OnClick is not working with checkbox created by PHP

Question

Well... I don't know what's going on. I have this code which works normally but the checkbox doesn't append anything or doesn't notify the script which really puzzles me. The php code creating the checkbox is:

echo '<div id ="school_content"><h3>School</h3>';
    while($row = mysqli_fetch_array($result))
   {
    echo '<p><input type="checkbox" onClick="ILike()" />'.$row["School"].'</p>';
   }
   echo '</div>';

Here is the plain HTML File:

<div id="container" class="row">
        <div id="School" class="col"></div>
     <div id="Department" class="col"></div>
        <div id="Level" class="col"></div>
        <div id="Source" class="col"></div>
        <div id="Coding" class="col"></div>
    </div>
    <input type='submit' value='Show Result' id='result' onClick=""/>
    <div id="dump_here">The dumping area:</div>

And here is the javascript code appending. I don't know why this is not working:

 $('#dump_here').append("test");

        function ILike(){
                $('#dump_here').append("test");
        }

It feels really weird. The first line is working fine. By the way... The function is enclosed inside a document ready but putting it outside document ready doesn't work.

jQuery imports properly as I am using AJAX there too.

helion3 · Accepted Answer · 2014-02-12 23:05:01Z

4

onClick("ILike()") isn't the right syntax. It should be:

onClick="ILike()"

answered Feb 12, 2014 at 23:05

helion3

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2 Comments

helion3 Over a year ago

@Gempio then you're missing something we're not seeing in the code. It works fine jsfiddle.net/8k5fE

Maciej Musialek Over a year ago

What I missed out is that the code needs to be in fact outside document ready before PHP runs. Works just fine now and I'm really happy :D Thanks for correcting my onclick :)

isJustMe · Accepted Answer · 2014-02-12 23:05:22Z

1

change it like this

echo '<div id ="school_content"><h3>School</h3>';
    while($row = mysqli_fetch_array($result))
   {
    echo '<p><input type="checkbox" onclick="ILike()" />'.$row["School"].'</p>';
   }
   echo '</div>';

answered Feb 12, 2014 at 23:05

isJustMe

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Comments

Jorhel Reyes · Accepted Answer · 2014-02-12 23:06:23Z

1

this is incorrect onClick("ILike()")

The right way is onClick onClick="ILike();"

answered Feb 12, 2014 at 23:06

Jorhel Reyes

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Comments

user3303864 · Accepted Answer · 2014-02-12 23:13:08Z

0

your syntax in the input element is incorrect. The correct is:

<input type="checkbox" onclick="ILike()" />

You can use your browser console to find issues like that one, which can greatly help you :D Also, I recommend you to put javascript code in the head of your HTML document:

<head>
    <script type="text/javascript">
        $().ready(function(){
            $(".checkboxILike").click(function(){
                ILike();
            });
        });
    </script>
</head>

Of course, you will need put a class named checkboxILike in every checkbox that will call the function

answered Feb 12, 2014 at 23:13

user3303864

1 Comment

helion3 Over a year ago

You could also use .click(ILike); to avoid wrapping a function in a function.

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