0

使用cordova-2.0.0 + jquery-1.10.1

尝试在我的 wordpress 博客上获得授权

提出这个要求

        $.ajax({
        url: "http://mydomain.ru/?json=auth.generate_auth_cookie",
        dataType: "jsonp",
        jsonpCallback: "myCallback",
        success: function (data) {
            alert('success');
        },
        error: function (data) {
            alert('error');
        }
    });

在我的本地主机上,此请求成功

myCallback({"status":"error","error":"You must include a 'username' var in your request."})

但是当我尝试在 Android Emulator 或 Android Device 中发出请求时出现错误

CordovaLog
SyntaxError: Parse error

Web Console
SyntaxError: Parse error at http://mydomain.ru/?json=auth.generate_auth_cookie&callback=myCallback&_=1370343811075:1

权限

<uses-permission android:name="android.permission.VIBRATE" />
<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_LOCATION_EXTRA_COMMANDS" />
<uses-permission android:name="android.permission.READ_PHONE_STATE" />
<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.RECEIVE_SMS" />
<uses-permission android:name="android.permission.RECORD_AUDIO" />
<uses-permission android:name="android.permission.MODIFY_AUDIO_SETTINGS" />
<uses-permission android:name="android.permission.READ_CONTACTS" />
<uses-permission android:name="android.permission.WRITE_CONTACTS" />
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" /> 
<uses-permission android:name="android.permission.GET_ACCOUNTS" />
<uses-permission android:name="android.permission.BROADCAST_STICKY" />

<access origin="http://mydomain.ru" subdomains="true" />
4

1 回答 1

2

试试下面的代码

 $.ajax({type : "POST",
                data : {country_key:key},//this is specify the prameters
                url : urlname,
                dataType : 'json',
                success : function(jd) {    
                        alert('success');
                },
                error : function(jd) {
                    alert('fail');
                }
            });
于 2013-06-04T13:35:47.633 回答