解析 JSON Issue 为 Android

Question

你好，Frnds 这是我在 iPhone 上使用 Xcode 并正确获得响应的代码，但问题是，当我使用相同的 url 和相同的 json 进行 android 开发时，我得到：-

{"code":"400","message":"Failed loading JSON. Special characters must not be included in the request. Please check the requested JSON."}

我认为这是我没有发现的任何小技术问题。 我是 android 开发的新手，在过去的 15 天里我尝试了 1000 次并且感到沮丧。 :-(:-(:-(:-(:-(:-(:-(:-(:-(:-( 谁能给我提供完整的解决方案和 android 的完整代码。我将非常感谢你。

iPhone 的代码是：

NSURL *url = [NSURL URLWithString:@"http://www.invoicera.com/app/api/check_json_api.php?token=7B92C122473A3D6F54E60D20AC5526D0"];


NSString *jsonRequest = [NSString stringWithFormat:@"&json_data=%@",[[NSString stringWithFormat:@"{\"listInvoice\":{\"client_id\":\"\",\"date_from\":\"\",\"date_to\":\"\",\"invoice_number\":\"\",\"invoice_record_status\":\"\",\"invoice_status\":\"\",\"page\":\"1\",\"per_page_record\":\"20\"}}"] stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

NSLog(@"%@",jsonRequest);

NSData *json_data = [NSData dataWithBytes:[jsonRequest UTF8String] length:[jsonRequest length]];

NSLog(@"%@",json_data);

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];


[request setHTTPMethod:@"POST"];
[request setHTTPBody: json_data];
NSLog(@"%@",json_data);
// [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
//[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:[NSString stringWithFormat:@"%d", [json_data length]] forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody:[[jsonRequest stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding]
                      dataUsingEncoding:NSUTF8StringEncoding 
                      allowLossyConversion:YES]];

// [NSURLConnection connectionWithRequest:[request autorelease] delegate:self];

NSURLConnection *nsUrlConnection=[[NSURLConnection alloc]
                                  initWithRequest:request 
                                  delegate:self];

// Successful connection.
if (nsUrlConnection) {

   // [self initSpinner];
   // [self spinBegin];

    NSMutableData *data = [[NSMutableData alloc] init];
    self.receivedData=data;
    [data release];
} 
// Unsuccessful connection.
else {

}  
// Clean up
[url release];
[request release];

编写此代码后，我可以从服务器获得完美的响应。 但是对于 android 开发，我怎样才能从服务器获得相同的响应。 请为我提供完整的解决方案，完整代码为 android。

{"@attributes":{"status":"200"},"invoices":{"@attributes":{"page":"1","per_page_record":"20","total_pages":"13","total_records":"246"},"invoice":[{"client":{"client_id":"421","organization":"max styles","address":"H-189,Vasundhara","billing_address":{"street":"H-189,Vasundhara","city":{},"state":{},"zip":........................etc etc etc.

这是 android 代码：

public class JSONParser {

    String s1 = "{\"listInvoice\":{\"client_id\":\"\",\"date_from\":\"\",\"date_to\":\"\",\"invoice_number\":\"\",\"invoice_record_status\":\"\",\"invoice_status\":\"\",\"page\":\"1\",\"per_page_record\":\"20\"}}";

    public String payload;

    public JSONParser(String[]array) {
        this.payload = null;
    }

    public void excuteHttpPost() throws Exception {
        BufferedReader in = null;

        try {
            // Creating HTTP client
            JSONObject listobj = new JSONObject ();
            JSONObject listInvoice = new JSONObject ();
            listInvoice.put("client_id","");
            listInvoice.put("date_from","");
            listInvoice.put("date_to","");
            listInvoice.put("invoice_number","");
            listInvoice.put("invoice_record_status","");
            listInvoice.put("invoice_status","");
            listInvoice.put("page","1");
            listInvoice.put("per_page_record","10");
            listobj.put("listInvoice", listInvoice);

            HttpClient client = new DefaultHttpClient();
            HttpPost request = new HttpPost(ConstantList.invoicelistURL);

            List<NameValuePair> postParameters = new ArrayList<NameValuePair>(1);
            postParameters.add(new BasicNameValuePair("json_data", s1));

            Log.d("Http Response:", postParameters.toString());

            // Build JSON string

            request.setHeader("Content-type", "application/x-www-form-urlencoded");
            request.setEntity(new UrlEncodedFormEntity(postParameters));
            HttpResponse response = client.execute(request);

            // writing response to log
            Log.d("Http Response:", response.toString());
            in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));

            StringBuffer sb = new StringBuffer("");
            String line = "";
            String NL = System.getProperty("line.separator");
            while ((line = in.readLine()) != null) {
                sb.append(line+NL);
            }
            in.close();
            payload = sb.toString();

        } finally {
            if (in != null) {
                try {
                    in.close();
                } catch (IOException e) {
                   e.printStackTrace();
                }
            }
        }
    }
}

Answer 1

问题是您根本不在请求中使用JSONObject listobj ，而是将s1添加到 post 参数。 使用此代码：

postParameters.add(new BasicNameValuePair("json_data", listobj.toString()));

此外，将 integer 值明确地放入 JSONObject： listInvoice.put("page", 1); 而不是listInvoice.put("page", "1");