用网页打开php登陆页面不提示登陆是否成功

晚风811 · 发表于 2023-11-29 21:12

用echo输出登陆成功或者失败的提示,但是打开网页登陆的时候点击登陆没有反应.

[PHP] 纯文本查看 复制代码

<?php
include_once 'init.php';

if($_POST['sub']){

    $username=filterstr($_POST['username']);
    $password=md5(filterstr($_POST['password']));
    $result=$conn->query("select * from users where username='$username' and password='$password'");
    if($result->num_rows>0){
        $row=$result->fetch_assoc();
        if ($row['password']==$password){
            echo "登录成功";
        }else{
            echo "账号或密码错误";
        }
    }else{
        echo "登录失败";
    }
}

?>
<html>
<head>
    <title>管理员登录</title>
    <style>
        .login{
            width: 400px;
            margin: 0px auto;
        }
    </style>
</head>
<body>
<div class="login">
    <form>
        <table>
            <tr><td><lable for="username">用户</lable></td><td><input type="text" name="username" id="username"></td></tr>
            <tr><td><lable for="password">密码</lable></td><td><input type="password" name="password" id="password"></td></tr>
            <tr><td colspan="2"><input type="submit" value="登录" name="sub"></td></tr>
        </table>
    </form>
</div>
</body>
</html>

dleo · 发表于 2023-11-29 23:59

form的method要标明是post才能拿到提交的内容吧

shipon · 发表于 2023-11-30 00:02

本帖最后由 shipon 于 2023-11-30 00:08 编辑

[PHP] 纯文本查看 复制代码

include_once 'init.php';
if ($_POST['sub']) {
    $username = filterstr($_POST['username']);
    $password = md5(filterstr($_POST['password']));
    $result = $conn->query("select * from users where username='$username' and password='$password'");
    if ($result->num_rows > 0) {
        $row = $result->fetch_assoc();
        if ($row['password'] == $password) {
            echo "登录成功";
        } else {
            echo "账号或密码错误";
        }
    } else {
        echo "登录失败";
    }
}
?>

<html>
<head>
    <title>管理员登录</title>
    <style>
        .login {
            width: 400px;
            margin: 0px auto;
        }
    </style>
</head>
<body>
<div class="login">
    <form method="POST">
        <table>
            <tr>
                <td><label for="username">用户</label></td>
                <td><input type="text" name="username" id="username"></td>
            </tr>
            <tr>
                <td><label for="password">密码</label></td>
                <td><input type="password" name="password" id="password"></td>
            </tr>
            <tr>
                <td colspan="2"><input type="submit" value="登录" name="sub"></td>
            </tr>
        </table>
    </form>
</div>
</body></html>

Maiz1888 · 发表于 2023-11-30 09:29

不会写的话建议看下别人开源的怎么写的登录，别自己瞎琢磨，不行网上搜搜也行。你的form的action去哪了

Do_zh · 发表于 2023-11-30 09:50

form表单是不是要表明post方式请求你才能那导值吧。

bhsmr · 发表于 2023-11-30 10:26

表单元素设置form的action属性

[PHP] 纯文本查看 复制代码

<?php
include_once 'init.php';

if (isset($_POST['sub'])) {

    $username = filterstr($_POST['username']);
    $password = md5(filterstr($_POST['password']));
    $result = $conn->query("select * from users where username='$username' and password='$password'");
    if ($result->num_rows > 0) {
        $row = $result->fetch_assoc();
        if ($row['password'] == $password) {
            echo "登录成功";
        } else {
            echo "账号或密码错误";
        }
    } else {
        echo "登录失败";
    }
    exit; // 结束脚本执行，确保只返回登录结果
}

?>
<html>
<head>
    <title>管理员登录</title>
    <style>
        .login{
            width: 400px;
            margin: 0px auto;
        }
    </style>
</head>
<body>
<div class="login">
    <form action="login.php" method="POST">
        <table>
            <tr><td><label for="username">用户</label></td><td><input type="text" name="username" id="username"></td></tr>
            <tr><td><label for="password">密码</label></td><td><input type="password" name="password" id="password"></td></tr>
            <tr><td colspan="2"><input type="submit" value="登录" name="sub"></td></tr>
        </table>
    </form>
</div>

<script>
    function submitForm(event) {
        event.preventDefault(); // 阻止表单默认提交行为

        var form = event.target;
        var formData = new FormData(form);

        var xhr = new XMLHttpRequest();
        xhr.open(form.method, form.action, true);
        xhr.onreadystatechange = function () {
            if (xhr.readyState === XMLHttpRequest.DONE && xhr.status === 200) {
                // 在页面中显示登录结果
                var response = xhr.responseText;
                // TODO: 在页面中显示响应
                console.log(response);
            }
        };
        xhr.send(formData);
    }
</script>

</body>
</html>

萌新表示是小白 · 发表于 2023-11-30 15:53

本帖最后由萌新表示是小白于 2023-11-30 15:54 编辑

[PHP] 纯文本查看 复制代码

<?php
include_once 'init.php';
 ?>

<html>
<head>
    <title>管理员登录</title>
    <style>
        .login{
            width: 400px;
            margin: 0px auto;
        }
    </style>
</head>
<body>
<?php
if($_POST['sub']){
 
    $username=filterstr($_POST['username']);
    $password=md5(filterstr($_POST['password']));
    $result=$conn->query("select * from users where username='$username' and password='$password'");
    if($result->num_rows>0){
        $row=$result->fetch_assoc();
        if ($row['password']==$password){
            echo "登录成功";
        }else{
            echo "账号或密码错误";
        }
    }else{
        echo "登录失败";
    }
}
?>
<div class="login">
    <form action="login.php" method="POST" style="<?php if($_POST['sub']){ disolay:none} ?>" >
        <table>
            <tr><td><lable for="username">用户</lable></td><td><input type="text" name="username" id="username"></td></tr>
            <tr><td><lable for="password">密码</lable></td><td><input type="password" name="password" id="password"></td></tr>
            <tr><td colspan="2"><input type="submit" value="登录" name="sub"></td></tr>
        </table>
    </form>
</div>
</body>
</html>

php的echo打印要写在body里面，method设置为POST，
这里我还帮你添加了个，如果登录了就将登录框隐藏的功能
{:17_1067:}

晚风811 · 发表于 2023-11-30 17:38

感谢大家,问题已经解决,就是form后面没加method="post",感谢.

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[求助] 用网页打开php登陆页面不提示登陆是否成功

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