php - 在 PHP 中从远程数据库检索数据时出错

Question

如何解决表中所示的问题？当我尝试从数据库检索数据时发生这种情况。在我的java代码中，我尝试调用的json数组为null。

db_connect.php

<?php 
class DB_CONNECT{
// constructor
function __construct() {
    // connecting to database
    $this->connect();
}

// destructor
function __destruct() {
    // closing db connection
    $this->close();
}

/**
 * Function to connect with database
 */
function connect() {
    // import database connection variables
    require_once __DIR__ . '/db_config.php';

    // Connecting to mysql database
    $con = mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die(mysql_error());

    // Selecing database
    $db = mysql_select_db(DB_DATABASE) or die(mysql_error()) or die(mysql_error());

    // returing connection cursor
    return $con;
}

/**
 * Function to close db connection
 */
function close() {
    // closing db connection
    mysql_close();
}


}

?>'

retEqp.php

<?php


/*
 * Following code will list all the products
 */

// array for JSON response
$response = array();

// include db connect class
require_once __DIR__ . '/db_connect.php';

// connecting to db
$db = new DB_CONNECT();

// get all products from products table
$result = mysql_query("SELECT *FROM facilities_equipments where item_Type='Equipment'") or die(mysql_error());

// check for empty result
if (mysql_num_rows($result) > 0) {
    // looping through all results
    // products node
    $response["equipments"] = array();

    while ($row = mysql_fetch_array($result)) {
        // temp user array
        $equipment = array();
        $equipment["item_ID"] = $row["item_ID"];
        $equipment["item_Name"] = $row["item_Name"];



        // push single product into final response array
        array_push($response["equipments"], $equipment);
    }
    // success
    $response["success"] = 1;

    // echoing JSON response
        echo json_encode($response);
    } else {
        // no products found
        $response["success"] = 0;
        $response["message"] = "No products found";


    // echo no users JSON
    echo json_encode($response);
    }
    ?>

Answer 1

替换您的代码

$con =  mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die(mysql_error());

与

$con =  mysqli_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die(mysql_error());

关闭所有已弃用的警告，包括来自 mysql_* 的警告:

error_reporting(E_ALL ^ E_DEPRECATED);