Javascript 正在寻找未定义索引PHP上出现此错误的解决方案
我正在尝试使用php创建一个功能性的登录页面,并且我在youtube上找到了一个教程。但是,当我运行代码时,我遇到了这个错误,我不知道如何修复它: 注意:第2行的C:\xampp\htdocs\PROJETOWEB\php\login.php中的未定义索引:tEmail 注意:第3行C:\xampp\htdocs\PROJETOWEB\php\login.php中未定义的索引:tSenha 这是密码Javascript 正在寻找未定义索引PHP上出现此错误的解决方案,javascript,php,html,css,Javascript,Php,Html,Css,我正在尝试使用php创建一个功能性的登录页面,并且我在youtube上找到了一个教程。但是,当我运行代码时,我遇到了这个错误,我不知道如何修复它: 注意:第2行的C:\xampp\htdocs\PROJETOWEB\php\login.php中的未定义索引:tEmail 注意:第3行C:\xampp\htdocs\PROJETOWEB\php\login.php中未定义的索引:tSenha 这是密码 <?php $tEmail = $_POST['tEmail']; $tS
<?php
$tEmail = $_POST['tEmail'];
$tSenha = $_POST['tSenha'];
$error = "";
$success = "";
if(isset($_POST['bLog'])){
if($tEmail == "admin@cyanmail.com"){
if($tSenha == "senha"){
$error = "";
$success = "Welcome";
}
else{
$error = "Incorrect Password";
$success = "";
}
}
else{
$error = "Invalid User";
$success = "";
}
}
?>
<!DOCTYPE html>
<html>
<head>
<script src="/PROJETOWEB/js/jquery-3.4.1.js"></script>
<script src="/PROJETOWEB/js/funcoes.js"></script>
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/css/bootstrap.min.css" integrity="sha384-ggOyR0iXCbMQv3Xipma34MD+dH/1fQ784/j6cY/iJTQUOhcWr7x9JvoRxT2MZw1T" crossorigin="anonymous">
<link rel="stylesheet" href="/PROJETOWEB/css/estilo.css">
</head>
<body>
<div class="divBox">
<div class="divBackground"></div>
<div class="divLogo"><img src="/PROJETOWEB/img/logo.png" alt="Logo" width="10%" height="8%"></div>
<div>
<h1><text class="cyan">Cyan</text>Mail</h1><br>
<h2>Bem vindo de volta!<br> Acesse sua conta para ver seus emails!</h2>
</div>
<p class="error"><?php echo $error; ?></p>
<p class="success"><?php echo $success; ?></p>
<div>
<input type="text" placeholder="E-mail" id="tEmail" class="divInput1"><br>
<input type="password" placeholder="Senha" id="tSenha" class="divInput2">
</div>
<div>
<button id="bLog" class="Botao">Log in</button>
</div>
<div class="divBotao2">
<button id="bSign" onclick="location.href='paginas/cadastro.html' "class="Botao2">Sign up</button>
</div>
</div>
</body>
</html>
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希望有人能让我知道我做错了什么。提前谢谢 尝试使用
$tEmail=$\u POST['tEmail'];
$tSenha=$_POST['tSenha']
if(isset($\u POST['bLog']){方法中的变量
因为如果不设置该值,则肯定会发生此错误
<?php
$error = "";
$success = "";
if(isset($_POST['bLog'])){
$tEmail = isset($_POST['tEmail']) ? $_POST['tEmail'] : '';
$tSenha = isset($_POST['tSenha']) ? $_POST['tSenha'] : '';
if($tEmail == "admin@cyanmail.com"){
if($tSenha == "senha"){
$error = "";
$success = "Welcome";
}
else{
$error = "Incorrect Password";
$success = "";
}
}
else{
$error = "Invalid User";
$success = "";
}
}
?>
好的。您需要一个表单,表单需要使用
POST
作为方法,并且您需要在调用$\u POST[var]
语句之前测试POST数组是否已被标注尺寸,即:isset($\u POST['tEmail']){etc