Why the javascript function is not getting called if I add an extra parameter

Question

I have a really weird situation, I am calling my javascript function like this...

window.top.window.stopUpload(<? echo $result; ?>,<? echo $file_name; ?>);

Javascript function looks like this,

function stopUpload(success,filePath){
          var result = '';
          if (success == 1){
             result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>';
          }
          else {
             result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>';
          }
          document.getElementById('f1_upload_process').style.visibility = 'hidden';
          document.getElementById('f1_upload_form').innerHTML = result + '<input name="image_file" type="file" class="browse" /><input type="submit" name="submit_button" value="Upload"  class="browse"/>';
          document.getElementById('f1_upload_form').style.visibility = 'visible';     
           
          return true;   
    }

Above code does't execute stopUpload function.

However If I do like this,

window.top.window.stopUpload(<? echo $result; ?>);

and javascript like this,

function stopUpload(success){
          var result = '';
          if (success == 1){
             result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>';
          }
          else {
             result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>';
          }
          document.getElementById('f1_upload_process').style.visibility = 'hidden';
          document.getElementById('f1_upload_form').innerHTML = result + '<input name="image_file" type="file" class="browse" /><input type="submit" name="submit_button" value="Upload"  class="browse"/>';
          document.getElementById('f1_upload_form').style.visibility = 'visible';     
           
          return true;   
    }

With one one param, it works!

Question

Why it works with one param and not with 2? I have tried sending normal string like 'hello' instead of $file_name but still it does't call.

have you echo'd $file_name outside of the function to ensure there is a value? — atmd
– atmd, Commented May 10, 2012 at 11:57
Load the page in your browser and check the source. Odds are that $result or $file_name is null, in which case it would lead to bad syntax (i.e. window.top.window.stopUpload(,'hello') — fivedigit
– fivedigit, Commented May 10, 2012 at 11:57
Make sure that you escape your arguments with quotes, i.e. window.top.window.stopUpload('<? echo $result; ?>','<? echo $file_name; ?>'); — VisioN
– VisioN, Commented May 10, 2012 at 11:57

Nuno Costa · Accepted Answer · 2012-05-10 11:58:56Z

3

Call your function like this:

window.top.window.stopUpload(<? echo $result; ?>,'<? echo $file_name; ?>');

Hope it helps.

answered May 10, 2012 at 11:58

Nuno Costa

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Comments

Asif · Accepted Answer · 2020-02-20 23:27:35Z

0

Try this :

window.top.window.stopUpload('<? echo $result; ?>','<? echo $file_name; ?>');

Remember, $result should not be any numerical value. User $result = '1' instead.

And change success == '1') in your if statement.

Hope it helps.

answered Feb 20, 2020 at 23:27

Asif

391 silver badge6 bronze badges

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