[英]obtain json data from jsonp
Currently, I have a url which renders data in json format. 目前,我有一个以json格式呈现数据的url。
url: 网址:
http://10.0.1.11/render?target=threshold(400,test)&from=-1mins&format=json&jsonp=?
when run in a browser gives me 在浏览器中运行时给我
?([{"target": "test", "datapoints": [[400, 1388755180], [400, 1388755190], [400, 1388755200], [400, 1388755210], [400, 1388755220], [400, 1388755230], [400, 1388755240]]}])
I would need the json result in a variable for further processing. 我需要将json结果放入变量中以进行进一步处理。 I tried the following
我尝试了以下
foo
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js"></script>
<script>
$.getJSON("http://10.0.1.11/render?target=threshold(400,test)&from=-1mins&format=json&jsonp=?", function(result){
//response data are now in the result variable
alert(result);
});
</script>
I ideally would need: 理想情况下,我需要:
var test = [{"target": "test", "datapoints": [[400, 1388755180], [400, 1388755190], [400, 1388755200], [400, 1388755210], [400, 1388755220], [400, 1388755230], [400, 1388755240]]}];
Where am i going wrong? 我要去哪里错了?
You need to interpret the request as jsonp rather than json. 您需要将请求解释为jsonp而不是json。 jsonp is like json, but it's json wrapped in a method-call.
jsonp类似于json,但它是通过方法调用包装的json。 (see: What is JSONP all about? )
(请参阅: JSONP的全部含义是什么? )
You can use something like: 您可以使用类似:
<script>
function myCallback(json_data){
//do something with json_data!
}
</script>
<script type="text/javascript" src="http://10.0.1.11/render?target=threshold(400,test)&from=-1mins&format=json&jsonp=myCallback"></script>
or 要么
<script>
$(document).ready(function(){
$.ajax({
url: 'http://10.0.1.11/render?target=threshold(400,test)&from=-1mins&format=json',
dataType: 'jsonp',
success: function(json_data){
//do something with json_data!
}
}
});
})
</script>
(examples adapted from the linked post) (示例摘自链接文章)
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