听了你的建议，我已经完成了计划。谢谢大家。给我们完美的密码。

$users = mysqli_query($con, "SELECT email, pass FROM users");
while ($row = mysqli_fetch_assoc($users)) {
    $email = $row['email'];
    $pass = $row['pass'];
    $user_pass3[] = array(
        $email => $pass,
    );

}
foreach($user_pass3 as $arr) {
            foreach($arr as $key => $data) {
                    $user_pass[$key]  =   $data;
                }
        }

   // print_r($user_pass);
$user_pass_final = $user_pass;

<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "test";
$conn = mysqli_connect($servername, $username, $password, $database); if (!($conn)) 
{
die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT email,pass FROM users";
$result = mysqli_query($conn, $sql);

$newArray = array();

if (mysqli_num_rows($result) > 0) 
{
    /* loop for return all the fetched data one by one */
while($row = mysqli_fetch_assoc($result)) 
{
    /* store the data from mysqli array in variable use same code for for as much as type of data you want*/
$email=$row["email"]; 
$password=$row["pass"];
    /* you can also add some value manually into the same array like this */
$newArray['admin']= 'adminpass';
$newArray['test']= 'testpass';
    /* You have to build one array variable which includes 2 type of data 1st one as array index and 2nd one as value */
$newArray[$email]= $password;
}
} 
else 
{
echo "0 results";
}
$conn->close();
    /* print the array or store in another variable to use elsewhere. */
print_r($newArray);
/* or print indiviual data one by one like data1, data2 */
echo "<br>";
foreach($newArray as $x => $x_value)
{
echo "Email=" . $x . ", Password=" . $x_value;
echo "<br>";
}

/* this example shows email and password in plain text just for the special requirement of the project. It's recomended to not to use passwords in plain text and inside the code. */

    //Credits gose to https://www.facebook.com/ansarali.molla.9
?>

如果您只需要登录名和密码：

$users = mysqli_query($con,"SELECT login, password FROM sughi_users"));

$user = array();

foreach ($users as $aUser) {
    $user[$aUser['login']] = $aUser['password'];
}

这个 [] 变量上的符号使其成为动态数组。所以不要使用 array() 之后。如果愿意，也可以在括号内传递唯一的键。

$users = mysqli_query($con, "SELECT email, pass FROM users");
while ($row = mysqli_fetch_assoc($users)) {
    $email = $row['email'];
    $pass = $row['pass'];
    $user_pass_final[$email] = $pass;
}
print_r($user_pass_final);

此外，您还将执行双重获取，并在数组上获取数据，该数组应失败，而不是：

$users=mysqli_fetch_assoc(mysqli_query($con,"SELECT email, pass FROM users"));
while ($row = mysqli_fetch_assoc($users)) {

你应该分配 query 返回 $users :

$users=mysqli_query($con,"SELECT email, pass FROM users");
while ($row = mysqli_fetch_assoc($users)) {

最后一点注意，密码不应以纯文本形式存储。看到了吗https://security.stackexchange.com/questions/120540/why-shouldnt-i-store-passwords-in-plaintext