How to pass PHP variable to JavaScript function?

Question

I have this JavaScript function which toggles the visibility of a certain div:

function showHideDiv(id){
var obj = document.getElementById(id);
if (obj.style.display=="none"){
  obj.style.display='block';
} else if(obj.style.display=="block"){
  obj.style.display='none';
}
}

I have a dynamic div that has the id="pm_message'.$row[0].'". I want to toggle the visibility of this div via the javascript function I mentioned above.

This is my link with the onclick-function:

echo '<h1><a onclick="showHideDiv('showHideDiv("message'.$row[0].'")" class="'.$link_class.'" href="'.$profile_url.'">'.$row['title'].'</a></h1>';

It doesn't work and I get this error message:

PHP Parse error:  syntax error, unexpected '?'

score 6 · Accepted Answer · answered Aug 02 '13 at 20:49

6

echo '<h1><a onclick="showHideDiv('.$row[0].')" class="'.$link_class.'" href="'.$profile_url.'">'.$row['title'].'</a></h1>';

It is already in php so you do not need to open another set of php tags.

answered Aug 02 '13 at 20:49

Danny

1,185
2
12
33

Should be `$row[0]`, not `echo $row[0]`. – theftprevention Aug 02 '13 at 20:50
@theftprevention you are correct, I have made the correction – Danny Aug 02 '13 at 20:51
Does not work. I updated my post. – drpelz Aug 02 '13 at 20:53
@drpelz `echo '
– Danny Aug 02 '13 at 20:57
1

You keep changing the post, just remember if you start out using single quotes you have to escape them if you want to use them literally – Danny Aug 02 '13 at 20:59

How to pass PHP variable to JavaScript function?

1 Answers1