Define javascript parameter in PHP

Question

During a while loop in PHP I want to create following link:

echo '<a onclick="getSolution('.ResultArray['qid'].')" style="color: red;">L&ouml;sung anzeigen</a>';

But there it is a mistake. It will not take the $ResultArray['qid'] as a parameter for the javascript function. I need this parameter for define which div it has to take when clicking on the link.

Here you get the complete while loop in php:

while($ResultArray = mysqli_fetch_array($getQuestions)) {
    echo '<p>';
    echo $ResultArray['question'];
    echo '<br />';          
            echo '<input type="text" style="width: 500px;"/>';
    echo '</p>';
    echo '<a onclick="getSolution('.ResultArray['qid'].')" style="color: red;">L&ouml;sung anzeigen</a>';
    echo '<div id="'.$ResultArray['qid'].'" style="visibility: hidden">';
    echo $ResultArray['answer'];
    echo '</div>';
}

How I can fix this problem?

Step one: [Enable error reporting](http://stackoverflow.com/a/6575502/1438393) — Amal Murali, Apr 24 '14 at 16:35
I get following error: Parse error: syntax error, unexpected '[', expecting ',' or ';' in ... /example.php on line 135 — AlbertoGiorgino, Apr 24 '14 at 16:37
Step two: fix it. If the ID is a text, then you are missing quotes. Also missing the dollar sign ($). — Kai, Apr 24 '14 at 16:38

score 3 · Accepted Answer · answered Apr 24 '14 at 16:39

3

You're trying to reference your ResultArray variable without prefacing it with your $ symbol.

Corrected:

echo '<a onclick="getSolution('.$ResultArray['qid'].')" style="color: red;">L&ouml;sung anzeigen</a>';

answered Apr 24 '14 at 16:39

esqew

42,425
27
92
132

Thank you. That's a stupid mistake. I haven't seen it. But thank you very much. – AlbertoGiorgino Apr 24 '14 at 16:41

score 2 · Answer 2 · answered Apr 24 '14 at 16:39

2

echo '<a onclick="getSolution('.ResultArray['qid'].')"
                                ^^^^^^^^^^^^---undefined constant, since it has no "$"

You must have $ResultArray in there. Without the $, it's a constant, and you cannot have constant arrays.

answered Apr 24 '14 at 16:39

Marc B

356,200
43
426
500

Thank you. That's a stupid mistake. I haven't seen it. But thank you very much. – AlbertoGiorgino Apr 24 '14 at 16:41

Define javascript parameter in PHP

2 Answers2