javascript function with 2 arguments is not working

Question

DEscription :

I have a php script that displays the div on an html page

echo '<div class = "star_box" id = "'.$each_status['post_id'].'" onmousedown = "star_it(this.id)"></div>';

Now when I click on this div the star_it function gets called and it works perfect...

The problem

I want to pass another argument in the star it function .. like this

echo '<div class = "star_box" id = "'.$each_status['post_id'].'" onmousedown = "star_it(this.id,'.$each_status['regno'].')"></div>';

or a simple alphabet if any like star_it(this.id,s)

but when I do so the function stops working as when I click the div the function does not gets called ....

I dont know why I am stuck and now my star rating system does not work... I have absolutely no idea what is wrong

Anyone ??

if the second attribute is a string or letter, add quotes around the parameter. otherwhise your php script will produce invalid syntax javascript. — dognose
– dognose, Commented Oct 24, 2014 at 17:35
What's the resulting markup of the second echo? Are you getting any JS errors? perhaps $each_status['regno'] is empty. — Carlos
– Carlos, Commented Oct 24, 2014 at 17:35
@BrianDriscoll thats not the point the star_it() if you really want to know should just alert me with two values passed for now .... it just doesn't do any this ... if I pass this.id it works good but as soon as the second argument comes I cant call the function — Rks Rock
– Rks Rock, Commented Oct 24, 2014 at 17:40
Does the star_it function accept a second argument? If it's not running at all, you're getting an error. Check your javascript console. — Leo Bedrosian
– Leo Bedrosian, Commented Oct 24, 2014 at 17:41

Xanarus · Accepted Answer · 2014-10-24 19:43:56Z

3

You should modify your string quote's like this :

<?php 
$status = "second test";
echo '<div class="star_box" id="'.$id.'" onmousedown="star_it(this.id,'."'".$status."'".');" >Hello</div>';
?>

<script>
    function star_it(foo1,foo2){
        console.log(foo1);
        console.log(foo2);
    }
</script>

This example works for me.

So with your code :

<?php 
echo '<div class="star_box" id="'.$id.'" onmousedown="star_it(this.id,'."'".$each_status['regno']."'".');" >Hello</div>';
?>

edited Oct 24, 2014 at 19:43

answered Oct 24, 2014 at 17:40

Xanarus

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2 Comments

Marc B Over a year ago

no. that's not "escaping". you're just quoting. for PHP->javascript insertion, you should ALWAYS use echo json_encode($var)

Rks Rock Over a year ago

You sir are a life saver :) ... I am not really familiar with escaping strings quotes ... thankyou so much :)

s4adev · Accepted Answer · 2014-10-24 17:53:50Z

2

Here you go

echo '<div class = "star_box" id = "'.$each_status['post_id'].'" onmousedown = "star_it(this.id,\''.$each_status['regno'].'\')"></div>';

you forget to add \' before and after your string , the JS engine on the browser will treat your string as an undefined variable and the function will not work because of that .

//my English is not so good.

answered Oct 24, 2014 at 17:53

s4adev

212 bronze badges

Comments

JuniarZ · Accepted Answer · 2014-10-24 17:36:38Z

-1

Try

echo '<div class = "star_box" id = "'.$each_status['post_id'].'" onmousedown = "star_it(this.id,\"'.$each_status['regno'].'\")"></div>';

answered Oct 24, 2014 at 17:36

JuniarZ

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