PHP & MySQL - 如果内容不在一个 table 中，请检查另一个

Question

我正在尝试设置一个 if/else 案例来查看一个值是否在一个 table 中，然后显示该值，如果它不在一个 table 中，以检查另一个 table.

我在 Cloud 9 中使用 PHP 和 MySQL。电子邮件是从未显示的表单中获取的。

使用正确的电子邮件搜索 table "Customer" 会显示 password_enc，但如果在客户 table 中找不到电子邮件，它会搜索 "Company" table, 就是不显示任何东西。我检查了 PHPMyAdmin，电子邮件存在且有效

我也检查了我的 table，"email" 和 "rep_email" 是有效的。

是否有更简单的方法来执行此操作，或者我当前的代码是否需要调整？感谢您的帮助。

$email              = !empty($_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

$email = !empty($_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;

Answer 2

您应该使用 num_rows 从结果集中检索行数。

$email              = !empty($_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 3

对于成功的 SELECT、SHOW、DESCRIBE 或 EXPLAIN 查询 mysqli_query() 将 return 一个 mysqli_result 对象

Php mysqli Manual

因此，您实际上可以像这样return检查行数

，而不是检查响应

$query  = "SELECT * FROM Customer WHERE email = '$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";
}

PHP & MySQL - 如果内容不在一个 table 中，请检查另一个

PHP & MySQL - If content is not in one table, check another

php

mysql

cloud9