Question

这是我的PHP代码，当我想做一个自定义方法if（字符串等于成功）但它不起作用。如果此方法返回一个字符串值的响应。如果响应成功，我想做一些方法

<?php
    define('HOST','mysql.hostinger.in');
    define('USER','example');
    define('PASS','example');
    define('DB','example');

    $con = mysqli_connect(HOST,USER,PASS,DB);

    $username = $_POST['PHONENO'];
    $password = $_POST['PASSWORD'];

    $sql = "SELECT * FROM Prago WHERE PHONENO='$username' and PASSWORD='$password'";

    $res = mysqli_query($con,$sql);

    $check = mysqli_fetch_array($res);

    if(isset($check)){
    echo 'success';
    }else{
    echo 'failure';
    }

    mysqli_close($con);
?>

我的凌空字符串请求类

   StringRequest stringRequest=new StringRequest(Request.Method.POST, REGISTER_URL, new Response.Listener<String>() {

        @Override
        public void onResponse(String response) {
            if(response.equals("success")){
                Toast.makeText(getApplicationContext(),"DONE",Toast.LENGTH_SHORT).show();
            }
            else
            {
                Toast.makeText(getApplicationContext(),"check username and password",Toast.LENGTH_SHORT).show();
            }
            callback.onSuccess(response);

        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(getApplication(),"error in volley",Toast.LENGTH_SHORT).show();
        }
    }){
        @Override
        protected Map<String, String> getParams() throws AuthFailureError {
            Map<String,String> map=new HashMap<>();
            map.put("PHONENO",Number);
            map.put("PASSWORD",Pass);
            return map;
        }
    };

    VolleySingleton.getInstance().addToRequestQueue(stringRequest);
}

Answer 1

正如@greenapps在评论中所指出的那样，将.equals()更改为.contains()是有效的，因为在我的情况下，回复是双引号，所以＆＃34;成功＆＃34;！=成功但是＆＃34;成功＆＃34;包含成功

Answer 2

首先检查您通过调试收到的回复，或者只是简单地祝贺您的回复

无法在volley响应方法中使用if语句

2 个答案: