mysql - Json mysql 连接和数组转换错误。

Question

我正在尝试使用 Android 应用程序的 JSON 从复选框列表中进行过滤。出于测试目的，我使用了 html 复选框。我收到的错误为

注意:数组到字符串的转换警告:mysql_fetch_array() 期望参数 1 为资源、 bool 值

我的代码是 需要(“config.inc.php”)； if (!empty($_POST)) {
$条件=“”； $值=“”； $response = array();
if(!empty($_POST['品牌'])){ $value = $_POST['品牌']; $value = 内爆("','", $value); $condition .= "PhoneMake IN "."('$valu')"; } if(!empty($_POST['ram'])){ $value = $_POST['ram']; $value = 内爆("','", $value); $condition .= "AND Os_ram IN "."('$value')";

}
 $query ="SELECT * FROM product INNTER JOIN specs WHERE "." $condition GROUP BY PhoneModel";

      $query_params = array(
        ':PhoneMake' => $_POST['brand'],
        ':Os_ram' => $_POST['ram']

    );

    try {
        $stmt   = $db->prepare($query);
        $result = $stmt->execute($query_params);
    }
    catch (PDOException $ex) {

        $response["success"] = 0;
        $response["message"] = "Database Error2. Please Try Again!";
        die(json_encode($response));
    }


 $a  = mysql_query($query);
    while($b = mysql_fetch_array($a))
    {

    $response["$b[PhoneMake]"][]["$b[PhoneModel]"] = array('OS' => $b['Os_name'], 'Display Size (Inch)' => $b['D_apps'], 'Ram' => $b['Os_ram']);

}
    echo json_encode($response);
    }
?>
    <h1>Filter</h1> 
    <form action="" method="post"> 
       <div class="panel-heading filterhead">Filter by brand</div>
            <label><input type="checkbox" name="brand[]" value="Motorola"> Motorola</label><br>
            <label><input type="checkbox" name="brand[]" value="Xolo"> Xolo</label><br>
          </div>

<div class="panel-heading filterhead">RAM</div>
            <label><input type="checkbox" name="ram[]" value="512"> 512 MB</label><br>
            <label><input type="checkbox" name="ram[]" value="1"> 1 GB</label><br>
          </div>

    </div>
        <input type="submit" value="Filter" /> 
    </form>

Config.inc.php

<?php 
    $host = "localhost"; 
    $dbname = "street";     
    $username = "root"; 
    $password = ""; 

    $options = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8'); 

    try 
    { 

        $db = new PDO("mysql:host={$host};dbname={$dbname};charset=utf8", $username, $password, $options); 
    } 
    catch(PDOException $ex) 
    { 
        die("Failed to connect to the database: " . $ex->getMessage()); 
    } 

    $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 


    $db->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC); 


    if(function_exists('get_magic_quotes_gpc') && get_magic_quotes_gpc()) 
    { 
        function undo_magic_quotes_gpc(&$array) 
        { 
            foreach($array as &$value) 
            { 
                if(is_array($value)) 
                { 
                    undo_magic_quotes_gpc($value); 
                } 
                else 
                { 
                    $value = stripslashes($value); 
                } 
            } 
        } 

        undo_magic_quotes_gpc($_POST); 
        undo_magic_quotes_gpc($_GET); 
        undo_magic_quotes_gpc($_COOKIE); 
    } 


    header('Content-Type: text/html; charset=utf-8'); 

    session_start();     

?>

我已经正确编码了所有内容，但仍然出现错误。主要问题是 mysql_fetch_array() 期望参数 1 是资源、 bool 值。帮助我走出这个错误。我想按原样使用 config.inc.php 文件。

Answer 1

您的查询中似乎存在拼写错误。 INNER JOIN 应为 INNER JOIN。