mysql_fetch_array接收3行，但返回4

我有下面的PHP代碼：mysql_fetch_array接收3行，但返回4

$queryBlueprint = "SELECT * FROM `plan_photos` WHERE project_id=" . $projectId; 
$resultBlueprints = mysql_query ($queryBlueprint); 
if ($resultBlueprints) { 
    // Add the results into an array 
    $blueprints [] = array(); 
    echo mysql_num_rows ($resultBlueprints); /* --- this returns: 3 */ 
    while ($row = mysql_fetch_array ($resultBlueprints)) { 
     $blueprints [] = $row; 
    } 
    echo "<br/>"; 
    echo count ($blueprints); /* --- this returns 4*/ 
    echo "<br/>"; 
    echo print_r ($blueprints [0]); 
    echo "<br/>"; 
    echo print_r ($blueprints [1]); 
    echo "<br/>"; 
    echo print_r ($blueprints [2]); 
    echo "<br/>"; 
    echo print_r ($blueprints [3]);

爲什麼mysql_num_rows回報3，但在我添加的每個結果到一個數組，數組包含4個項目？第一個（[0]）是「null」，接下來的3個（[1]，[2]和[3]）是它們應該是的（又名它們包含數據）

回顯數據：

3 
4 
Array () 1 /* <------ what is this?!?! */ 
Array ([id] => 8 [project_id] => 2 [photo] => http://webja5309b6cf8a525.jpg [title] => first) 1 
Array ([id] => 9 [project_id] => 2 [photo] => http://webja7ee76.jpg [title] => second) 1 
Array ([id] => 10 [project_id] => 2 [photo] => http://webj022d3.jpg [title] => third blueprint) 1

的表，如果它可以幫助：

CREATE TABLE IF NOT EXISTS `plan_photos` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `project_id` int(11) NOT NULL, 
    `photo` text NOT NULL, 
    `title` varchar(50) NOT NULL, 
    KEY `project_id` (`id`), 
    KEY `project_id_2` (`project_id`)

來源

2014-03-01 AndreiBogdan

請在新代碼使用'mysql_ *'不要，它的正式棄用PHP 5.5.0的。相反，您應該使用[mysqli_ *]（http://codular.com/php-mysqli）或[PDO]（http://uk1.php.net/PDO）。查看所述API的比較[這裏]（http://www.php.net/manual/en/mysqlinfo.api.choosing.php）。 –

的原因是因爲你是在變量的聲明中加入的元素。而不是做的：

$blueprints [] = array();

做

$blueprints = array();

來源

2014-03-01 18:38:21 Mike

哈哈哈......我知道這是愚蠢的：D謝謝 – AndreiBogdan

$blueprints [] = array(); // <------ this is that

來源

2014-03-01 18:38:15

哈哈哈...我知道這是愚蠢的：D謝謝 – AndreiBogdan

mysql_fetch_array接收3行，但返回4

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