Xcode报错问题如下: 解决办法如下:0x1 ->请求数据时加上缺少的类型 AFHTTPSessionManager *manager = [selfAFHTTPSessionManager];
// manager.responseSerializer.acceptableContentTypes = [NSSet setWithObject:@”text/html”];
[manager GET:URLString parameters:parameters progress:nilsuccess:^(NSURLSessionDataTask * _Nonnull task, id _Nullable responseObject) {
if (success) {//并且code = 正确
success(responseObject);
}
} failure:^(NSURLSessionDataTask * _Nullable task, NSError * _Nonnull error) {
if (failure) {
failure(error);
} }];
0x2->服务器后台指定数据返回类型<?php header(“Content-Type:application/json;charset = utf-8”); // 请求–get // 参数:userName passWord $userName = $_GET[“userName”]; $passWord = $_GET[“passWord”]; $respond = array(“code”=>200,”userName”=>$userName,”title”=>”success”); // 返回结果 echo(json_encode($respond)); // http://127.0.0.1/Login/Login2.php?userName=fa&passWord=123?>