OnClick无法使用PHP创建的复选框

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.

Answer 1

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

onClick="ILike()"