How to apply if-else on css class in php code?

Question

I am trying to apply if else condition in bootstrap class with php variable but its not working. My variable is getting result. Below is my tbody where i am applying php code:

<div class="<?php '#count_rows#' <= '2')
        { 
            echo 'middle_menu';
        } 
    else 
        { 
            echo 'middle_menu1';
        } ?>">
        <table class="table_4">
            <thead>
                <tr>
                    <th class="th_11">Quantity</th>
                </tr>
            </thead>
            <tbody>
                #TABLE_BODY#
            </tbody>
        </table>
</div>

Below is my 2 css classes:

.middle_menu
    {
        margin-top: 40px;       
        padding-bottom: 200px;
    }
.middle_menu1
    {
        margin-top: 40px;
    }

I am fetching my variable from another page where i set this type of opening and closing variable with #. Below is my code for your reference but i dont think that is an issue for me because i check this #count_rows# variable with print_r in my current page and it showing me correct result.

foreach ($form_field as $key => $value){
    $htmlContent = str_replace('#'.$key.'#', $value, $htmlContent);
}

Maybe there's some sort of shorthand I've never used, but I don't see an if there at all, nor an opening parentheses — git-e-up
– git-e-up, Commented May 10, 2018 at 15:17
that is an typo mistake i added if as you can see else condition — amit sutar
– amit sutar, Commented May 10, 2018 at 15:19
Ehm, missing if ( aside, are you comparing the strings '#count_rows#' and '2'? Neither of those are variables. — Mr Lister
– Mr Lister, Commented May 10, 2018 at 15:20
Your code doesn't make any sense, and is invalid php. First off you are missing an 'if'. Next you are missing an opening paren. Then you are comparing a string constant '#count_rows#' with another string constant '2'. That is always going to evaluate to the same thing. Last and certainly not least, you can't evaluate a javascript variable that is changed or runs in the browser in a serverside php script. — gview
– gview, Commented May 10, 2018 at 15:20

Fernando Avalos · Accepted Answer · 2018-05-10 17:05:48Z

2

<?php $className = $count_rows <= 2 ? 'middle_menu' : 'middle_menu1'; ?>
<div class="<?php echo $className; ?>" />

You can create a test.php then run php test.php and you should see the result

<?php

// $count_rows = 1;
$count_rows = 3;
$className = $count_rows <= 2 ? 'middle_menu' : 'middle_menu1';
echo "<div class=\"$className\"/>\n";

edited May 10, 2018 at 17:05

answered May 10, 2018 at 15:33

Fernando Avalos

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Comments

symcbean · Accepted Answer · 2018-05-10 15:29:58Z

 <div class="<?php '#count_rows#' <= '2')

You are missing a keyword here and also some brackets (which should cause a fatal syntax error), Also you are comparing two different literal strings - surely one of the terms should be a PHP variable?

You can't read back stuff you've previously written to the output stream - you are confusing what is happening in HTML, CSS and PHP.

I think you mean...

 <div class="<?php if ($form_fields['count_rows'] <= '2')

Comparing a number by casting it to a string is rather dangerous.

Personally I would do this:

 <div class="<?php echo (2 <= $form_fields['count_rows'])
        ? 'middle_menu' : 'middle_menu1'; ?>">

Since you didn't mention that the script blew up in your face, I suspect there may be a lot of other issues.

 foreach ($form_field as $key => $value){
   $htmlContent = str_replace('#'.$key.'#', $value, $htmlContent);
 }

This is very innefficient. Consider:

 $find=array_keys($form_field);
 foreach ($find as $index=>$term) {
     $find[$index]='#' . $term . '#';
 }
 $htmlContent = str_replace($find, $form_feld, $htmlContent);

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