Question

<form name="login" action=""<?php echo $_SERVER['PHP_SELF']; ?>"" method="post" accept-charset="utf-8">

            <label for="usermail">Username</label>
            <input type="text" name="nume" placeholder="username" required>

            <label for="password">Password</label>
            <input type="password" name="password" placeholder="password" required>

            <input type="submit" value="Login">
        </form>
        <?php
        $connection = mysqli_connect("127.0.0.1", "root", "", "agentie");
        if (isset($_POST["nume"]) && isset($_POST["password"])) {
            if (($_POST["nume"] == "admin") && ($_POST["password"] == "admin123")) {
                echo "ati intrat in panoul de control";
            }

            $query1 = mysqli_query($connection, "SELECT user, password FROM user u WHERE u.user='$_POST[nume]'") or die("Error in the consult.." . mysqli_error($connection));
            while ($row = mysqli_fetch_assoc($query1)) {
                if (($row["user"] == $_POST["nume"]) && ($row["password"] == $_POST["password"]))
                    echo "Bine ati venit" . $row["user"];
            }
        }
        ?>

我不知道为什么我的2个套装永远不会成真。如果我删除if条件，我它们出现在simply print $_POST["nume"]和$_POST["password"]。

提前致谢。

Answer 1

你的isset工作但是嵌套的if语句不匹配尝试在first if语句中回显它。

if (isset($_POST["nume"]) && isset($_POST["password"])) {
      echo $_POST["nume"]; 
      echo $_POST["password"];
            if (($_POST["nume"] == "admin") && ($_POST["password"] == "admin123")) {
                echo "ati intrat in panoul de control";
            }

Answer 2

<form name="login" method="post" accept-charset="utf-8">

<label for="usermail">Username</label>
<input type="text" name="nume" placeholder="username" required>

<label for="password">Password</label>
<input type="password" name="password" placeholder="password" required>

<input type="submit" value="Login">
</form>
<?php

if (isset($_POST["nume"], $_POST["password"])) {
    echo "It works";
    if (($_POST["nume"] == "admin") && ($_POST["password"] == "admin123")) {
        echo "ati intrat in panoul de control";
    }
}
?>

Answer 3

未评估您的$_POST变量。改变这个： $query1 = mysqli_query($connection, "SELECT user, password FROM user u WHERE u.user='$_POST[nume]'") or die("Error in the consult.." . mysqli_error($connection));

到

$query1 = mysqli_query($connection, "SELECT user, password FROM user u WHERE u.user='{$_POST[nume]}'") or die("Error in the consult.." . mysqli_error($connection));

我的情况永远不会成真。这是为什么？

3 个答案: