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I'm trying to make my Image Input pass a URL into a Javascript function using OnClick, however when I click on the input it does nothing. The code I am using is in PHP, and it echo's out a string with HTML tags in it.

echo "<input type=\"image\" onclick=\"imgp(\'" . $url .  "\')\" src=\"" . $url . "\" style=\"max-width: 20%; max-height: 20%\" /><br />";

The function that is getting called is as follows: 

function imgp(url){
            if(url != null){
                var text = document.getElementById("body").value + "[img]" + url + "[/img]";
                document.getElementById("body").value = text;
                imgpopup();
            }
        }
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before using echo to print the html test the function using normal html –  Pradyut Bhattacharya Mar 7 '14 at 19:24

1 Answer 1

up vote 0 down vote accepted

As I can't test with the code you posted, try to do the following:

echo '<input type="image" onclick="imgp(\" . $url . '\')" src="' . $url . '" style="max-width: 20%; max-height: 20%" /><br />';

If you use simple quote ('), inside the echo, you can use the double quote (") without using (), but instead, you need to use only with ('), and vice-versa.

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The {$url} thing doesn't seem to be outputting what it's supposed to... –  ProgrammingTurtle Mar 7 '14 at 19:35
    
Using echo '<input type="image" onclick="imgp(\'' . $url . '\')" src="' . $url . '" style="max-width: 20%; max-height: 20%" /><br />'; Seemed to work, update your code and I'll mark it as an answer. –  ProgrammingTurtle Mar 7 '14 at 19:37
    
There you go, done! Glad to help –  Leandragem Mar 7 '14 at 19:42
    
Merci, helped a lot! –  ProgrammingTurtle Mar 7 '14 at 19:43

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