php validation. Can't quite wrap my head around the logic I need to happen
I have a page, where there are a 3 values passed and the rest of the page will respond based on any of those 3.
if 1 and 2 have values from a form, but 3 and 4 don't { do a } else warn the user to fill out both 1 and 2
if 1 and 2 don't have values, 3 does, 4 doesn't { do b } else warn the user to fill in 3, disregard the state of 1,2 and 4
if 1 and 2 and 3 don't have values, but 4 does { do c } else warn the user to fill in 4, disregard the state of 1,2, and 3
if ($fname != '' and $lname != '') {
querytype = "name";
} else {
echo "Please enter a first and last name.";
}
if ($reg != '' and $fname == '' and $lname == '') {
$querytype = "reg";
} else {
echo "Please enter a Registration Number.";
}
if ($num != '' and $reg == '' and $fname == '' and $lname == '') {
$querytype = "number";
} else {
echo "Please enter a Phone Number.";
}
switch ($querytype) {
#name is 1 and 2.. it's a first and last name
case "name": {
break;
}
#reg is 3
case "reg":
{
break;
}
#number is 4
case "number":
{
break;
}
default: echo "Something broke.. idk";
}
php conditional
edited Nov 13 '18 at 1:03
Barmar
419k34244344
asked Nov 12 '18 at 23:50
Ken Gordon
94
|
show 8 more comments
I have a page, where there are a 3 values passed and the rest of the page will respond based on any of those 3.
if 1 and 2 have values from a form, but 3 and 4 don't { do a } else warn the user to fill out both 1 and 2
if 1 and 2 don't have values, 3 does, 4 doesn't { do b } else warn the user to fill in 3, disregard the state of 1,2 and 4
if 1 and 2 and 3 don't have values, but 4 does { do c } else warn the user to fill in 4, disregard the state of 1,2, and 3
if ($fname != '' and $lname != '') {
querytype = "name";
} else {
echo "Please enter a first and last name.";
}
if ($reg != '' and $fname == '' and $lname == '') {
$querytype = "reg";
} else {
echo "Please enter a Registration Number.";
}
if ($num != '' and $reg == '' and $fname == '' and $lname == '') {
$querytype = "number";
} else {
echo "Please enter a Phone Number.";
}
switch ($querytype) {
#name is 1 and 2.. it's a first and last name
case "name": {
break;
}
#reg is 3
case "reg":
{
break;
}
#number is 4
case "number":
{
break;
}
default: echo "Something broke.. idk";
}
php conditional
edited Nov 13 '18 at 1:03
Barmar
419k34244344
asked Nov 12 '18 at 23:50
Ken Gordon
94
What is the problem with your logic now?
– Jeff
Nov 12 '18 at 23:59
if I input fields 1 and 2.. I still get the errors for 3 and 4
– Ken Gordon
Nov 13 '18 at 0:00
Because theifcondition from the first case is the same as theelsefrom 2nd and 3rd case.
– Jeff
Nov 13 '18 at 0:01
You'd need to skip the following tests if one is done.
– Jeff
Nov 13 '18 at 0:03
querytype = "name";is missing the$beforequerytype.
– Barmar
Nov 13 '18 at 1:04
|
show 8 more comments
I have a page, where there are a 3 values passed and the rest of the page will respond based on any of those 3.
if 1 and 2 have values from a form, but 3 and 4 don't { do a } else warn the user to fill out both 1 and 2
if 1 and 2 don't have values, 3 does, 4 doesn't { do b } else warn the user to fill in 3, disregard the state of 1,2 and 4
if 1 and 2 and 3 don't have values, but 4 does { do c } else warn the user to fill in 4, disregard the state of 1,2, and 3
if ($fname != '' and $lname != '') {
querytype = "name";
} else {
echo "Please enter a first and last name.";
}
if ($reg != '' and $fname == '' and $lname == '') {
$querytype = "reg";
} else {
echo "Please enter a Registration Number.";
}
if ($num != '' and $reg == '' and $fname == '' and $lname == '') {
$querytype = "number";
} else {
echo "Please enter a Phone Number.";
}
switch ($querytype) {
#name is 1 and 2.. it's a first and last name
case "name": {
break;
}
#reg is 3
case "reg":
{
break;
}
#number is 4
case "number":
{
break;
}
default: echo "Something broke.. idk";
}
php conditional
edited Nov 13 '18 at 1:03
Barmar
419k34244344
asked Nov 12 '18 at 23:50
Ken Gordon
94
I have a page, where there are a 3 values passed and the rest of the page will respond based on any of those 3.
