Question

我正在使用此代码将数据发布到我的服务器

NSURL *mainurl = [NSURL URLWithString:@"http://xxxxxxxxxx/api/PhonePaymentApi/Transaction/"];

    NSString * postdata = [[NSString alloc]initWithFormat:@"UniqueId=%@&jsonProduct=%@&BranchId=%@&OrderToTime=%@",GETUnicidentifire,JsonOrderDetail,BranchId,OrderToTime];

    ASIFormDataRequest *requestt = [ASIFormDataRequest requestWithURL:mainurl];

    [requestt setRequestMethod:@"POST"];
    [requestt addRequestHeader:@"application/x-www-form-urlencoded" value:@"Content-Type"];
    [requestt appendPostData:[postdata dataUsingEncoding:NSUTF8StringEncoding]];

    NSString * theurl = [NSString stringWithFormat:@"%@",mainurl];
    NSURLRequest *thereqest = [[NSURLRequest alloc] initWithURL:[NSURL URLWithString:theurl]];
    [NSURLConnection connectionWithRequest:thereqest delegate:self];

方法

- (void)connection:(NSURLConnection *)connection didFailWithError:(NSError *)error {

    NSLog(@"%@",error); 
}

imting： {消息：请求的资源不支持http方法'GET'。}

我做错了什么？

Answer 1

你在这里混合了什么。您构建了ASIFormDataRequest，但实际上并未发送它。你发送的是NSURLConnection。

自从我使用ASI以来已经很长时间了，但这可能有所帮助：

NSURL *mainurl = [NSURL URLWithString:@"http://xxxxxxxxxx/api/PhonePaymentApi/Transaction/"];

ASIFormDataRequest *requestt = [ASIFormDataRequest requestWithURL:mainurl];

[requestt addPostValue:GETUnicidentifire forKey:@"UniqueId";
[requestt addPostValue:JsonOrderDetail   forKey:@"jsonProduct";
[requestt addPostValue:BranchId          forKey:@"BranchId";
[requestt addPostValue:OrderToTime       forKey:@"OrderToTime";

[requestt setCompletionBlock:^{
    // Process the response
}];
[requestt setFailedBlock:^{
    NSError *error = [requestt error];
    NSLog(@"%@",error);
}];

[requestt startAsynchronous];

作为建议，请用AFNetworking之类的东西替换ASI。 ASI已不再开发。

ASIFormDataRequest POST请求问题

1 个答案: