Javascript function onclick

Question

I am trying to call a jquery script that will post data to a PHP form. Been googling this for 5h now but i cant seems to solve it on my own. The Script looks like this.

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script language="JavaScript" type="text/javascript">
    function postscrape(igid, iguser, groupid) {

        alert(igid);
        $.post("scrapeadder.php", {
                instagramid: igid,
                username: iguser,
                groupid: groupid
            },
            function(data, status) {
                alert("Data: " + data + "\nStatus: " + status);
            });
    };
</script>

And my HTML/PHP code to call this function looks like:

  echo '<td><button type="button" class="btn btn-success datascraper" name="addscrape" id="addscrape" onclick="postscrape('.$row["instagram_id"].', '.$row["user_name"].', 1);">Add to Scrape</button></td>';

The error i get is: Uncaught ReferenceError: X is not defined

Where X = The iguser (which gets from $row["user_name"] If i check the HTML code on the homepage it list all the $rows correct. So my guess is i messed up in the function.

Maybe you should post the code that contains the error instead ? — adeneo
– adeneo, Commented Jun 8, 2015 at 23:46
are you able to print out the $row["instagram_id"].', '.$row["user_name"] and sure that there valid — ShapCyber
– ShapCyber, Commented Jun 8, 2015 at 23:53
Yes, when i check the source code in my webbrowser i can see the correct values in all those variables. — Zara Rebecka Elisabeth Zentio
– Zara Rebecka Elisabeth Zentio, Commented Jun 8, 2015 at 23:53
Always look at the clientside code generated by the serverside. Do not debug from looking at server code when it is a JS error. — epascarello
– epascarello, Commented Jun 8, 2015 at 23:53

Barmar · Accepted Answer · 2015-06-08 23:51:12Z

2

You need to put the user name in quotes.

onclick="postscrape('.$row["instagram_id"].', \''.$row["user_name"].'\', 1);"

answered Jun 8, 2015 at 23:51

Barmar

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Comments

Eric · Accepted Answer · 2015-10-31 19:20:12Z

2

You didn't quote the user name. The

echo 'postscrape('.$row["user_name"].')'

yields

postscrape(X)

...where X is undefined.

edited Oct 31, 2015 at 19:20

Eric

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answered Jun 8, 2015 at 23:50

Christoph Sommer

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