PHP Ajax回调从我的PHP指定的数据 [英] PHP Ajax callback specified data from my php
问题描述
好了,这里是我的问题;
OK , here is my issue;
我的问题是,当我回调我的PHP文件AJAX调用我的文件的所有内容
My issue is that when i callback my php file with ajax it calls all the content of my file
这是我的code
包括我view.php文件我source.php文件,该文件
my source.php file which including my view.php file
<?php
if(isset($_POST['test'])){
echo "OK";
}
include_once "view.php";
?>
我的view.php文件
my view.php file
<!DOCTYPE html>
<html>
<head>
<title>Hello!</title>
</head>
<body>
<h1>Form</h1>
<form action="" method="post" id="form">
<input type="text" name="test" id="test" /><br>
<input type="submit" id="sub" name="sub" />
</form>
<div id="result"></div>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#sub").click(function(){
$.ajax({
type: "POST",
data: $('#form').serialize(),
success: function(slider_data){
$("#result").text(slider_data);
}
});
return false;
});
});
</script>
</body>
</html>
现在,当我提出我的表格 我#result这里是
Now when i submit my form my #result here is
OK<!DOCTYPE html>
<html>
<head>
<title>Hello!</title>
</head>
<body>
<h1>Form</h1>
<form action="" method="post" id="form">
<input type="text" name="test" id="test" /><br>
<input type="submit" id="sub" name="sub" />
</form>
<div id="result"></div>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#sub").click(function(){
$.ajax({
type: "POST",
data: $('#form').serialize(),
success: function(slider_data){
$("#result").text(slider_data);
}
});
return false;
});
});
</script>
</body>
</html>
我想只显示单词OK或什么的,我想和prevet发送的所有数据,其中包括文件
i want to show only the word "OK" or whatever i want and prevet sending all data and included file
推荐答案
因为你有包括
在 source.php
这就是为什么你得到 view.php
了。
because you have include
in source.php
that's why you are getting view.php
again.
更改code这样
source.php
source.php
<?php
if(isset($_POST['test'])){
echo "OK";
}
?>
&LT;脚本&GT;
在view.php会是这样
<script>
in view.php will be like this
<script>
$(document).ready(function(){
$("#sub").click(function(event){
event.prevantDefault();
$.ajax({
url:"source.php" // url to source.php
type: "POST",
data: $('#form').serialize(),
success: function(slider_data){
$("#result").text(slider_data);
}
});
return false;
});
});
</script>
这篇关于PHP Ajax回调从我的PHP指定的数据的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!