我正在设置完全有组织的注册表单,我正在尝试将信息发送到我的 MySQL 数据库服务器中。我的代码不起作用,无法弄清楚弹出消息您已注册!我尝试了几个选项,但没有一个在服务器上工作,<?phpif (array_key_exists('email', $_POST) or array_key_exists('password', $_POST)) { // So I'm guessing here need to some additions of based my own inclinatios.我的代码<?php if (array_key_exists('email', $_POST) or array_key_exists('password', $_POST)) { $link = mysqli_connect("localhost", "id8955440_bigebro", "sunshine1987**", "id8955440_bigebro"); if (mysqli_connect_error()) { die("There was an error connecting to the database"); } if ($_POST['email'] == '') { echo "<p>Email address is required.</p>"; } else if ($_POST['password'] == '') { echo "<p>Password is required.</p>"; } else { $query = "SELECT `id` FROM `bigebro` WHERE email = '" . mysqli_real_escape_string($link, $_POST['email']) . "'"; $result = mysqli_query($link, $query); if (mysqli_num_rows($result) > 0) { echo "<p>That email address has already been taken.</p>"; } else { $query = "INSERT INTO `bigebro` (`email`, `password`) VALUES ('" . mysqli_real_escape_string($link, $_POST['email']) . "', '" . mysqli_real_escape_string($link, $_POST['password']) . "')"; if (mysqli_query($link, $query)) { echo "<p>You have been signed up!"; } else { echo "<p>There was a problem signing you up - please try again later.</p>"; } } }}?>HTML<form method="post"> <input name="email" type="text" placeholder="Email address"> <input name="password" type="password" placeholder="Password"> <input type="submit" value="Sign up"></form>$link = mysqli_connect("localhost", "id8955440_bigebro", "sunshine1987**", "id8955440_bigebro");当我注册表格时只弹出“注册您时出现问题 - 请稍后再试”我希望从结果中得到“您已注册”。请看报名表
2 回答
翻翻过去那场雪
TA贡献2065条经验 获得超14个赞
使用mysqli_error查询跟踪误差。
$query = "INSERT INTO `bigebro` (`email`, `password`) VALUES ('" . mysqli_real_escape_string($link, $_POST['email']) . "', '" . mysqli_real_escape_string($link, $_POST['password']) . "')";
if (mysqli_query($link, $query)) {
echo "<p>You have been signed up!";
} else {
echo "Error: " . mysqli_error($conn);
}
还要检查你的数据库结构,比如 id 应该自动递增,电子邮件和密码应该是 varchar。
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