insert data to mysql table using jquery ajax [duplicate] - php

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What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
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Closed 5 years ago.
i want to insert data to mysql table called vacancy using ajax.ajax successfully function alert run.but data don't save in table.and errors are occer.i am new for php and ajax
html code
<form class="form-horizontal" action="postvaccode2.php" type="post">
<fieldset>
<!-- Form Name -->
<legend>Post Vacancy</legend>
<!-- Select Basic -->
<div class="form-group">
<div class="col-md-5">
<select id="cato" name="cato" class="form-control" id="cato">
<option value="IR">IT</option>
<option value="USA">Finance</option>
</select>
</div>
</div>
<!-- Text input-->
<div class="form-group">
<div class="col-md-6">
<input id="comid" name="comid" type="text" placeholder="comid" class="form-control input-md" required="" >
</div>
</div>
<!-- Text input-->
<div class="form-group">
<div class="col-md-6">
<input id="loc" name="loc" type="text" placeholder="city or town" class="form-control input-md" required="">
</div>
</div>
<!-- Text input-->
<div class="form-group">
<div class="col-md-8">
<input id="qul" name="qul" type="text" placeholder="qulification" class="form-control input-md">
</div>
</div>
<!-- Text input-->
<div class="form-group">
<div class="col-md-8">
<input id="indate" name="indate" type="date" placeholder="indate" class="form-control input-md">
</div>
</div>
<!-- Text input-->
<div class="form-group">
<div class="col-md-4">
<input id="expdate" name="expdate" type="date" placeholder="expdate" class="form-control input-md" required="">
</div>
</div>
<div class="form-group">
<div class="col-md-8">
<input id="gpa" name="gpa" type="text" placeholder="gpa" class="form-control input-md">
</div>
</div>
<div class="form-group">
<div class="col-md-8">
<input id="des" name="des" type="text" placeholder="description" class="form-control input-md">
</div>
</div>
<div class="form-group">
<div class="col-md-8">
<input id="title" name="title" type="text" placeholder="title" class="form-control input-md">
</div>
</div>
<div class="form-group">
<div class="col-md-8">
<input id="comname" name="comname" type="text" placeholder="name" class="form-control input-md">
</div>
</div>
<div class="form-group">
<div class="col-md-8">
<input id="vacpost" name="vacpost" type="submit" class="form-control input-md">
</div>
</div>
</fieldset>
</form>
script
</script>
<script>
$('#vacpost').click(function () {
$.ajax({
type: "POST",
url: "postvaccode2.php",
data:{id:$('#comid').val(),cato:$('#cato').val(),loc:$('#loc').val(),quli:$('#qul').val(),indate:$('#indate').val(),expdate:$('#expdate').val(),gpa:$('#gpa').val(),des:$('#des').val(),title:$('#title').val(),comname:$('#comname').val()},
success:function () {
alert('Sccessfuly Post the Job vacency');
location.reload(true);
}
});
});
</script>
postvaccode2.php
<?php
$conn=mysqli_connect("localhost","root","","internship");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id=$_POST['id'];
$cato=$_POST['cato'];
$loc=$_POST['loc'];
$quli=$_POST['quli'];
$indate=$_POST['indate'];
$expdate=$_POST['expdate'];
$gpa=$_POST['gpa'];
$des=$_POST['des'];
$title=$_POST['title'];
$comname=$_POST['comname'];
$sql="insert into vacancy(companyid,catogary,location,qulification,indate,expectgpa,description,title,expdate,companyname) values('$id','$cato','$loc','$quli','$indate','$gpa','$des','$title','$expdate','$comname',)";
$conn->query($sql);
?>