Question

如何使用get方法在具有6个值和json格式的api链接上发布数据？

我真的很费力。我这样做好几天了。是否最好有6个用于写入值的文本字段和一个用于指示发布数据的按钮？

查看我的代码并请求帮助。

-(IBAction)postDataPressed
{

    NSString *urlString = @"http://192.168.18.8/apisample2/friendb.php?fm=jsn";
    NSURL *url = [NSURL URLWithString:urlString];
    ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:url];

    [request setPostValue:user.text forKey:@"un"];
    [request setPostValue:pass.text forKey:@"pd"];
    [request setPostValue:gender.text forKey:@"gd"];
    [request setPostValue:age.text forKey:@"ag"];
    [request setPostValue:status.text forKey:@"st"];
    [request setPostValue:lookfor.text forKey:@"lf"];

    [request setRequestMethod:@"GET"];
    [request setCompletionBlock:^{
    NSString *responseString = [request responseString];
    NSLog(@"Response: %@", responseString);
    }];
    [request setFailedBlock:^{
        NSError *error = [request error];
        NSLog(@"Error: %@", error.localizedDescription);
    }];
    [request setDidFinishSelector:@selector(requestFinished:)];
    [request setDidFailSelector:@selector(requestFailed:)]; 
    [request startAsynchronous];


}

我希望有人会帮忙。

Answer 1

NSString *request_url = [NSString stringWithFormat: @"http://192.168.18.8/apisample2/friendb.php?fm=jsn&un=%@&pd=%@&gd=%@&ag=%@&st=%@&lf=%@",
    user.text, pass.text, gender.text, age.text,status.text,lookfor.text];
NSURL *url = [NSURL URLWithString:request_url];
ASIHttpRequest *request = [ASIHttpRequest requestWithURL:url];
request.delegate = self;
....

HTTP Get方法只接受url中的数据，称为查询字符串，因此您必须自己构建查询字符串。

发布数据，但GET方法json -ios

1 个答案: