I have been looking into a problem, but I can't figure out how to solve it or what the solution is. I am working on my website using
So now I am even more puzzled as to why it doesn't work when I try to insert it into the table.
CodePudding user response:
I know this isn't the right answer for the question, but it is an alternative and maybe you will find it somewhat helpful. The solution I came up with was to take the value given by your jquery and set it to an invisible input value inside the sumbit form.
<form id="reviewForm" method="POST" action="">
<textarea type="text" id="reviewTextarea"name="uploadReview</textarea>
<div id="submitReview">
<input type="hidden" id="jqValue" name="jqueryValue" value=""/>
<button id="submitButton" type="submit" form="reviewForm">Submit</button>
</div>
Then take the variable ratedIndex
and give it to your inputjqValue
value like this
$('.fa-star').on('click', function (){
ratedIndex = parseInt($(this).data('index'));
$("#jqValue").val(ratedIndex);
});
Lastly call in your php
$ratedIndex=$_POST["jqueryValue"];
Like I said this just an alternate way to solve your problem if you are stuck and cant make ajax
work you can use this. I posted because you might have not thought about doing it this way.