CODEWITHSUNDEEP
phpmysqlajaxjson
Donny
Donny

Reputation: 699

Auto Fill second textbox with first textbox input

auto fill second textbox with the input of first textbox. Like if I choose Jane as first name I want the second textbox to retrieve the last name from database and auto fill it with Doe. I am getting no auto input on second textbox not sure what I am doing wrong.

Here is the test.php code

<?php   

?>
<script src="jquery-1.11.1.min.js"></script>
<script>
$(document).ready(function(){
	  $("#firstName").change(function(){
		  var firstName=$(this).val();
		   if(firstName != ''){
		 	      $.ajax({
                              type:"post",
                              url:"insert_process.php",
                              data:"firstName="+firstName,
							  datatype:"json",
				              success:function(data){ $("#lastname").val(data);
							    $('#ename').css( "background-color","#B3CBD6"  ) 
							 $('#ename').animate({backgroundColor: "#fff",});
							   }
                                
	                                                       });
		   }
		   else{
			   $("#lastname").val("");
			   }
		  });
	 });
</script>
<!DOCTYPE html>
<html>
<head>
<title>Systems Request </title>

</head>
<body>

<div align="center">

<form action = "insert.php" method ="post" class="form" style="width: 285px; height: 192px">

<br><br>First Name<br>
<input type="text" id="firstName" name="firstName">

<br>&nbsp;Last Name<br>
<input type="text" id="lastname" name="lastname" ><br><br>

<input type="submit" value= "Submit Name "><br>

</form>
</div>
</body>
</html>

and then here is insert_proccess.php code

<?php
header('Content-Type: application/json');
$host = "localhost";
$user = "root";
$db_name= "people";
$pass= "systems399";
$con = mysql_connect($host, $user, $pass);
$db_conx=mysql_select_db("people", $con);
$fname=  $_POST["firstName"];
	   $sql="SELECT * FROM  names WHERE firstName='$fname'   ";	
		$query= mysqli_query($db_conx, $sql);
	$row = mysqli_fetch_array($query2 MYSQLI_ASSOC);
	$rc	= $row["lastname"];

echo json_encode ($rc);
 
?>

Upvotes: 3

Views: 3773

Answers (1)

Typhomism
Typhomism

Reputation: 284

Is the request actually being sent (you can checking the networking tab)? Can you alert the returned data? Basically to be more helpful I would need more information on where exactly the process is failing, at a quick glance, try setting up your data structure more like this:

data: { "firstName": firstName }

Upvotes: 1

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