AJAX request callback using jQuery

2018-12-31 03:00发布

I am new to the use of jQuery for handling AJAX, and have written a basic script to get the basics down. Currently I am POSTing an AJAX request to the same file, and I wish to do some additional processing based on the results of that AJAX call.

Here is my code:

**/*convertNum.php*/**

$num = $_POST['json'];
if (isset($num))
 echo $num['number'] * 2;
?>

<!DOCTYPE html>
<html>
 <head>
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
 <style type="text/css">
 td {border:none;}

 </style>
 </head>
 <body>

 <table width="800" border="1">
 <tr>
 <td align="center">Number To Send<br /><input type="text" id="numSend" size="40%" style="border:2px solid black;"></td>
 <td align="center">Number Returned<br /><input type="text" id="numReturn" size="40%" readonly></td>
 </tr>

 <tr><td align="center" colspan="4"><input type="button" value="Get Number" id="getNum" /></td></tr>
 </table>



 <script>
 $(document).ready(function () {
 $('#getNum').click(function () {
 var $numSent = $('#numSend').val();
 var json = {"number":$numSent};

 $.post("convertNum.php", {"json": json}).done(function (data) 
 {
 alert(data);
 }
 );

 });
 });
 </script>
 </body>
</html>

Here is the response I get if I submit the number '2':

4

<!DOCTYPE html>

<html>

<head>

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

 <style type="text/css">

td {border:none;}

</style>

</head>

<body>



<table width="800" border="1">

 <tr>

 <td align="center">Number To Send<br /><input type="text" id="numSend" size="40%" style="border:2px solid black;"></td>

 <td align="center">Number Returned<br /><input type="text" id="numReturn" size="40%" readonly></td>

 </tr>

 <tr><td align="center" colspan="4"><input type="button" value="Get Number" id="getNum" /></td></tr>

</table>

<script>

$(document).ready(function () {

 $('#getNum').click(function () {

 var $numSent = $('#numSend').val();

 var json = {"number":$numSent};

 $.post("convertNum.php", {"json": json}).done(function (data) 

 {

 alert(data);

 }

 );



 });

});

</script>
</body>
</html>

Obviously I'm only interested in receiving and using the number '4', hence my question: What is the best way to specify exactly what data I want returned?

Some thoughts I've had:

  • wrapping all my HTML inside a if statement (i.e., if $num isset, do NOT output html; else output HTML) but then I'm echoing HTML, and I'd rather not do that.
  • Setting up a separate PHP script to receive my AJAX call: That was what I did originally, and it works just fine. However, I am interested in keeping everything inside one file, and wanted to explore possible ways of doing so.

I'm sure there is an elegant way to do this. Thanks for any suggestions!

标签: php jquery ajax
5条回答
流年柔荑漫光年
2楼-- · 2018-12-31 03:07

The elegant way would be to have a separate(!) PHP file that will only output the number times 2 part of your current PHP. Then you generate an AJAX request from your current PHP to the new separate PHP.

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情到深处是孤独
3楼-- · 2018-12-31 03:14

I believe this post answers one aspect of what you are seeing - the echoing of the entire page in your alert box.

Next, here are some good posts for getting the basics of AJAX:

A simple example

More complicated example

Populate dropdown 2 based on selection in dropdown 1

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步步皆殇っ
4楼-- · 2018-12-31 03:14

You could add die(); after you echo the number if you really want to keep it on the same page. This will terminate the execution of the script. Don't forget to add brackets around the if statement:

if (isset($num)) {
  echo $num['number'] * 2;
  die();
}

It is, however, not the most elegant solution.

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与风俱净
5楼-- · 2018-12-31 03:25

I don't know how efficient this is but you can do something like:

<?php
 /*convertNum.php*/
 $num = isset($_POST['json']) ? $_POST['json'] : NULL; //you have errors in this line
 if (!is_null($num)){
  echo $num['number'] * 2;
  exit(); //exit from script
 }
?>
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姐姐魅力值爆表
6楼-- · 2018-12-31 03:33

NOTE *The better approach is to keep things like this in a separate file, makes it easier to read and easier to understand, especially if you use a good naming conversion.

It is a bad procedural approach but here I go :) :

<?php
$num = $_POST['json'];
if (isset($num))
 echo ($num['number'] * 2);
 die();
?>

or better yet:

<?php
$num = $_POST['json'];
if (isset($num))
 die($num['number'] * 2); //In case of error you could put them in brackets
?>

PS As alternative to die(); you could use exit();, they both do pretty much the same: terminating further execution

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