Undefined constant error PHP

Question

I am getting an undefined constant error with the following code, for "success".

    if(success)
    {
        echo "<script type=\"text/javascript\">".
            "alert('Form submitted successfully.');".
            "</script>";
    }

Can anyone help me with resolving this issue?

`success` is no variable you have to `$success` and check if that variable is true — Blueblazer172, Dec 07 '16 at 12:30
By adding $success == TRUE; above it then gives me an unidentified variable. — Nicholas Nicolaou, Dec 07 '16 at 12:34
Success is just there so when I submit a form a success message pops up. — Nicholas Nicolaou, Dec 07 '16 at 12:35
@NicholasNicolaou then post your whole code so we can look on it — Blueblazer172, Dec 07 '16 at 12:36
You may want to read about the [PHP syntax](http://php.net/manual/en/langref.php), more specifically [variables](http://php.net/manual/en/language.variables.basics.php) and [constants](http://php.net/manual/en/language.constants.php). Please note this has absolutely nothing to do with JavaScript. — Álvaro González, Dec 07 '16 at 12:41

score 2 · Accepted Answer · answered Dec 07 '16 at 12:34

2

Please read doc http://php.net/manual/en/language.constants.php

Your code must be like this (if you want use constants)

define('success', true);
if(success)
{
    echo "<script type=\"text/javascript\">".
        "alert('Form submitted successfully.');".
        "</script>";
}

Or use variables http://php.net/manual/en/language.variables.basics.php

answered Dec 07 '16 at 12:34

Artem Ilchenko

990
8
20

What's the use case of having such construct? Being able to define `success` as `false` to avoid code execution? ;-P – Álvaro González Dec 07 '16 at 12:44

Jalpa · Answer 2 · 2016-12-07T12:40:20.933

1

Please try below code.

    $success = "test"; // set your value that you can get.
    if(isset($success) && $success != "")
        {
            echo "<script type=\"text/javascript\">".
                "alert('Form submitted successfully.');".
                "</script>";
        }

edited Dec 07 '16 at 12:40

answered Dec 07 '16 at 12:35

Jalpa

687
3
13

Alert message does not appear – Nicholas Nicolaou Dec 07 '16 at 12:37
@NicholasNicolaou Do you have set $success variable ? – Jalpa Dec 07 '16 at 12:38
@NicholasNicolaou check now my ans again i have made changes. – Jalpa Dec 07 '16 at 12:40

score 0 · Answer 3 · edited Dec 07 '16 at 12:45

0

change success to $success

if($success)
{
    echo "<script type=\"text/javascript\">".
    "alert('Form submitted successfully.');".
    "</script>";
}

edited Dec 07 '16 at 12:45

Manoj Sharma

1,457
2
13
20

answered Dec 07 '16 at 12:33

Shanu k k

1,215
1
16
37

undefined variable – Nicholas Nicolaou Dec 07 '16 at 12:36
@nicholas Nicolaou Can you please show us whole code? – Shanu k k Dec 07 '16 at 12:40

score 0 · Answer 4 · answered Dec 07 '16 at 12:35

0

"success" without $ is a constant, not a variable. And this constant is not initialized with a value to be checked in the if.

answered Dec 07 '16 at 12:35

João Paulo de Lima

98
1
8

score 0 · Answer 5 · answered Dec 07 '16 at 12:55

your variable has to be set befor you go to that if-clause. in your case you want to check if the variable is true and if so it should alert() with the content Form submitted successfully.

there are 2 ways to set a variable.

define a constant with the define("success", true)- function. Then you can let your code as it is.
or define a variable lik $success = true;

then you have to change your code to this:

<?php 

$success = true; // define the variable and set it to true

if ($success) {

    echo "<script type=\"text/javascript\">" .
        "alert('Form submitted successfully.');" .
        "</script>";

} 

?>

or

<?php

$success = true; // define the variable and set it to true

if (isset($success) && !empty($success) && $success != false) {

    echo "<script type=\"text/javascript\">" .
        "alert('Form submitted successfully.');" .
        "</script>";

}

?>

Undefined constant error PHP

5 Answers5