简单的 jQuery、PHP 和 JSONP 示例？
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Simple jQuery, PHP and JSONP example?
提问by Jeff
I am facing the same-origin policy problem, and by researching the subject, I found that the best way for my particular project would be to use JSONP to do cross-origin requests.
我面临着同源策略问题，通过研究该主题，我发现我的特定项目的最佳方法是使用 JSONP 进行跨域请求。
I've been reading this article from IBM about JSONP, however I am not 100% clear on what is going on.
我一直在阅读IBM 关于 JSONP 的这篇文章，但是我并不是 100% 清楚发生了什么。
All I am asking for here, is a simple jQuery>PHP JSONP request (or whatever the terminology may be ;) ) - something like this (obviously it is incorrect, its just so you can get an idea of what I am trying to achieve :) ):
我在这里要求的只是一个简单的 jQuery>PHP JSONP 请求（或任何可能的术语；）） - 像这样（显然它是不正确的，它只是为了让您了解我想要实现的目标:) ):
jQuery:
jQuery：
$.post('http://MySite.com/MyHandler.php',{firstname:'Jeff'},function(res){
alert('Your name is '+res);
});
PHP:
PHP：
<?php
$fname = $_POST['firstname'];
if($fname=='Jeff')
{
echo 'Jeff Hansen';
}
?>
How would I go about converting this into a proper JSONP request? And if I were to store HTML in the result to be returned, would that work too?
我将如何将其转换为正确的 JSONP 请求？如果我将 HTML 存储在要返回的结果中，那也行吗？
回答by Liam Bailey
When you use $.getJSON on an external domain it automatically actions a JSONP request, for example my tweet slider here
当您在外部域上使用 $.getJSON 时，它会自动执行 JSONP 请求，例如这里的推文滑块
If you look at the source code you can see that I am calling the Twitter API using .getJSON.
如果您查看源代码，您会发现我正在使用 .getJSON 调用 Twitter API。
So your example would be: THIS IS TESTED AND WORKS (You can go to http://smallcoders.com/javascriptdevenvironment.htmlto see it in action)
所以你的例子是：这已经过测试并且有效（你可以去http://smallcoders.com/javascriptdevenvironment.html查看它的实际效果）
//JAVASCRIPT
$.getJSON('http://www.write-about-property.com/jsonp.php?callback=?','firstname=Jeff',function(res){
alert('Your name is '+res.fullname);
});
//SERVER SIDE
<?php
$fname = $_GET['firstname'];
if($fname=='Jeff')
{
//header("Content-Type: application/json");
echo $_GET['callback'] . '(' . "{'fullname' : 'Jeff Hansen'}" . ')';
}
?>
Note the ?callback=? and +res.fullname
注意 ?callback=? 和 +res.fullname
回答by Ewout Kleinsmann
First of all you can't make a POST request using JSONP.
首先，您不能使用 JSONP 发出 POST 请求。
What basically is happening is that dynamically a script tag is inserted to load your data. Therefore only GET requests are possible.
基本上发生的是动态插入脚本标记以加载您的数据。因此只有 GET 请求是可能的。
Furthermore your data has to be wrapped in a callback function which is called after the request is finished to load the data in a variable.
此外，您的数据必须包装在回调函数中，该函数在请求完成后调用以将数据加载到变量中。
This whole process is automated by jQuery for you. Just using $.getJSON on an external domain doesn't always work though. I can tell out of personal experience.
整个过程由 jQuery 为您自动化。只是在外部域上使用 $.getJSON 并不总是有效。我可以从个人经历中说出来。
The best thing to do is adding &callback=? to you url.
最好的办法是添加 &callback=? 给你网址。
At the server side you've got to make sure that your data is wrapped in this callback function.
在服务器端，您必须确保您的数据包含在此回调函数中。
ie.
IE。
echo $_GET['callback'] . '(' . $data . ')';
EDIT:
编辑：
Don't have enough rep yet to comment on Liam's answer so therefore the solution over here.
没有足够的代表对 Liam 的回答发表评论，因此这里的解决方案。
Replace Liam's line
替换利亚姆的线路
echo "{'fullname' : 'Jeff Hansen'}";
with
和
echo $_GET['callback'] . '(' . "{'fullname' : 'Jeff Hansen'}" . ')';
回答by Meni Samet
More Suggestion
更多建议
JavaScript:
JavaScript：
$.ajax({
url: "http://FullUrl",
dataType: 'jsonp',
success: function (data) {
//Data from the server in the in the variable "data"
//In the form of an array
}
});
PHP CallBack:
PHP回调：
<?php
$array = array(
'0' => array('fullName' => 'Meni Samet', 'fullAdress' => 'New York, NY'),
'1' => array('fullName' => 'Test 2', 'fullAdress' => 'Paris'),
);
if(isset ($_GET['callback']))
{
header("Content-Type: application/json");
echo $_GET['callback']."(".json_encode($array).")";
}
?>
回答by mckoch
To make the server respond with a valid JSONP array, wrap the JSON in brackets ()and preprend the callback:
要使服务器使用有效的 JSONP 数组进行响应，请将 JSON 括在方括号中()并预先添加callback：
echo $_GET['callback']."([{'fullname' : 'Jeff Hansen'}])";
Using json_encode()will convert a native PHP array into JSON:
使用json_encode()会将原生 PHP 数组转换为 JSON：
$array = array(
'fullname' => 'Jeff Hansen',
'address' => 'somewhere no.3'
);
echo $_GET['callback']."(".json_encode($array).")";
回答by Mayur S
Simple jQuery, PHP and JSONP example is below:
简单的 jQuery、PHP 和 JSONP 示例如下：
window.onload = function(){
$.ajax({
cache: false,
url: "https://jsonplaceholder.typicode.com/users/2",
dataType: 'jsonp',
type: 'GET',
success: function(data){
console.log('data', data)
},
error: function(data){
console.log(data);
}
});
};<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>回答by user4000483
$.ajax({
type: "GET",
url: '<?php echo Base_url("user/your function");?>',
data: {name: mail},
dataType: "jsonp",
jsonp: 'callback',
jsonpCallback: 'chekEmailTaken',
success: function(msg){
}
});
return true;
In controller:
在控制器中：
public function ajax_checkjp(){
$checkType = $_GET['name'];
echo $_GET['callback']. '(' . json_encode($result) . ');';
}
回答by Atul Sharma
Use this ..
用这个 ..
$str = rawurldecode($_SERVER['REQUEST_URI']);
$arr = explode("{",$str);
$arr1 = explode("}", $arr[1]);
$jsS = '{'.$arr1[0].'}';
$data = json_decode($jsS,true);
Now ..
现在 ..
use $data['elemname']to access the values.
用于$data['elemname']访问值。
send jsonp request with JSON Object.
使用 JSON 对象发送 jsonp 请求。
Request format :
请求格式：
$.ajax({
method : 'POST',
url : 'xxx.com',
data : JSONDataObj, //Use JSON.stringfy before sending data
dataType: 'jsonp',
contentType: 'application/json; charset=utf-8',
success : function(response){
console.log(response);
}
})
