简单if语句不起作用，尽管相等[关闭]

I have a PHP page for checking username/password:

foreach($_REQUEST as $k=>$v){
    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    $sql = "SELECT password FROM authentication WHERE username='$k'";
    $result = $conn->query($sql);
    $sql2 = "SELECT owner FROM listofowners WHERE owner='$k'";
    $result2 = $conn->query($sql2);

    if ($result->num_rows > 0) {
        // output data of each row
        $row = mysqli_fetch_assoc($result);
        if ($row["password"] == '$v') {
            if ($result2->num_rows > 0) {
                echo "successfulowner";
                break;
            } else {
                echo "successfulguest";
                break;
            }
        } elseif($row["password"] != '$v') {
            echo "wrong password";
        }
    } else {
        echo "No such email as:" .$k;
    }
    $conn->close();
}
?>

Despite entering the correct password, I keep getting the wrong password message.

What am I doing wrong?

You are comparing it wrong

$row["password"]!='$v'

Try like this:-

$row["password"]!= $v

you are mentioning $v string itself

And you are also not comparing username also.Follow same approach for it.

as stated in the comments, it's the single quotes around your variable '$v'

inside of a longer string enclosed by double quotes, as in your query line, the single quotes won't effect the variable from parsing, but on their own, they make it no longer a variable, but the static string $v so that condition will only pass as true if the password is literally $v

Remove single quotes from your variable $v. It's a variable not a string which is holding your value. So remove quotes from your code wherever you wrote like this '$v' in your conditional statement.

Try this:

$row["password"]!=$v

And

$row["password"]==$v

Assuming that you $_REQUEST have something like:

"username" = "name_of_the_user"
"password" = "123456"

This code can validate them:

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$stAuth = $conn->prepare("SELECT password FROM authentication WHERE username=?");
$stAuth->bind_param("s", $_REQUEST["username"]);
$stAuth->execute();
$stAuth->bind_result($password);
if ($stAuth->fetch()) {
    if ($_REQUEST["password"] == $password) {
        $stOwner = $conn->prepare("SELECT owner FROM listofowners WHERE owner=?");
        $stOwner->bind_param("s", $_REQUEST["username"]);
        $stOwner->execute();
        $stOwner->bind_result($owner);
        if ($stOwner->fetch()) {
            echo "successfulowner";
            break;
        } else {
            echo "successfulguest";
            break;
        }
        $stOwner->close();      
    } else {
        echo "wrong password";
    }
} else {
    echo "No such email as:" . $_REQUEST["username"];
}
$stAuth->close();
$conn->close();

Your code is vulnerable for SQL injection and my code can be better to. Take a look how SQL injection and input validation work to make a better code!!

Your original code was assuming that the $_REQUEST has the username as a key, that's not usual so a think you code has the wrong logic.