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(){
$('#dump_here').append("test");
}
It feels really weird. The first line is working fine. By the way... The function is enclosed inside a document ready but putting it outside document ready doesn't work.
jQuery imports properly as I am using AJAX there too.
onClick("ILike()") isn't the right syntax. It should be:
onClick="ILike()"
change it like this
echo '<div id ="school_content"><h3>School</h3>';
while($row = mysqli_fetch_array($result))
{
echo '<p><input type="checkbox" onclick="ILike()" />'.$row["School"].'</p>';
}
echo '</div>';
this is incorrect onClick("ILike()")
The right way is onClick onClick="ILike();"
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