Creating a javascript alert with php that has a php variable inside?

Question

I'm making a form that is supposed to create a javascript alert when some fields aren't filled out or filled out properly. I want to be able to take the error messages I've put in a php variable and display them in the javascript alert window.

The following code does not work:

function died($error) {
    echo '<script type="text/javascript"> alert('.$error.')</script>';
    die();
}

How can I add the string contained in $error between the two "script" strings so it will output properly as a javascript alert?

Thank you!

Why not use javascript to detect if all the forms are filled in? — David Corbin
– David Corbin, Commented Apr 28, 2014 at 2:03
Sorry, I screwed up and added an old version of the code I was using to figure out what was wrong. The $string variable is completely unnecessary. I took it out. Thank you! — MarvinLazer
– MarvinLazer, Commented Apr 28, 2014 at 2:03
You just can do a client checking if you just want to 'alert' the error.. but then again the best way to validate is in the server side. — Eddie
– Eddie, Commented Apr 28, 2014 at 2:09

hakre · Accepted Answer · 2023-07-04 10:42:14Z

20

You only forgot quotations that are required for the JavaScript alert.

If you passed 'hello' to the function, your current code would create alert as:

alert(hello)

instead of doing:

alert("hello")

Therefore, change your code to the following (json_encode() the string and create a javascript variable out of it):

function died(string $message) {
    if ('' === trim($message)) {
        $message = 'died.';
    }
?>
    <script type="text/javascript">
      var message = <?php echo json_encode($message); ?>;
      alert(message);
    </script>
<?php
    die();
}

and you can use your function:

died('error on whatever');

edited Jul 4, 2023 at 10:42

hakre

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answered Apr 28, 2014 at 2:02

user1978142

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5 Comments

MarvinLazer Over a year ago

Thanks! It works, although I don't really understand why. What if I wanted to add other text in the alert that would be concatenated with the variable? Would I have to do something like this? echo '<script>alert('.'Here is some other text!\\n"'.$error.'");</script>';

user1978142 Over a year ago

as @frankgorman clearly stated, what happens is that, after you concatenated, you missed the quotes inside rendering the error on javascript. If you want to add, just concatenate another string on $error. Example: $error = "Another Line \n" . $error

Eon Over a year ago

You know you are very new to a language when you look for ways to alert/log variables :) (happens that I came to search for this answer too)

Scarl Over a year ago

This displays nothing. As in it does not display the php variable's value for some reason

Crickets Over a year ago

Your answer did not work for me, but this did: echo '<script>alert("' . $str . '");</script>'; -source

antelove · Accepted Answer · 2017-09-25 04:48:24Z

6

Display variable php in alert javascript

   <?php 
          function died($error) { ?>

            <script>alert("<?php echo $error; ?>")</script>

    <?php   die(); 
          } ?>

edited Sep 25, 2017 at 4:48

answered Sep 25, 2017 at 3:12

antelove

3,37830 silver badges20 bronze badges

Comments

user3571318 · Accepted Answer · 2014-04-28 02:11:43Z

3

You can use function follow this:

function died($error) {
    echo '<script> alert("'.$error.'")</script>';
    die();
}

answered Apr 28, 2014 at 2:11

user3571318

311 bronze badge

Comments

SASIDHARAN L · Accepted Answer · 2020-08-15 12:13:36Z

0

<?php
 echo "<script type='text/javascript'>alert('{$_SESSION["success"]}');</script>";
 unset($_SESSION["success"]);
?>

Use this code it would work correctly

answered Aug 15, 2020 at 12:13

SASIDHARAN L

11

1 Comment

Giovanni Esposito Over a year ago

Ciao, try to add a small explanation. Code-only answer discourages.

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