PHP Syntax causing results to fail

Question

I am currently writing a dummy comment/answer system using PHP/SQL/AJAX and have come an issue on my output divs due to my coding style syntax.

To simplify the code I have just included the while loop: //prep query $sqlQuery = "SELECT * FROM Questions ORDER BY QuestionID DESC";

    //exec query
    $result = mysql_query($sqlQuery) or die(mysql_error());

    //loop for multipule results
        while ($row = mysql_fetch_array($result))
    {
        echo '<div class="test" id="test_$questionID">'; //<--doesn't work
        //code here
        echo '</div>';

        echo "<div class='test' id='test_$questionID'>"; //<--works
        //code here
        echo '</div>';
    }

Output of failed code line:

<div class="test" id="test_$questionID">

Output of successful code line:

<div class='test' id='test_21'>

The code above works when I start an echo using " instead of ' my PHP style is to start each outer with ' and innners with ".

Should I change my div echos to start with " or is there another way?

Edit: Thanks for all the quick replies all!

I was taught to start using singles instead of doubles without the actual reason behind it.

I didn't know the conflict was caused because of the text being interpreted incorrectly and the environment wasn't throwing any errors to guide. Escaping them as @PressingOnAlways or changing them as @santhy recommend.

It's not that I was asking the difference between singles and doubles since I didn't know that was the area of error. I can see now that it was, but not sure I would have know to look for that stack article first sorry.

in PHP the quotes matter - stuff between single quotes is not parsed for any values and is taken to be as-is. I.e. `$var = 'some $text';` is taken to literally be `some $text`, where as `$var = "some $text";` is parsed using whatever value is defined for `$text` variable at the time of parsing. — Jhecht, Aug 04 '17 at 03:40

score 2 · Answer 1 · answered Aug 04 '17 at 03:36

2

You need to use double quotes, but escape the inside double quotes with backslash. For example:

 echo "<div class=\"test\" id=\"test_$questionID\">";

answered Aug 04 '17 at 03:36

PressingOnAlways

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score 1 · Answer 2 · answered Aug 04 '17 at 03:40

Try this:

Replace echo '<div class="test" id="test_$questionID">'; //<--doesn't work line:

 <?php
 //exec query
    $result = mysql_query($sqlQuery) or die(mysql_error());

    //loop for multipule results
    while ($row = mysql_fetch_array($result))
    {
        echo '<div class="test" id="test_'. $questionID .'">'; //<--now work
        //code here
        echo '</div>';

        echo '<div class="test" id="test_'. $questionID .'">'; //<--now work
        //code here
        echo '</div>';
    }

score 1 · Accepted Answer · answered Aug 04 '17 at 03:41

We can use different ways, see some of them

while ($row = mysql_fetch_array($result))
{
    echo '<div class="test" id="test_'.$questionID.'">';
    //code here
    echo '</div>';

    echo "<div class='test' id='test_{$questionID}'>";
    //code here
    echo '</div>';

    echo "<div class=\"test\" id=\"test_{$questionID}\">";
    //code here
    echo '</div>';

    ?>
    <div class="test" id="test_<?=$questionID;?>"> <?php
    //code here
    ?>
    </div> <?php
}

score 0 · Answer 4 · answered Aug 04 '17 at 03:38

PHP doesn't do variable interpolation in strings enclosed by single quotes. You must use double quotes. Also, I make it a practice when interpolating a variable that combines with nearby characters, I wrap the variable in brackets for clarity.

So

 echo '<div class="test" id="test_$questionID">';

becomes:

 echo "<div class=\"test\" id=\"test_{$questionID}\">";

PHP Syntax causing results to fail

4 Answers4