Question

所以我有一个代码块，在数据库中查找用户名，所以我不会有重复项。问题是即使有一个名为“swane”的用户名，它也说没有这样的东西。这是java调用：

params.put("name", "swane");
JSONObject json = jsonParser.makeHttpRequest(urlNameLookup, "POST", params);

这是解析器：

package com.example.denis.onthego;
import android.content.ContentValues;
import android.util.Log;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.IOException;
import okhttp3.MediaType;
import okhttp3.OkHttpClient;
import okhttp3.Request;
import okhttp3.RequestBody;
import okhttp3.Response;

public class JSONParser {

static JSONObject jObj;
static String json;

// constructor
public JSONParser() {
}

// function get json from url
// by making HTTP POST or GET mehtod
public static JSONObject makeHttpRequest(String url, String method, ContentValues params) {
    // Making HTTP request
    try {
        final OkHttpClient client = new OkHttpClient();
        Request request;
        // check for request method
        if (method.equals("POST")) {
            // request method is POST

            MediaType contentType = MediaType.parse("application/x-www-form-urlencoded; charset=UTF-8");
            String content = "";
            for (String key : params.keySet())
            {
                if ( !content.isEmpty())
                    content += "&";

                content += key + "=" + params.get(key);
            }

            RequestBody body = RequestBody.create(contentType, content);
            request = new Request.Builder().url(url).post(body).build();
        }
        else  {
            // request method is GET
            request = new Request.Builder().url(url).build();
        }
        final Response response = client.newCall(request).execute();
        json = response.body().string();

    } catch (IOException e) {
        e.printStackTrace();
    }
    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e ){
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }
    // return JSON String
    return jObj;
}
}

这是php文件：

<?php
$response = array();
require_once __DIR__ . '/db_connect2.php';
$db = new DB_CONNECT();
if (isset($_POST['name'])) {
$name = $_POST['name'];
$result = $db->query("SELECT * FROM users WHERE Name = $name");
if (!empty($result)) {
if (mysqli_num_rows($result) > 0) {
$result = mysqli_fetch_array($result);
$user = array();
$user["name"] = $result["name"];
$response["success"] = 1;
$response["users"] = array();
array_push($response["users"], $user);
echo json_encode($response);
} else {
$response["success"] = 0;
$response["message"] = "No user found";
echo json_encode($response);
}
} else {
$response["success"] = 0;
$response["message"] = "No user found";
echo json_encode($response);
}
} else {
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
echo json_encode($response);
}
?>

我收到错误：Create Response: {"success":0,"message":"No user found"} 我该如何解决？

使用PHP和Android Studio在数据库中找不到行

0 个答案: