使用PDO運行mySQL查詢時出現問題

Question

下面的代碼不會將任何值返回到我的$result變量中。數據庫連接是好的。但它似乎沒有返回$result變量的任何值。

我確實有player表中的數據，並且在所述表中存在1和2的值player_id。

運行時打印回聲是這樣的。當將整數1和2傳遞給函數時。

數據庫連接

SELECT * 
FROM player 
WHERE player_id = 1 

結果不是陣列。

數據庫連接

SELECT * 
FROM player 
WHERE player_id = 2 

結果不是陣列。

function displayName($player_id) 
{ 
    // setting database variables 
    $dsn = 'mysql:dbname=dbName;host=localhost'; 
    $username = "username"; 
    $password = "password"; 

    // try to make connection 
    try 
    { 
     $conn = new PDO($dsn, $username, $password); 
     $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
     echo "Database connected <br>"; 
    } 
    catch(PDOException $e) 
    { 
     echo "Error: " . $e->getMessage(); 
    } 

    // create, print, and execute the query 
    $sql = "SELECT * FROM player WHERE player_id = $player_id"; 
    echo "$sql <br>"; 
    $result = $conn->query($sql); 

    //if the query returned an array 
    if (is_array($result)) 
    { 
     // if the array has data 
     if ($result->num_rows > 0) 
     { 
      Echo "Result is an array and has values. <br>"; 
      Echo "$result[0] $result[1]. <br>"; 
      return $result; 
     } 
     else 
     { 
      echo "No results. <br>"; 
     } 
    } 
    else 
    { 
     echo "Results not an array. <br>"; 
    } 
    //close the database connection. 
    $conn = null; 
} 

Answer 1

PDO查詢返回一個對象，而不是數組。

http://php.net/manual/en/pdo.query.php

它會返回失敗虛假所以只檢查

if ($result) { 
    ...