Update div in html using Ajax function failed - php

Question

I'm trying to display the search result of my page under the the search area. So I used AJAX to display the result in a div. but I could'nt get it work.
I have three main pieces, the div, the searchResult page and the ajax function
<input type="text" name="studentName">
<button type="submit" name="searchByName" onclick='get_info();'>بحث</button>
<div id="searchResult"><b></b></div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script type="text/javascript">
function get_info() { // Call to ajax function
$.ajax({
type: "POST",
url: "NameSearchResult.php", // Name of the php files
data: {name: <?php echo $_POST['studentName']; ?>},
success: function(html)
{
$("#searchResult").html(html);
}
});
}
and my search Page:
<?php
include_once 'dbConfigBDO.php';
$studentName = $_POST["name"];
$counter=0;
$emptyString = "لايوجد";
$sql = "SELECT * FROM Student";
$result = $conn->query($sql);
$row_count = $result->rowCount();
if ($row_count > 0){
.......... }
Now when I search nothing appears, although it works when I put all the code in one page (which would be messy in term of the appearance of the result!).

Answer 1

From function return the output as per below:
return json_encode($result);
In ajax call use dataType:"json" and show your html
Example ajax call:
$.ajax({
type: "POST",
dataType:"json",
url: "NameSearchResult.php", // Name of the php files
data: {name: $("#studentName").val()},
success: function(html)

Answer 2

change code like this
<input type="text" name="studentName" id="studentName">
<button type="submit" name="searchByName" onclick='get_info();'>بحث</button>
<div id="searchResult"><b></b></div>
<script>
$.ajax({
type: "POST",
url: "NameSearchResult.php", // Name of the php files
data: {name: $("#studentName").val()},
success: function(html)
{
$("#searchResult").html(html);
}
});
}
</script>

Answer 3

inside ajax success method try catch what you are getting
success: function(html)
{
console.log(html);
}
if you getting something then your code must be work.