objective-c HTTP POST 以表单形式发送请求

Posted 2023-03-11

【中文标题】objective-c HTTP POST 以表单形式发送请求

【英文标题】：objective-c HTTP POST send request as form

【发布时间】：2017-08-24 09:12:25

【问题描述】：

我已经使用 POST 方法在我的应用程序中调用带有标头值和参数的 API。 服务器只接受格式为

的表单

"form": 
        "action" : "login",
        "user" : "311"
,

当我们使用代码时

NSString *urlString = [NSString stringWithFormat:@"%@", url_string];

NSURL *url = [NSURL URLWithString:urlString];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
NSError *error;

NSDictionary *parameters = @@"action": @"login", @"user": @"311";
NSString *params = [self makeParamtersString:parameters withEncoding:NSUTF8StringEncoding];
NSData *jsonData2 = [params dataUsingEncoding:NSUTF8StringEncoding];

[request setHTTPMethod:@"POST"];
        [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody: jsonData2];
        [NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:ourBlock];

我的表单是这样的

form = 
        action = login;
        user = 311;
;

你能产生你想要的结果吗？ 你能帮我解决这个问题吗？

【参考方案1】：

试试

NSString *urlString = [NSString stringWithFormat:@"%@", url_string];

NSURL *url = [NSURL URLWithString:urlString];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
NSError *error;

NSDictionary *parameters = @@"action": @"login", @"user": @"311";
NSData *jsonData = [NSJSONSerialization dataWithJSONObject: parameters options:0 error:&error];

[request setHTTPMethod:@"POST"];
        [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody: jsonData];
        [NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:ourBlock];

【参考方案2】：

这样改参数怎么样。

NSDictionary *parameters = @@"form":@@"action": @"login", @"user": @"311";
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:parameters options:NSJSONWritingPrettyPrinted error:nil];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request addValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody: jsonData];
[NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:ourBlock];

【讨论】：

然后我有错误异常：-[__NSDictionaryI stringValue]: unrecognized selector sent to instance

您的服务器似乎需要json格式的正文。如果是这样，请尝试更新的答案。

【参考方案3】：

如果你需要 base64 编码试试这个

NSMutableDictionary *param = [@@"form":@@"action": @"login", @"user": @"311" mutableCopy];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:serviceURL];
NSString *strEncoded = [self encodeParameters:param];
NSData *requestData = [strEncoded dataUsingEncoding:NSUTF8StringEncoding];
[request setHTTPBody:requestData];
[request setValue:[NSString stringWithFormat:@"%lu",(unsigned long)requestData.length] forHTTPHeaderField:@"Content-Length"];

[request setHTTPMethod:@"POST"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];


// Function encodeParameters 

+(NSString *)encodeParameters:(NSDictionary *)dictEncode

// Encode character set as per BASE64
NSCharacterSet *URLBase64CharacterSet = [[NSCharacterSet characterSetWithCharactersInString:@"/+=\n"] invertedSet];

NSMutableString *stringEncode = [[NSMutableString alloc] init];
NSArray *allKeys = [dictEncode allKeys];
for (int i = 0;i < allKeys.count; i++) 
    NSString *key = [allKeys objectAtIndex:i];
    if([dictEncode valueForKey:key])
    
        [stringEncode appendFormat:@"%@=%@",key,[[dictEncode valueForKey:key] stringByAddingPercentEncodingWithAllowedCharacters:URLBase64CharacterSet]];
    

    if([allKeys count] > i+1)
    
        [stringEncode appendString:@"&"];
    


return stringEncode;

【参考方案4】：

试试这个

NSURL * url = [NSURL URLWithString:@"%@",url_string];

NSURLSessionConfiguration * config = [NSURLSessionConfiguration defaultSessionConfiguration];

NSURLSession * session = [NSURLSession sessionWithConfiguration:config];

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

request.HTTPMethod = @"POST";

NSDictionary * paramters = [NSDictionary dictionaryWithObjectsAndKeys:@"login",@"action",@"311",@"user", nil]; // [NSDictionary dictionaryWithObjectsAndKeys:@"value",@"key", nil];


NSDictionary *params = @@"form": paramters;
NSError *err = nil;
NSData *jsonData2 = [NSJSONSerialization dataWithJSONObject:params options:0 error:&err];

【讨论】：

然后我的请求：form = "\"action\":\"login\",\"user\":\"311\"" = ""; ;

@Albert ，我已经更新了答案。请检查并告诉我。

现在请求是：form = "\"form\":\"action\":\"login\",\"user\":\"311\"" = ""; ;

你试过这个吗？你能告诉我你的更新代码吗？

【参考方案5】：

试试这个，

NSString *parameters = @"\"form\":\"action\" : \"login\", \"user\" : \"311\"";
NSData *jsonData2 = [parameters dataUsingEncoding:NSUTF8StringEncoding];

[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody: jsonData2];
[NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:ourBlock];

【参考方案6】：

 NSError *error;
 NSDictionary *parameters = @@"form": @@"action": @"login", @"user": @"311";
 NSData *jsonData = [NSJSONSerialization dataWithJSONObject:parameters 
      options:NSJSONWritingPrettyPrinted error:&error];
 request.HTTPBody = jsonData

 //Using NSURLSession is better option than using NSURLConnection
 NSURLSession *session = [NSURLSession sharedSession];
 NSURLSessionDataTask *dataTask = [session dataTaskWithRequest:request completionHandler:^(NSData * _Nullable data, NSURLResponse * _Nullable response, NSError * _Nullable error) 
  NSHTTPURLResponse* respHttp = (NSHTTPURLResponse*) response;

  if (!error && respHttp.statusCode == 200) 

    NSDictionary* respondData = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:&error];
    NSLog(@"%@", respondData);

   else
    NSLog(@"%@", error);
  
];

[dataTask resume];

【讨论】：

我的结果表单 = "\n \"action\" : \"login\",\n \"user\" : \"311\"\n" = ""; ;

【参考方案7】：

试试AFNetwoking

   NSString *urlString = [NSString stringWithFormat:@"URL"];
   NSDictionary *para=  @@"action": @"login", @"user": @"311";
   AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
   manager.responseSerializer = [AFJSONResponseSerializer serializer];
   manager.responseSerializer.acceptableContentTypes = [NSSet setWithObject:@"text/html"];

   [manager POST:urlString parameters:para success:^(AFHTTPRequestOperation *operation, id responseObject) 
                       NSLog(@"JSON: %@", responseObject);

     failure:^(AFHTTPRequestOperation *operation, NSError *error) 
        NSLog(@"Error: %@", error);

    ];