PHP和MySQL-如果內容不在一個表中，請檢查另一個表

Question

我試圖設置一個if / else情況，看一個值是否在一個表中，然后顯示該值，如果它不在一個表中，則檢查另一個表。

我正在Cloud 9中使用PHP和MySQL。電子郵件是從未顯示的表單中獲得的。

使用正確的電子郵件搜索“客戶”表將顯示password_enc，但是如果在“客戶”表中找不到該電子郵件，並且搜索“公司”表，則不會顯示任何內容。 我檢查了PHPMyAdmin，電子郵件存在並且有效

我還檢查了表，“ email”和“ rep_email”有效。

有沒有更簡單的方法可以執行此操作，或者我當前的代碼需要調整嗎？ 謝謝您的幫助。

$email              = !empty($_POST ['email']) ? $_POST['email'] : "";
$email2             = $email;


$query  = "SELECT * FROM Customer WHERE email = '$email'";
$result = $db->query($query);

if ($result){
    $row            = $result->fetch_assoc(); /* These lines query the dataabse when user enters email */
    $password_enc   = $row['password_enc'];
}
elseif(!$result){
    $query2         = "SELECT * FROM Company WHERE rep_email = '$email2'";
    $result2        = $db->query($query2);
    $row2           = $result2->fetch_assoc(); /* These lines query the dataabse when user enters email */
    $password_enc   = $row2['password_enc'];
}

echo "password_enc: ".$password_enc;

Answer 1

您應該使用num_rows從結果集中檢索行數。

$email              = !empty($_POST ['email']) ? $_POST['email'] : "";
$email2             = $email;


$query  = "SELECT * FROM Customer WHERE email = '$email'";
$result = $db->query($query);

if ($result->num_rows > 0){ //check if results is bigger then 0 
    $row            = $result->fetch_assoc(); /* These lines query the dataabse when user enters email */
    $password_enc   = $row['password_enc'];
} else {
    $query2         = "SELECT * FROM Company WHERE rep_email = '$email2'";
    $result2        = $db->query($query2);
    $row2           = $result2->fetch_assoc(); /* These lines query the dataabse when user enters email */
    $password_enc   = $row2['password_enc'];
}

echo "password_enc: ".$password_enc;

Answer 2

對於成功的SELECT，SHOW，DESCRIBE或EXPLAIN查詢，mysqli_query（）將返回mysqli_result對象

PHP MySQLi手冊

因此，除了檢查響應之外，您實際上可以檢查返回的行數，如下所示

$query  = "SELECT * FROM Customer WHERE email = '$email'";
$result = $db->query($query);

if ($result->num_rows > 0){
   $row            = $result->fetch_assoc(); /* These lines query the dataabse when user enters email */
   $password_enc   = $row['password_enc'];
}else{
   $query2         = "SELECT * FROM Company WHERE rep_email = '$email2'";
   $result2        = $db->query($query2);
   if($result2->num_rows>0){
      $row2           = $result2->fetch_assoc(); /* These lines query the dataabse when user enters email */
      $password_enc   = $row2['password_enc'];
   }
}

if(isset($password_enc)){
   echo "password_enc: ".$password_enc;
}else{
   echo "Password Not Available";
}

Answer 3

$email = !empty($_POST ['email']) ? $_POST['email'] : "";

$query = "SELECT result.* FROM( SELECT IFNULL((SELECT password_enc FROM Customer WHERE email = '$email'), (SELECT password_enc FROM Company WHERE rep_email = '$email')) as password_enc) as result";

$result = $db->query($query);
$password_enc = '';

if ($result->num_rows > 0){
    $row            = $result->fetch_assoc();
    $password_enc   = $row['password_enc'];
}

echo "password_enc: ".$password_enc;