當$id是abc@ab 時，它返回true。這正如預期的那樣！但是當$id是1234567 時，它返回false。這并不 像預期的那樣！我的數據庫：+----+---------+---------+-----------+| Id | Email | Mobile | Password |+----+---------+---------+-----------+| 15 | [email protected] | 1234567 |$2y$10$ix..|+----+---------+---------+-----------+我的代碼：public function customer_login($id,$password) { try { $stmt = $this->conn->prepare("SELECT * FROM customer WHERE Email=:_email"); $stmt->execute(array(":_email"=>$id)); $userRow=$stmt->fetch(PDO::FETCH_ASSOC); $id = $userRow['Id']; $stmtM = $this->conn->prepare("SELECT * FROM customer WHERE Mobile=:_mobile"); $stmtM->execute(array(":_mobile"=>$id)); $userRowM=$stmtM->fetch(PDO::FETCH_ASSOC); $id = $userRowM['Id']; if (password_verify($password, $userRow['Password']) || password_verify($password, $userRowM['Password'])) { $_SESSION['customer'] = $id; return true; } else { return false; } } catch(PDOException $ex) { echo $ex->getMessage(); } }當$id是abc@ab 時，它返回true。這正如預期的那樣！但是當$id是1234567 時，它返回false。這并不 像預期的那樣！
2 回答
回首憶惘然
TA貢獻1847條經驗 獲得超11個贊
// Change this:
$stmt = $this->conn->prepare("SELECT * FROM customer WHERE Email=:_email");
$stmt->execute(array(":_email"=>$id));
$userRow=$stmt->fetch(PDO::FETCH_ASSOC);
$id = $userRow['Id'];
$stmtM = $this->conn->prepare("SELECT * FROM customer WHERE
Mobile=:_mobile");
$stmtM->execute(array(":_mobile"=>$id));
$userRowM=$stmtM->fetch(PDO::FETCH_ASSOC);
$id = $userRowM['Id'];
//to this :
$stmt = $this->conn->prepare("SELECT * FROM customer WHERE Email=:_email or
Mobile=:_mobile limit 0,1 ");
$stmt->execute(array(":_email"=>$id , ":_mobile"=>$id ));
$userRow=$stmt->fetch(PDO::FETCH_ASSOC);
$id = $userRow['Id'];
GCT1015
TA貢獻1827條經驗 獲得超4個贊
在下一行中，您將更改 $id 的值。$id 變量獲取表的“Id”字段的值，并將其輸入到基于手機號碼的第二個查詢中。這就是原因，您基于手機號碼的查詢不返回任何結果并返回 false。
$id = $userRow['Id'];
將上述分配給不同的變量并保持 $id 不變作為初始值。
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