越来越拉JSON的错误从URL

-1

我得到错误说：SyntaxError: Unexpected token ':'. Parse error."越来越拉JSON的错误从URL

我是否做

url = 'http://api.openweathermap.org/v3/uvi/20,77/current.json?appid=c0d8761ca979157a45651a5c7f12a6be'; 
 
function getJSONP(url, success) { 
 

 
    var ud = '_' + +new Date, 
 
     script = document.createElement('script'), 
 
     head = document.getElementsByTagName('head')[0] 
 
       || document.documentElement; 
 

 
    window[ud] = function(data) { 
 
     head.removeChild(script); 
 
     success && success(data); 
 
    }; 
 

 
    script.src = url.replace('callback=?', 'callback=' + ud); 
 
    head.appendChild(script); 
 

 
} 
 

 
getJSONP(url, function(data){ 
 
    console.log(data); 
 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
 
<html><body>

，或者如果我这样做：

url = 'http://api.openweathermap.org/v3/uvi/20,77/current.json?appid=c0d8761ca979157a45651a5c7f12a6be'; 
 

 
function CallURL() { 
 
    $.ajax({ 
 
    url: url, 
 
    type: "GET", 
 
    dataType: "jsonp", 
 
    async: false, 
 
    success: function(msg) { 
 
     console.log(msg); 
 
     JsonpCallback(msg); 
 
    }, 
 
    error: function() { 
 
     // ErrorFunction(); 
 
     // break ; 
 
    } 
 
    }); 
 
} 
 

 
function JsonpCallback(json) { 
 
    alert(json); 
 
    //document.getElementById('summary').innerHTML = json.result; 
 
} 
 

 
CallURL();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
 
<html> 
 

 
<body> 
 

 
</body> 
 

 
</html>

来源

2017-02-26 Ciasto piekarz

此网址没有返回JSONP，它的回报JSON：只为几个电话

{"time":"2017-02-25T12:00:00Z","location":{"latitude":18.75,"longitude":76.75},"data":10.8}

OpeanWeather API支持JSONP，像http://api.openweathermap.org/data/2.5/weather?q=London,uk&callback=test&appid=c0d8761ca979157a45651a5c7f12a6be，其他电话只支持JSON，XML或HTML（见https://openweathermap.org/current详细信息）。

所以使用JSON 在 AJAX调用类型：

url = 'http://api.openweathermap.org/v3/uvi/20,77/current.json?appid=c0d8761ca979157a45651a5c7f12a6be'; 
 

 
function CallURL() { 
 
    $.ajax({ 
 
    url: url, 
 
    type: "GET", 
 
    dataType: "json", 
 
    async: false, 
 
    success: function(msg) { 
 
     console.log(msg); 
 
    }, 
 
    error: function() { 
 
     // ErrorFunction(); 
 
     // break ; 
 
    } 
 
    }); 
 
} 
 

 
CallURL();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
 
<html> 
 

 
<body> 
 

 
</body> 
 

 
</html>

来源

2017-02-26 10:48:19 kb0

越来越拉JSON的错误从URL

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