JavaScript: missing ) after argument list

Question

This javascript produces an error:

missing ) after argument list

In firebug with the code:

<script type=\"text/javascript\">
function add( answer )
{   
  $.post('../page.php?cmd=view&id=3523', 
    {user_id: 3523, other_user_id: 2343}, function(d)
      $(answer).after(\"<span>Done!</span>\").remove();
    });
  }
}
</script>

What am I doing wrong?

David Hedlund · Accepted Answer · 2009-12-08 10:01:33Z

9

function d misses an opening bracket, {

$(answer).after( should not be escaped \", just a regular quote will do "

answered Dec 8, 2009 at 10:01

David Hedlund

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2 Comments

Lea Over a year ago

I knew it was something simple, Im just too noob:) Thanks for your help!

David Hedlund Over a year ago

since you're using firebug; when you encounter things like this, you can always paste the code into the firebug console and reproduce the error, and from there, if you don't find the syntax error, you can try and remove things, bit by bit, and see what removed line causes the error to go away.

Amirhossein Mehrvarzi · Accepted Answer · 2015-10-19 13:27:23Z

4

Close post() function. third string from bottom should be ), not }.

EDIT: sorry, should be like this:

<script type=\"text/javascript\">
function add( answer )
{   
    $.post('../page.php?cmd=view&id=3523', {user_id: 3523, other_user_id: 2343}, function(d) {
        $(answer).after(\"<span>Done!</span>\").remove();
    });
}

edited Oct 19, 2015 at 13:27

Amirhossein Mehrvarzi

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answered Dec 8, 2009 at 10:01

Kuroki Kaze

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Comments

yoda · Accepted Answer · 2009-12-08 10:02:02Z

2

Why are you escaping quotes? The problem is here :

$(answer).after(\"<span>Done!</span>\").remove();

change to

$(answer).after("<span>Done!</span>").remove();

or

$(answer).after('<span>Done!</span>').remove();

Also, you're missing a { after the post() function (probably you missed the right spot, since there's another in the wrong place), so the final output :

<script type=\"text/javascript\">
function add( answer )
{   
$.post('../page.php?cmd=view&id=3523', {user_id: 3523, other_user_id: 2343}, function(d) {
            $(answer).after("<span>Done!</span>").remove();
        });
}
</script>

answered Dec 8, 2009 at 10:02

yoda

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Comments

markmywords · Accepted Answer · 2009-12-08 10:04:35Z

2

function add( answer )
{   
$.post('../page.php?cmd=view&id=3523', 
       {user_id: 3523, other_user_id: 2343}, 
       function(d){
         $(answer).after("<span>Done!</span>").remove()
       });
};

answered Dec 8, 2009 at 10:04

markmywords

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