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

    – ChartProblems
    Nov 15 '18 at 22:26











  • Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

    – miken32
    Nov 15 '18 at 22:28














1












1








1







It's a bit tough figuring out what your goal is based on the code you provided, but it seems far more complicated than it needs to be. I think what I've got here will replicate your desired output.



You should also be using prepared statements; they are safer, and take a lot of the overhead out of database queries when you're going to be repeating the same query multiple times. I've implemented them here.



$idArray = ["List1", "List2"];
$sth = $conn->prepare("SELECT * FROM `$tableName` WHERE ID = ?");

foreach($idArray as $val) {
$data = ;
$sth->bind_param("s", $val);
$sth->execute();
$result = $sth->get_result();
while ($r = $result->fetch_array()) {
$data = array_merge($data, array_splice($r, 1, $CI_NOYEARS));
}
$compData[$val]["data"] = $data;
}

foreach($compData["List1"]["data"] as $result) {
echo $result;
}





share|improve this answer















It's a bit tough figuring out what your goal is based on the code you provided, but it seems far more complicated than it needs to be. I think what I've got here will replicate your desired output.



You should also be using prepared statements; they are safer, and take a lot of the overhead out of database queries when you're going to be repeating the same query multiple times. I've implemented them here.



$idArray = ["List1", "List2"];
$sth = $conn->prepare("SELECT * FROM `$tableName` WHERE ID = ?");

foreach($idArray as $val) {
$data = ;
$sth->bind_param("s", $val);
$sth->execute();
$result = $sth->get_result();
while ($r = $result->fetch_array()) {
$data = array_merge($data, array_splice($r, 1, $CI_NOYEARS));
}
$compData[$val]["data"] = $data;
}

foreach($compData["List1"]["data"] as $result) {
echo $result;
}






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 15 '18 at 22:22

























answered Nov 15 '18 at 22:11









miken32miken32

24.1k95073




24.1k95073













  • This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

    – ChartProblems
    Nov 15 '18 at 22:26











  • Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

    – miken32
    Nov 15 '18 at 22:28



















  • This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

    – ChartProblems
    Nov 15 '18 at 22:26











  • Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

    – miken32
    Nov 15 '18 at 22:28

















This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

– ChartProblems
Nov 15 '18 at 22:26





This works pretty much perfectly! Thank you very much. However, for some reason the echo includes the "List1" ID name as the first entry. For instance the echo currently is: "List1, 1, 2, 3..". Would you be able to tell me how I can prevent that?

– ChartProblems
Nov 15 '18 at 22:26













Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

– miken32
Nov 15 '18 at 22:28





Unless it's in the database, my code isn't outputting that value anywhere. Try print_r($compData) to see what's actually in the array.

– miken32
Nov 15 '18 at 22:28













0














I don't understand why you're using that curly-brace notation at all when you're just copying it over into a multidimensional compData array anyway? Why not just directly put the data into the array:



for ($n = 1; $n <= $CI_NOYEARS; $n++) {
$compdata[$val]['data'] = $r[$n];
}


(You'll also need to declare it as an array at the top level of the foreach of course.)



Side note: I would suggest not interpolating data directly into your query string like that, and instead using bound parameters: http://php.net/manual/en/mysqli-stmt.bind-param.php It's much more secure.






share|improve this answer
























  • Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

    – ChartProblems
    Nov 15 '18 at 21:06











  • Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

    – IceMetalPunk
    Nov 16 '18 at 18:35
















0














I don't understand why you're using that curly-brace notation at all when you're just copying it over into a multidimensional compData array anyway? Why not just directly put the data into the array:



for ($n = 1; $n <= $CI_NOYEARS; $n++) {
$compdata[$val]['data'] = $r[$n];
}


(You'll also need to declare it as an array at the top level of the foreach of course.)



Side note: I would suggest not interpolating data directly into your query string like that, and instead using bound parameters: http://php.net/manual/en/mysqli-stmt.bind-param.php It's much more secure.






share|improve this answer
























  • Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

    – ChartProblems
    Nov 15 '18 at 21:06











  • Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

    – IceMetalPunk
    Nov 16 '18 at 18:35














0












0








0







I don't understand why you're using that curly-brace notation at all when you're just copying it over into a multidimensional compData array anyway? Why not just directly put the data into the array:



for ($n = 1; $n <= $CI_NOYEARS; $n++) {
$compdata[$val]['data'] = $r[$n];
}


(You'll also need to declare it as an array at the top level of the foreach of course.)



Side note: I would suggest not interpolating data directly into your query string like that, and instead using bound parameters: http://php.net/manual/en/mysqli-stmt.bind-param.php It's much more secure.






share|improve this answer













I don't understand why you're using that curly-brace notation at all when you're just copying it over into a multidimensional compData array anyway? Why not just directly put the data into the array:



for ($n = 1; $n <= $CI_NOYEARS; $n++) {
$compdata[$val]['data'] = $r[$n];
}


(You'll also need to declare it as an array at the top level of the foreach of course.)



Side note: I would suggest not interpolating data directly into your query string like that, and instead using bound parameters: http://php.net/manual/en/mysqli-stmt.bind-param.php It's much more secure.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 15 '18 at 20:58









IceMetalPunkIceMetalPunk

954716




954716













  • Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

    – ChartProblems
    Nov 15 '18 at 21:06











  • Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

    – IceMetalPunk
    Nov 16 '18 at 18:35



















  • Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

    – ChartProblems
    Nov 15 '18 at 21:06











  • Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

    – IceMetalPunk
    Nov 16 '18 at 18:35

















Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

– ChartProblems
Nov 15 '18 at 21:06





Thank you for helping out. I've tried your solution but my echo of the results in $compData['List1'] for instance doesn't return anything. The code above works just fine when I do execute it for every ID but due to the large amounts of ID's I just wanted to do this dynamic by using an array and foreach statement instead? I hope this makes sense. I will have a look at the link you provided as well.

– ChartProblems
Nov 15 '18 at 21:06













Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

– IceMetalPunk
Nov 16 '18 at 18:35





Did you declare the value as an array before the code I gave you? If so, and it's still not working, I'd suggest printing the value of $val inside that loop, so you can make sure there is, in fact, a "List1" in there. (I'd also suggest using either print_r or var_dump when outputting arrays, instead of echo, since those will give you the full values of the array entries.)

– IceMetalPunk
Nov 16 '18 at 18:35


















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