ASIFormDataRequest POST返回网站源代码？

NSURL *url = [NSURL URLWithString:@"http://example.com"];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:url];

// 添加需要提交的表单数据
[request setPostValue:@"value1" forKey:@"key1"];
[request setPostValue:@"value2" forKey:@"key2"];

// 发送请求
[request startSynchronous];

// 获取响应数据
NSError *error = [request error];
if (!error) {
    NSString *responseString = [request responseString];
    NSLog(@"网站的源代码：%@", responseString);
} else {
    NSLog(@"请求失败：%@", error.localizedDescription);
}