Php 仅在填写表单后显示信息
我有以下代码:Php 仅在填写表单后显示信息,php,Php,我有以下代码: <html> <head> </head> <body> <?php if (!isset($_COOKIE["loggedin"])){ ?> <form action="index.php" method="POST" name="name_form"> Username <input type=
<html>
<head>
</head>
<body>
<?php
if (!isset($_COOKIE["loggedin"])){
?>
<form action="index.php" method="POST" name="name_form">
Username <input type="text" name="username">
<br/>
Password <input type="text" name="password">
<br/>
Remember Me <input type ="checkbox" name="remember_me" value="1">
<br/>
<input type="submit" name="submit" value="Log in">
</form>
<?php
if(preg_match("/<|>/", $_POST["username"])){
echo "do not log in";
}
else if(preg_match("/<|>/", $_POST["password"])){
echo "do not log in";
}
else {
//Open/create passwords.txt
$passwordsFile = fopen("passwords.txt", "a");
//write users username and password to passwords.txt
$text_written = fwrite($passwordsFile, $_POST["username"] . "," . $_POST["password"] . "\r\n");
fclose($passwordsFile);
setcookie("loggedin", $_POST["username"]);
setcookie("loggedintime", time());
echo "<h1>Welcome " . $_COOKIE["loggedin"] . "</h1>";
echo "You have been logged in for " . $_COOKIE["loggedintime"] . " seconds.";
echo "<nav>
<ul>
<li>Browse books in store</li>
<li>Analytics</li>
<li>Logout</li>
</ul>
</nav>";
}
}
?>
</body>
用户名
密码
记得我吗
但是在用户正确填写表单之前,它会显示“欢迎,您已登录…”部分,我想知道如果我只想在用户正确填写表单之后显示它,该怎么办
谢谢 好吧,请按逻辑逐步完成代码。如果用户未登录,则为真:
if (!isset($_COOKIE["loggedin"])){
如果未发布任何表单值,则为false:
if(preg_match("/<|>/", $_POST["username"])){
else if(preg_match("/<|>/", $_POST["password"])){
这将检查在评估其余代码之前是否按下了“提交”按钮。您似乎正在验证表单并显示欢迎消息,而没有提交表单。请检查下面的代码。这可能对你有帮助。仅在提交表单后验证表单
<?php
if (!isset($_COOKIE["loggedin"])) {
?>
<form action="index.php" method="POST" name="name_form">
Username <input type="text" name="username">
<br/>
Password <input type="text" name="password">
<br/>
Remember Me <input type ="checkbox" name="remember_me" value="1">
<br/>
<input type="submit" name="submit" value="Log in">
</form>
<?php
if($_POST) {
if(preg_match("/<|>/", $_POST["username"])){
echo "do not log in";
}
else if(preg_match("/<|>/", $_POST["password"])){
echo "do not log in";
}
else {
//Open/create passwords.txt
$passwordsFile = fopen("passwords.txt", "a");
//write users username and password to passwords.txt
$text_written = fwrite($passwordsFile, $_POST["username"] . "," . $_POST["password"] . "\r\n");
fclose($passwordsFile);
setcookie("loggedin", $_POST["username"]);
setcookie("loggedintime", time());
echo "<h1>Welcome " . $_COOKIE["loggedin"] . "</h1>";
echo "You have been logged in for " . $_COOKIE["loggedintime"] . " seconds.";
echo "<nav>
<ul>
<li>Browse books in store</li>
<li>Analytics</li>
<li>Logout</li>
</ul>
</nav>";
}
}
}
?>
用户名
密码
记得我吗
“但它只在用户正确填写表单后才显示[…],如果我想在用户正确填写表单后显示它,我想知道该怎么做”–所以你在问如何让它完成它已经完成的工作…?哦,在*哈哈哈之前!嗨,Gokul,它解决了这个问题,但现在我发现了两个错误:第42行的C:\xampp\htdocs\assignment1\index.php中的未定义索引:loggedintime和第43行的C:\xampp\htdocs\assignment1\index.php中的未定义索引:loggedintime。除此之外,表单在登录后继续显示。谢谢你,我想你是在尝试设置cookie并立即使用它。它有什么目的吗?
<?php
if (!isset($_COOKIE["loggedin"])) {
?>
<form action="index.php" method="POST" name="name_form">
Username <input type="text" name="username">
<br/>
Password <input type="text" name="password">
<br/>
Remember Me <input type ="checkbox" name="remember_me" value="1">
<br/>
<input type="submit" name="submit" value="Log in">
</form>
<?php
if($_POST) {
if(preg_match("/<|>/", $_POST["username"])){
echo "do not log in";
}
else if(preg_match("/<|>/", $_POST["password"])){
echo "do not log in";
}
else {
//Open/create passwords.txt
$passwordsFile = fopen("passwords.txt", "a");
//write users username and password to passwords.txt
$text_written = fwrite($passwordsFile, $_POST["username"] . "," . $_POST["password"] . "\r\n");
fclose($passwordsFile);
setcookie("loggedin", $_POST["username"]);
setcookie("loggedintime", time());
echo "<h1>Welcome " . $_COOKIE["loggedin"] . "</h1>";
echo "You have been logged in for " . $_COOKIE["loggedintime"] . " seconds.";
echo "<nav>
<ul>
<li>Browse books in store</li>
<li>Analytics</li>
<li>Logout</li>
</ul>
</nav>";
}
}
}
?>