if 1 and 2 have values from a form, but 3 and 4 don't { do a } else warn the user to fill out both 1 and 2
if 1 and 2 don't have values, 3 does, 4 doesn't { do b } else warn the user to fill in 3, disregard the state of 1,2 and 4
if 1 and 2 and 3 don't have values, but 4 does { do c } else warn the user to fill in 4, disregard the state of 1,2, and 3
if ($fname != '' and $lname != '') {
querytype = "name";
} else {
echo "Please enter a first and last name.";
}
if ($reg != '' and $fname == '' and $lname == '') {
$querytype = "reg";
} else {
echo "Please enter a Registration Number.";
}
if ($num != '' and $reg == '' and $fname == '' and $lname == '') {
$querytype = "number";
} else {
echo "Please enter a Phone Number.";
}
switch ($querytype) {
#name is 1 and 2.. it's a first and last name
case "name": {
break;
}
#reg is 3
case "reg":
{
break;
}
#number is 4
case "number":
{
break;
}
default: echo "Something broke.. idk";
}
php conditional
php conditional
edited Nov 13 '18 at 1:03
Barmar
419k34244344
asked Nov 12 '18 at 23:50
Ken Gordon
94
edited Nov 13 '18 at 1:03
Barmar
419k34244344
asked Nov 12 '18 at 23:50
Ken Gordon
94
edited Nov 13 '18 at 1:03
Barmar
419k34244344
edited Nov 13 '18 at 1:03
Barmar
419k34244344
edited Nov 13 '18 at 1:03
Barmar
419k34244344
419k34244344
asked Nov 12 '18 at 23:50
Ken Gordon
94
asked Nov 12 '18 at 23:50
Ken Gordon
94
asked Nov 12 '18 at 23:50
Ken Gordon
94
94
What is the problem with your logic now?
– Jeff
Nov 12 '18 at 23:59
if I input fields 1 and 2.. I still get the errors for 3 and 4
– Ken Gordon
Nov 13 '18 at 0:00
Because theifcondition from the first case is the same as theelsefrom 2nd and 3rd case.
– Jeff
Nov 13 '18 at 0:01
You'd need to skip the following tests if one is done.
– Jeff
Nov 13 '18 at 0:03
querytype = "name";is missing the$beforequerytype.
– Barmar
Nov 13 '18 at 1:04
|
show 8 more comments
What is the problem with your logic now?
– Jeff
Nov 12 '18 at 23:59
if I input fields 1 and 2.. I still get the errors for 3 and 4
– Ken Gordon
Nov 13 '18 at 0:00
Because theifcondition from the first case is the same as theelsefrom 2nd and 3rd case.
– Jeff
Nov 13 '18 at 0:01
You'd need to skip the following tests if one is done.
– Jeff
Nov 13 '18 at 0:03
querytype = "name";is missing the$beforequerytype.
– Barmar
Nov 13 '18 at 1:04
What is the problem with your logic now?
– Jeff
Nov 12 '18 at 23:59
What is the problem with your logic now?
– Jeff
Nov 12 '18 at 23:59
if I input fields 1 and 2.. I still get the errors for 3 and 4
– Ken Gordon
Nov 13 '18 at 0:00
if I input fields 1 and 2.. I still get the errors for 3 and 4
– Ken Gordon
Nov 13 '18 at 0:00
Because the
if condition from the first case is the same as the else from 2nd and 3rd case.– Jeff
Nov 13 '18 at 0:01
Because the
if condition from the first case is the same as the else from 2nd and 3rd case.– Jeff
Nov 13 '18 at 0:01
You'd need to skip the following tests if one is done.
– Jeff
Nov 13 '18 at 0:03
You'd need to skip the following tests if one is done.
– Jeff
Nov 13 '18 at 0:03
querytype = "name"; is missing the $ before querytype.– Barmar
Nov 13 '18 at 1:04
querytype = "name"; is missing the $ before querytype.– Barmar
Nov 13 '18 at 1:04
|
show 8 more comments
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What is the problem with your logic now?
– Jeff
Nov 12 '18 at 23:59
if I input fields 1 and 2.. I still get the errors for 3 and 4
– Ken Gordon
Nov 13 '18 at 0:00
Because the
ifcondition from the first case is the same as theelsefrom 2nd and 3rd case.– Jeff
Nov 13 '18 at 0:01
You'd need to skip the following tests if one is done.
– Jeff
Nov 13 '18 at 0:03
querytype = "name";is missing the$beforequerytype.– Barmar
Nov 13 '18 at 1:04