Android MySQL + PHP + JSON

Question

我一直在使用MySQL-> PHP和JSON开发这个Android应用，用于学校作业。 我对PHP没有经验，并且在此php文件中找不到错误。

<?php

require("db_config.php");

//initial query
$query = "SELECT * FROM GAME G, CONSOLE C WHERE G.consoleid = C.consoleid";

//execute query
try 
{
    $stmt   = $db->prepare($query);
    $result = $stmt->execute($query);
}
catch (PDOException $ex) 
{
    $response["success"] = 0;
    $response["message"] = "Database Error!";
    die(json_encode($response));
}

$rows = $stmt->fetchAll();


if ($rows) 
{
    $response["success"] = 1;
    $response["message"] = "Games Available!";
    $response["games"]   = array();

foreach ($rows as $row) 
{
    $gameslist          = array();
    $games["gameid"]        = $row["gameid"];
    $games["gamename"]          = $row["gamename"];
    $games["gamevalue"]         = $row["gamevalue"];
    $games["gamerarity"]        = $row["gamerarity"];
    $games["gamedescription"]   = $row["gamedescription"];
    $games["consolename"]       = $row["consolename"];
    $games["gameimgstring"]     = $row["gameimgstring"];

    //update our repsonse JSON data
    array_push($response["games"], $gameslist);
}

    // echoing JSON response
    echo json_encode($response);   
} 
else 
{
    $response["success"] = 0;
    $response["message"] = "No games Available!";
    die(json_encode($response));
}

?>

PHP脚本应返回一个填充有游戏数据的数组。 查询不需要任何数据。

我的IDE在此部分给出了错误

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

由于失败，因此不会返回值。

Answer 1

可能是

$stmt->execute();

在这里检查： http : //php.net/manual/en/pdostatement.execute.php