javascript 使用 JQuery 从 PHP 获取 JSON 响应
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Using JQuery to get JSON response from PHP
提问by WelshJohn
I have accumulated and chopped up about 5 or 6 different tutorials of this now, and I still can't find out what's wrong!
我现在已经积累和切碎了大约 5 到 6 个不同的教程，但我仍然无法找出问题所在！
Using JQuery Mobile (phonegap) to send and receive data to a PHP server. I cannot seem to get the JQuery script to pick up a response. PHP on server:
使用 JQuery Mobile (phonegap) 向 PHP 服务器发送和接收数据。我似乎无法让 JQuery 脚本获取响应。服务器上的 PHP：
<?php
// Set up associative array
$data = array('success'=> true,'message'=>'Success message: worked!');
// JSON encode and send back to the server
echo json_encode($data);
?>
JQuery Function (Which is being called):
JQuery 函数（被调用）：
<script type="text/javascript">
$(function () {
$('#inp').keyup(function () {
var postData = $('#inp').val();
$.ajax({
type: "POST",
dataType: "json",
data: postData,
beforeSend: function (x) {
if (x && x.overrideMimeType) {
x.overrideMimeType("application/json;charset=UTF-8");
}
},
url: 'removedmyurlfromhere',
success: function (data) {
// 'data' is a JSON object which we can access directly.
// Evaluate the data.success member and do something appropriate...
if (data.success == true) {
alert('result');
$('.results').html(data.message);
} else {
$('.results').html(data.message);
}
}
});
});
});
</script>
Sorry for the formatting, it all went pair shaped copying it across. Removed my url.
抱歉格式化，这一切都是成对复制的。删除了我的网址。
I know that the function is firing, because if I remove the alert('here'); and put it above the ajax call it displays.
我知道该函数正在触发，因为如果我删除 alert('here'); 并将其放在它显示的 ajax 调用上方。
Anybody know why the success function isn't calling? nothing shows in the div with class results on any browser.
有人知道为什么成功函数没有调用吗？div 中没有任何内容显示任何浏览器上的类结果。
Thanks!
谢谢！
采纳答案by ivar ajet
Hey i had the same problem
嘿，我有同样的问题
Firstly i used $.getJSON blah blah.... but didn't worked for some reason...
首先我使用了 $.getJSON blah blah .... 但由于某种原因没有工作......
But the normal ajax call worked.. MY Page returns a json output as you.. try removing the "datatype"
但是正常的ajax调用有效..我的页面会像你一样返回一个json输出..尝试删除“数据类型”
I Used code copied from JQUERYsite which is below i pasted
我使用了从JQUERY站点复制的代码，该代码位于我粘贴的下方
$.ajax({
type: "POST",
url: "some.php",
data: { name: "John", location: "Boston" }
})
.done(function( msg ) {
alert( "Data Saved: " + msg );
});
Below is the code how i used and worked with json output i got...
下面是我如何使用和处理我得到的 json 输出的代码...
$.ajax({
type: "GET",
url: "json.php",
data: { searchword: q, projid:prid }
})
.done(function( jsonResult ) {
var str='';
for(var i=0; i<jsonResult.length;i++)
{
//do something here.. with jsonResult."yournameof "
alert(jsonResult[i].id+jsonResult[i].boqitem);
}
});
回答by E-comm
Keep same domain origins in mind here, use jsonp with a callback has always been helpful for me, that does mean adding a $_GET['callback']to your php script and wrap the json_encode with '('.json_encode($Array).')'for proper formatting.
在这里记住相同的域起源，将 jsonp 与回调一起使用对我来说一直很有帮助，这确实意味着$_GET['callback']在您的 php 脚本中添加 a并用 json_encode 包装'('.json_encode($Array).')'正确的格式。
http://api.jquery.com/jQuery.getJSON/
http://api.jquery.com/jQuery.getJSON/
The last demo on that page is the best example I have found.
该页面上的最后一个演示是我找到的最好的例子。
回答by byJeevan
Looks like cross domain request.
看起来像跨域请求。
Try it by changing :
尝试通过更改：
dataType : "json"
to
到
dataType : "jsonp"
回答by kazimt9
I know it is very late. But it can simply be handled like
我知道已经很晚了。但它可以简单地处理
var postData = {'inp': $('#inp').val()};
$.ajax({
type: "POST",
data: postData,
url: 'removedmyurlfromhere', // your url
success: function (response) {
// use exception handling for non json response
try {
response = $.parseJSON(response);
console.log(response); // or whatever you want do with that
} catch(e) {}
},
error: function( jqXHR, textStatus, errorThrown ) {},
complete: function( jqXHR, textStatus ) {}
});
回答by Stefan
In PHP, add the JSON content-type header:
在 PHP 中，添加 JSON 内容类型标头：
header('Content-Type: application/json');
also, try listening for complete as well to see if you doget any response back:
另外，尝试收听 complete 以查看您是否收到任何回复：
$.ajax({
// ....
complete: function (xhr, status) {
$('.results').append('Complete fired with status: ' + status + '<br />');
$('.results').append('XHR: ' + (xhr ? xhr.responseText : '(undefined)'));
}
// ...
});
