java - 必应 API 错误 1002

Question

首先，我对我的英语感到抱歉。 我正在用java开发一个应用程序，我想使用搜索Bing API，所以我打开了Bing的以用户为中心的开发( http://www.bing.com/dev/en-us/dev-center )并接受 key 号码，然后我编写了以下代码来获取结果Bing

String q = "http://api.bing.net/json.aspx?Appid=MyClientId=girls&sources=web&web.count=40&web.offset=41";

URL searchURL;
try {
    searchURL = new URL(q);
    HttpURLConnection httpURLConnection = (HttpURLConnection) searchURL.openConnection();

    if(httpURLConnection.getResponseCode() == HttpURLConnection.HTTP_OK){
        InputStreamReader inputStreamReader = new InputStreamReader(httpURLConnection.getInputStream());
        BufferedReader bufferedReader = new BufferedReader(inputStreamReader, 8192);

        String line = null;
        String result = "";
        while((line = bufferedReader.readLine()) != null){
            result += line;
        }

        bufferedReader.close();
    }
} catch (MalformedURLException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

为什么我会收到以下错误 1002？

{"SearchResponse":{
    "Version":"2.2",
    "Query":{"SearchTerms":"girls"},
    "Errors":[
        {"Code":1002,
         "Message":"Parameter has invalid value.",
         "Parameter":"SearchRequest.AppId",
         "Value":"MyClientId",
         "HelpUrl":"http:\/\/msdn.microsoft.com\/en-us\/library\/dd251042.aspx"}]
}}

Answer 1

您的地址似乎有误 这看起来很可疑:

Appid=MyClientId=girls

您应该查看文档 http://msdn.microsoft.com/en-us/library/dd250882.aspx ，但我猜您需要将 MyClientId 替换为某些内容，而且您还没有分离查询和 clientId，即 &q=girls

编辑:您需要在某处获取 AppId Steps of creating appid for bing search

这里有一些问题可以帮助您: Bing search API and Azure