php sees, for example, 11 as 1











up vote
-2
down vote

favorite












function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1










share|improve this question




















  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30

















up vote
-2
down vote

favorite












function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1










share|improve this question




















  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30















up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1










share|improve this question















function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1







javascript php mysql arrays






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 0:26









Jeff

6,12911024




6,12911024










asked Nov 12 at 0:18









lysekk

1




1








  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30
















  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30










2




2




until id is a one-digit number What is id? There is no id in your code
– CertainPerformance
Nov 12 at 0:23




until id is a one-digit number What is id? There is no id in your code
– CertainPerformance
Nov 12 at 0:23












"php sees" - but we don't see any php!?
– Jeff
Nov 12 at 0:24




"php sees" - but we don't see any php!?
– Jeff
Nov 12 at 0:24












sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
– lysekk
Nov 12 at 0:24




sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
– lysekk
Nov 12 at 0:24












if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
– lysekk
Nov 12 at 0:30




if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
– lysekk
Nov 12 at 0:30




1




1




did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
– Jeff
Nov 12 at 0:30






did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
– Jeff
Nov 12 at 0:30














1 Answer
1






active

oldest

votes

















up vote
0
down vote













I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



Anyway, try this:



if(isset($_POST['zwrot'])) { 

$zwrot=$_POST['zwrot'];

$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

}


Or if you want to pass by array:



function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;

var zwrot_array = ;

for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
zwrot_array[i] = zwrot[i].value;
}
}

var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: JSON.stringify(zwrot_array)},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę"
});

setTimeout(function() {
location.reload()
}, 2000
);
}
});
}


PHP:



if(isset($_POST['zwrot'])) { 

$zwrots = json_decode($_POST['zwrot']);
$data=date("d-m-Y");

foreach ($zwrots as $zwrot) {
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
}
}





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    up vote
    0
    down vote













    I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



    Anyway, try this:



    if(isset($_POST['zwrot'])) { 

    $zwrot=$_POST['zwrot'];

    $data=date("d-m-Y");
    $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
    $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

    }


    Or if you want to pass by array:



    function zwrot() {
    var zwrot = document.getElementsByClassName("zwroc");
    var i;

    var zwrot_array = ;

    for (i = 0; i < zwrot.length; i++) {
    if(zwrot[i].checked){
    zwrot_array[i] = zwrot[i].value;
    }
    }

    var ajax = $.ajax({
    url: 'php/request.php',
    type: 'POST',
    data: {zwrot: JSON.stringify(zwrot_array)},
    success: function(data)
    {
    $.growl.notice({
    title: "INFO",
    message: "Oddano książkę"
    });

    setTimeout(function() {
    location.reload()
    }, 2000
    );
    }
    });
    }


    PHP:



    if(isset($_POST['zwrot'])) { 

    $zwrots = json_decode($_POST['zwrot']);
    $data=date("d-m-Y");

    foreach ($zwrots as $zwrot) {
    $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
    $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
    }
    }





    share|improve this answer



























      up vote
      0
      down vote













      I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



      Anyway, try this:



      if(isset($_POST['zwrot'])) { 

      $zwrot=$_POST['zwrot'];

      $data=date("d-m-Y");
      $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
      $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

      }


      Or if you want to pass by array:



      function zwrot() {
      var zwrot = document.getElementsByClassName("zwroc");
      var i;

      var zwrot_array = ;

      for (i = 0; i < zwrot.length; i++) {
      if(zwrot[i].checked){
      zwrot_array[i] = zwrot[i].value;
      }
      }

      var ajax = $.ajax({
      url: 'php/request.php',
      type: 'POST',
      data: {zwrot: JSON.stringify(zwrot_array)},
      success: function(data)
      {
      $.growl.notice({
      title: "INFO",
      message: "Oddano książkę"
      });

      setTimeout(function() {
      location.reload()
      }, 2000
      );
      }
      });
      }


      PHP:



      if(isset($_POST['zwrot'])) { 

      $zwrots = json_decode($_POST['zwrot']);
      $data=date("d-m-Y");

      foreach ($zwrots as $zwrot) {
      $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
      $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
      }
      }





      share|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



        Anyway, try this:



        if(isset($_POST['zwrot'])) { 

        $zwrot=$_POST['zwrot'];

        $data=date("d-m-Y");
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

        }


        Or if you want to pass by array:



        function zwrot() {
        var zwrot = document.getElementsByClassName("zwroc");
        var i;

        var zwrot_array = ;

        for (i = 0; i < zwrot.length; i++) {
        if(zwrot[i].checked){
        zwrot_array[i] = zwrot[i].value;
        }
        }

        var ajax = $.ajax({
        url: 'php/request.php',
        type: 'POST',
        data: {zwrot: JSON.stringify(zwrot_array)},
        success: function(data)
        {
        $.growl.notice({
        title: "INFO",
        message: "Oddano książkę"
        });

        setTimeout(function() {
        location.reload()
        }, 2000
        );
        }
        });
        }


        PHP:



        if(isset($_POST['zwrot'])) { 

        $zwrots = json_decode($_POST['zwrot']);
        $data=date("d-m-Y");

        foreach ($zwrots as $zwrot) {
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
        }
        }





        share|improve this answer














        I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



        Anyway, try this:



        if(isset($_POST['zwrot'])) { 

        $zwrot=$_POST['zwrot'];

        $data=date("d-m-Y");
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

        }


        Or if you want to pass by array:



        function zwrot() {
        var zwrot = document.getElementsByClassName("zwroc");
        var i;

        var zwrot_array = ;

        for (i = 0; i < zwrot.length; i++) {
        if(zwrot[i].checked){
        zwrot_array[i] = zwrot[i].value;
        }
        }

        var ajax = $.ajax({
        url: 'php/request.php',
        type: 'POST',
        data: {zwrot: JSON.stringify(zwrot_array)},
        success: function(data)
        {
        $.growl.notice({
        title: "INFO",
        message: "Oddano książkę"
        });

        setTimeout(function() {
        location.reload()
        }, 2000
        );
        }
        });
        }


        PHP:



        if(isset($_POST['zwrot'])) { 

        $zwrots = json_decode($_POST['zwrot']);
        $data=date("d-m-Y");

        foreach ($zwrots as $zwrot) {
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
        }
        }






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        edited Nov 12 at 1:22

























        answered Nov 12 at 1:13









        ACD

        714111




        714111






























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