Form not updating using jquery ajax PHP

Question

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Have a form where if one enters a number, it calls back data from the mysql table based on that number, then writes the data to the page w/out reloading the whole page. Even tho the ajax is working and showing "Success", the PHP isn't getting any value.
All the code in one the same page (ajax and PHP), and the URL for the Ajax is the page itself.

What I have tried:

HTML

<pre><!DOCTYPE html>

<html>
	
	<script src="Js/jquery-3.4.1.js"></script>

	<body>
	
		<form method="post" id="myForm" name="myForm" action="" enctype="multipart/form-data" > 	
				
			<input type="Submit" value="submit"  style="visibility:hidden"/>				
			<input type="text" name="square" id="square" size="2" onchange="show()">			
			<input type="hidden" name="submitType" id="submitType">
	
		</form>   
		
		<script type="text/javascript">
		
			function show() 
			{	
				document.getElementById("submitType").value="Show";	

				$(document).ready(function()
				{
					var square=$("#square").val();	
					alert(square)
					var submitType=$("#submitType").val();
			  					
					$.ajax
					({		
						url: "help.php",									
						type: "POST",			
										
						data:
						{
							submitType:"Show",
							square:square		
						},
						
						cache:"false",
						dataType:"html",
						
						error: function (xhr, ajaxOptions,errorThrown)
						{						
							alert("ERROR");
						}, 
				
						success: function (data)
						{            
							alert("Success");
						}					
					});
					
					return false; 
				});						 
			}
		</script>

		<?php
			
			if(isset($_POST["submitType"])=="Show")
			{
				include "mysql.php";  	
							
				$square=$_POST["square"];	

				print_r($square);				
				
				$sqlQuery="Select links from myTable where number='$square'";
				mysqli_query($conn, $sqlQuery) or die(mysqli_error($conn));					
				$result = mysqli_query($conn,$sqlQuery);
				$array = mysqli_fetch_row($result); 					
				print_r("<script type=\"text/javascript\">");	
				print_r("currentSquare=".$square.";");
				print_r("info='".$array[0]."';");			
				print_r("</script>");					
			}

		?>
		
	</body>

</html>

Posted 15-Dec-19 8:44am

Member 13358514

Updated 16-Dec-19 2:07am

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Member 13358514 15-Dec-19 14:51pm

FYI, have tried:

1)document.getElementById("myForm").submit(); in place of the document.ready
- no change

2)$("#myForm").on("submit", function (even); in place of the document.ready
and added event.preventDefault();
- still no change

3) document.getElementById("myForm").submit
- reloaded the whol page

1 solution

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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

ZurdoDev · Answer 1 · 2019-12-16T02:08:00

Solution 1

You say that the message "Success" is displayed which means your success function is running. However, you don't do anything else in the success function. That would be where you change a value on the page.

Posted 16-Dec-19 2:08am

ZurdoDev

Form not updating using jquery ajax PHP

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Solution 1

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Form not updating using jquery ajax PHP

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