顯示全部我似乎無法訪問我的表。我再次確信它忽略了一些簡單的東西。我沒有收到任何錯誤。沒有什麼被添加到表中,沒有任何顯示。我已經重新命名了幾次,確保連接良好,確保桌子存在。我看不到任何錯誤。不能插入SQL表或從
將數據添加到表
// Valid Data
require_once('resources/php/db.php');
$id = uniqid();
//die ($id . $name . $sex. $age. $hair. $eye. $skin. $body. $pf);
// Insert Data to the Table
$statement1 = $db->prepare('INSERT INTO player VALUES(:player_id, :name, :sex, :age, :hair, :eye, :skin, :body, :pf)');
$result1 = $statement1 -> execute(array(
':player_id' =>$id,
':name' =>$name,
':sex' =>$sex,
':age' =>$age,
':hair' =>$hair,
':eye' =>$eye,
':skin' =>$skin,
':body' =>$body,
':pf' =>$pf
));
// Make Sure Everything Worked
if($result1 == false)
{
die('Update Failed, Please Check Your Database.');
}
header("Location: ../../new_success.php?id=$id;");
exit();
成功頁
// Start the Load
$query1 = "SELECT *
FROM player
WHERE player_id = :player_id";
$statement1 = $db->prepare($query1);
$statement1 -> execute(array(
':player_id' =>$id
));
// Make Sure the Data Exists
if($statement1->rowCount() == 0)
{
die('Please Enter a Valid ID Tag - (id)');
}
else
{
$notEmpty = true;
}
while($row = $statement1->fetch())
{
$name = $row['name'];
$sex = $row['sex'];
$age = $row['age'];
$hair = $row['hair'];
$eye = $row['eye'];
$skin = $row['skin'];
$body = $row['body'];
$pf = $row['pf'];
}
你檢查是否有查詢錯誤,即從MySQL? – MrCode 2012-03-05 19:31:13
您使用的是PDO嗎? – 2012-03-05 19:33:28
是的,即時通訊使用pdo。數據庫文件沒有任何改變。我將它用於所有數據庫連接。至於查詢錯誤,我沒有找到。 – 2012-03-05 19:35:26