8

我想使用以下代码将数据传递给 POST JSON Web 服务。

NSString *urlString = @"http://3.10.204.99:9090/service/outage/data/getShortFormData";
NSDictionary *inputData = [[NSDictionary alloc] initWithObjectsAndKeys:
                           @"1020", @"outageId",
                           @"Alex", @"customerName",
                           nil];

NSError *error = nil;
NSData *jsonInputData = [NSJSONSerialization dataWithJSONObject:inputData options:NSJSONWritingPrettyPrinted error:&error];
NSString *jsonInputString = [[NSString alloc] initWithData:jsonInputData encoding:NSUTF8StringEncoding];
[self checkWithServer:urlString jsonString:jsonInputString];    


-(void)checkWithServer:(NSString *)urlname jsonString:(NSString *)jsonString{

    NSURL *url1 = [NSURL URLWithString:urlname];
    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url1];
    [request setHTTPMethod:@"POST"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
    [request setValue:@"charset=utf-8" forHTTPHeaderField:@"Content-Type"];
    [request setHTTPBody:[jsonString dataUsingEncoding:NSUTF8StringEncoding]];
    NSURLResponse *response;
    NSError *err;
    NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&err];
    id jsonResponseData = [NSJSONSerialization JSONObjectWithData:responseData options:kNilOptions error:nil];

    NSDictionary *jsonResponseDict;
    if ([jsonResponseData isKindOfClass:[NSDictionary class]]) {
        jsonResponseDict = jsonResponseData;
    } else {
        // Error-handling code
    }
    jsonResponseData = [jsonResponseDict objectForKey:@"d"];
    if (jsonResponseData == nil) {
        // Server may have returned a response containing an error
        // The "ExceptionType" value is returned from my .NET server used in sample
        id jsonExceptioTypeData = [jsonResponseDict objectForKey:@"ExceptionType"];
        if (jsonExceptioTypeData != nil) {
            NSLog(@"%s ERROR : Server returned an exception", __func__);
            NSLog(@"%s ERROR : Server error details = %@", __func__, jsonResponseDict);
        }
    }
}

执行此代码时，它没有给出错误，但数据没有得到更新。

对此的任何帮助都会感到遗憾。

谢谢你们。

4

2 回答 2

8

试试下面的代码

+(NSData *)postDataToUrl:(NSString*)urlString:(NSString*)jsonString
{
    NSData* responseData = nil;
    NSURL *url=[NSURL URLWithString:[urlString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
    responseData = [NSMutableData data] ;
    NSMutableURLRequest *request=[NSMutableURLRequest requestWithURL:url];
    NSString *bodydata=[NSString stringWithFormat:@"data=%@",jsonString];

    [request setHTTPMethod:@"POST"];
    NSData *req=[NSData dataWithBytes:[bodydata UTF8String] length:[bodydata length]];
    [request setHTTPBody:req];
    NSURLResponse* response;
    NSError* error = nil;
    responseData = [NSURLConnection sendSynchronousRequest:request     returningResponse:&response error:&error];
    NSString *responseString = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];

    NSLog(@"the final output is:%@",responseString);

    return responseData;
}
于 2013-11-06T08:59:10.620 回答
1

尝试这个：

[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
于 2013-11-06T08:55:58.103 回答