Using AJAX

Question

Trying to figure out Ajax. Help me find the file :( It should work like this: enter text in one input, output to another. And in my second page content is stupidly written (

<script language="javascript" type="text/javascript"> <!-- function ajaxFunction1(){ $.post("task12.php", { "in" : $("#in1").val()}, function(data){ $("#out1").val(data); }, "text"); } --> </script> Demo post metod<br> Input text: <input type="text" onkeyup="ajaxFunction1();" name="in1" id="in1" /> ajax result: <input type="text" name="out1" id="out1" /> </form>

Added.

task12.php:

 <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <title> Task12</title> <link href="style7.css" rel="stylesheet"> <script type="text/javascript" src="jquery.min.js"></script> </head> <script language="javascript" type="text/javascript"> <!-- function ajaxFunction1(){ $.post("task10.php", { "in" : $("#in1").val()}, function(data){ $("#out1").val(data); }, "text"); } --> </script> <body> <form> Demo post metod<br> Input text: <input type="text" onkeyup="ajaxFunction1();" name="in1" id="in1" /> ajax result: <input type="text" name="out1" id="out1" /> </form> </div> </body> </html>

$ .post ("task12.php", {"in": $ ("# in1"). val ()} Why then do you have HTML in the task12.php file? You should simply process the data in this file and send them back.
@Shrek, just about, but here you don’t even know what it has) task12.php -> task10.php
@shrek, @Boolean I thought that there is writing the name of the file of the page in which the box.
There's just bare PHP, which gets data from you and then sends it back, no HTML should be there!

angry 5,855 15 56 171 · Accepted Answer · 2011-11-29T13:12:56

On the client side:

 $.post( "/ajaxtest.php", { param1: "param1", param2: 2 }, onAjaxSuccess ); function onAjaxSuccess(data) { // Здесь мы получаем данные, отправленные сервером и выводим их на экран. alert(data); }

On the server side, no HTML:

 <?php // файл http://hostname/ajaxtest.php echo "I get param1 = ".$_POST['param1']." and param2 = ".$_POST['param1']; ?>

In response, you will get the string I get param1 = param1 and param2 = 2 .

ling ling 14k one 18 45 · Answer 2 · 2011-11-29T13:00:07

Apparently, in task12.php you have the whole page displayed, and not just the result. This can happen, for example, when using a template engine. Make sure that you have displayed only the desired value, without any <html> -influx. Or extract the value from the item from the answer page:

 data = $(data).find('#content').html(); $("#out1").val(data);

@ Timmi55, there should be nothing at all, except for the desired value.

Using AJAX

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