我有PHP代码,该代码应该在PHP代码中具有访问代码。当您进入页面时,您输入访问代码,如果正确,则将继续到另一页,如果不是,则会提供错误消息。
这是我的代码:
<?php
$code = $_POST[$code];
$code = '7613';
if($code == '7613') {
echo " it worked! ";
} else {
echo "it did not work !";
}
?>
<form action="login.php" method="post">
<input type="password" name="code" />
<input type="submit" name="submit" value="Submit" />
</form>
这应该使您入门。
对于起动器而言,两个变量不同。它们可以匹配。查看ISSET功能和空功能。我检查值是否存在,如果有,则对其进行验证。如果没有,请显示表格,不要试图什么都不验证。
<?php
$code = isset( $_POST['code'] ) ? $_POST['code'] : NULL;
$codeToMatch = '7613';
if ( ! empty( $code ) )
{
if( $code == $codeToMatch ) {
echo "it worked!";
} else {
echo "it did not work !";
}
}
?>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<input type="password" name="code" />
<input type="submit" name="submit" value="Submit" />
</form>
我修复了
代码:
<?php
if(isset($_POST['submit'])) {
$code = $_POST['code'];
if($code == '7613') {
echo " it worked! ";
} else {
echo "it did not work !";
}
}
?>
<form action="login.php" method="post">
<input type="password" name="code" maxlength="4" />
<input type="submit" name="submit" value="Submit" />
</form>