检查变量是否设置然后加密用户帖子不工作

Question

<?php

$password = $_POST["password"];

if (isset($password) == true){
    echo "your password = ".$password."<hr>";
    echo "Encrypted password = ".md5($password)."<hr>";
}else{
    echo "please enter your password";
};

echo'
<form action="index.php" method="post" >
password : <input type="password" name="password">
<input type="submit" />
';

?>

在此代码中我想使用md5函数加密用户密码
但是当用户输入任何内容时它回显

  

d41d8cd98f00b204e9800998ecf8427e

并使用isset函数来解决此问题，但它无法正常工作

Answer 1

尝试这种方式：

$scope.quoteToAdd = {
    author : $scope.quoteAuthor,
    text: $scope.quoteText
  };
 $scope.addQuote = quoteService.addData;
}

Answer 2

更改此

$password = $_POST["password"];

if (isset($password) == true){
    echo "your password = ".$password."<hr>";
    echo "Encrypted password = ".md5($password)."<hr>";
}else{
    echo "please enter your password";
};

到这个

if (isset($_POST["password"]) == true){
    $password = $_POST["password"];
    echo "your password = ".$password."<hr>";
    echo "Encrypted password = ".md5($password)."<hr>";
}else{
    echo "please enter your password";
};

因为，当您第一次加载页面时，$_POST['password']未设置。所以加载你的表格会遇到麻烦。

Answer 3

此代码有效

<?php

$password = $_POST["password"];

if ($password == ""){ 
    echo "please enter your password";
}else{
    echo "your password = ".$password."<hr>";
    echo "Encrypted password = ".md5($password)."<hr>";
};

echo'
<form action="index.php" method="post" >
password : <input type="password" name="password">
<input type="submit" />
';

?>

Answer 4

实际上$_POST解释here是通过HTTP POST传递给当前脚本的关联变量数组。因此，直接使用$_POST本身检查任何预期变量的存在以及使用

之前的来检查是否存在任何预期变量

e.g。 array_key_exists('password', $_POST);，isset($_POST['password']);等

换句话说，脚本中的条件可以更改为：

if (array_key_exists('password', $_POST)){
    $password = $_POST["password"];
    echo "your password = ".$password."<hr>";
    echo "Encrypted password = ".md5($password)."<hr>";
}else{
    echo "please enter your password";
};