Display Javascript Code In PHP

Question

I Have A problem In displaying JS Code In PHP My Code Is

<?php
    if($result)
    {
        echo '<script language="javascript">';
        echo "$('#Err').hide()</script>";
    }
    else
    {
        echo '<script language="javascript">';
        echo "$('#Err').show()</script>";
    }
?>

I Am Using XAMPP For running Code When I Run This Code It Will Not Display I Don't know what's the problem plz tell me Thanks!

The way you have written your code tells me you don't know about escaping quotes in strings. If you don't know such a basic thing, I'm not hopeful that you tried View Source to see what PHP is outputting. — Niet the Dark Absol
– Niet the Dark Absol, Commented May 5, 2012 at 6:12
When you 'view source' of the resulting page in the browser is the <script> tag in there at all and just not working like you expect or is it completely missing? — nvuono
– nvuono, Commented May 5, 2012 at 6:13
The script tag is there for me in FF12, and I get a JS error, "$ is not defined." Have you included your jQuery dependency file? Also, if you're not already, load up Firefox and add the 'Firebug' extension, it will make debugging Javascript much easier. — danjah
– danjah, Commented May 5, 2012 at 6:14
The given answers are correct. However, my personal preference is to always have static JavaScript that acts on some dynamic bit of data via a hidden field. That way you can easily cache the JS into external files. — Matthew
– Matthew, Commented May 5, 2012 at 6:18
I put an answer below. Try not to use echo when throwing html or scripts, it's a good practice for PHP development, and also you won't need to worry about escaping (and so far, hardcoding) other languages. — lu1s
– lu1s, Commented May 5, 2012 at 6:21

Nadh · Accepted Answer · 2012-05-05 06:12:33Z

2

You have to escape the $ sign as follows:

<?php
    if($result)
    {
        echo '<script language="javascript">';
        echo "\$('#Err').hide()</script>";
    }
    else
    {
        echo '<script language="javascript">';
        echo "\$('#Err').show()</script>";
    }
?>

answered May 5, 2012 at 6:12

Nadh

7,2633 gold badges23 silver badges21 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

hakre Over a year ago

@user1375948: No, this is not your issue and this is just an unrelated suggestion to prevent errors in the future with your code when you change it.

hakre Over a year ago

@NADH: Technically this answer is wrong because for that usage of $ it does not have to be escaped. (But your good intentions a clear, just FYI).

user1375948 Over a year ago

I have Got It But There Is A little problem i am using this code in post with action on default page when submit is clicked and i print the Err Tag Than There Will Be A very Short delay The Page Reloads Again And It Is Hidden Again As By default It Is Hidden when i print it through alert() delay occurs and i can see the printed tag but i want to do it without alert() function Kindly tell Me

Bryan · Accepted Answer · 2012-05-05 06:12:06Z

0

Use single quotations, your double quotations are rendering $('#Err') as a PHP variable

<?php
    if($result)
    {
        echo '<script language="javascript">';
        echo '$(\'#Err\').hide()</script>';
    }
    else
    {
        echo '<script language="javascript">';
        echo '$(\'#Err\').show()</script>';
    }
?>

answered May 5, 2012 at 6:12

Bryan

6,7522 gold badges19 silver badges21 bronze badges

Comments

Moslem Shahsavan · Accepted Answer · 2012-05-05 06:17:07Z

0

<?php
    if($result)
    {
        echo '<script type="text/javascript" language="javascript">
             $("#Err").hide()</script>';
    }
    else
    {
        echo '<script type="text/javascript" language="javascript">
              $("#Err").show()</script>';
    }
?>

answered May 5, 2012 at 6:17

Moslem Shahsavan

1,34719 silver badges19 bronze badges

Comments

lu1s · Accepted Answer · 2012-05-05 06:18:43Z

0

This way is more understandable and uses less processing:

<!-- this is your php script with your html layout -->

<script type="text/javascript">
    <?php if($result === true): ?>
        $("#Err").hide();
    <?php else: ?>
        $("#Err").show();
    <?php endif; ?>
</script>

Try not to use echo. It's a good practice for PHP. If you really want to increase your php productivity, I highly recommend the Codeigniter framework. Give it a try =).

answered May 5, 2012 at 6:18

lu1s

5,9014 gold badges24 silver badges37 bronze badges

Comments

hakre · Accepted Answer · 2012-05-05 08:15:24Z

0

You need to do some debugging, look into the source of your HTML that get's generated:

<?php
    // debug code:
    var_dump($result); die('This place has been reached');

    // normal code:
    printf("<script language=\"javascript\">\n\$('#Err').%s()\n</script>"
           , $result ? 'hide' : 'show');
?>

When you execute this, things will go wrong, but you can see what $result contains so to debug your script. If it does not display that inside the HTML source, it means that your code is not executed at all and your problem lies somewhere else.

answered May 5, 2012 at 8:15

hakre

200k55 gold badges454 silver badges865 bronze badges

Collectives™ on Stack Overflow

Display Javascript Code In PHP

5 Answers 5

3 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

3 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related