将HTML变量转换为PHP变量

Question

I'm trying to save a variable to a text file using PHP code. 我正在尝试使用PHP代码将变量保存到文本文件。 Everything else works apart from line 33 of my code, where I try and pass an HTML variable into PHP , in order to save that variable to my text file. 其他所有工作都与代码的第33行不同，我尝试将HTML变量传递给PHP ，以便将该变量保存到我的文本文件中。 Wouldn't like to use forms as I don't want to change my earlier code round too much as it's taken me ages to get it to work. 我不想使用表格，因为我不想过多更改我的早期代码，因为花了我很多时间才能使它起作用。

Have already tried, $_GET and $_REQUEST 已尝试$_GET和$_REQUEST

<DOCTYPE.html>
<html>

<body>

<h3>Please enter your UN and PW</h3>

<p id="demo"></p>
<p id="demo1"></p>
<p id="demo2"></p>

<script>
var username, password, x, INput;
username = prompt("Username: ");
password = prompt("Password: ");
document.getElementById("demo").innerHTML= "UN: " + username;
document.getElementById("demo1").innerHTML = "PW: " + password;

x = (password<=0) ? "No":"Yes";
document.getElementById("demo2").innerHTML = "Typed password: " + x;

INput= String(username + ", " + x + "\n" )
alert(INput);

</script>


<?php

    // Open the text file
    $f = fopen("textfile.txt", "a");

    // Write text line
    fwrite($f, #######); 

    // Close the text file
    fclose($f);

?>


</body>
</html>

The "##s" are where I would like "INput" to go, but it doesn't work as it is an html variable. 我想在“ ## s”中输入“ INput”，但由于它是一个html变量，所以它不起作用。

Expected Output: INput will save to the "textfile.txt" file 预期输出： INput将保存到“ textfile.txt”文件

Actual Output: "Notice: Use of undefined constant INput - assumed 'INput' in /storage/ssd1/856/9172856/public_html/UNPW.php on line 33" 实际输出： “注意：使用未定义的常量INput-在第33行的/storage/ssd1/856/9172856/public_html/UNPW.php中假定为'INput'”

EDIT: Using 000webhost.com, if this changes anything 编辑：使用000webhost.com，如果这更改任何内容

Answer 1

The easiest way to make AJAX in Javascript is to use the Fetch function. 用Java语言制作AJAX的最简单方法是使用Fetch函数。 There are lots of examples available on the web, but here's what I use: 网络上有很多可用的示例，但这是我使用的示例：

side1.html side1.html

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <title>Document</title>
</head>
<body>
  <h3>Please enter your UN and PW</h3>
  <p id="demo"></p>
  <p id="demo1"></p>
  <p id="demo2"></p>

  <script>
    var username, password, x, INput;

    username = prompt("Username: ");
    password = prompt("Password: ");

    document.getElementById("demo").innerHTML = "UN: " + username;
    document.getElementById("demo1").innerHTML = "PW: " + password;

    /* -----------------------------------------------------------------------
    x = (password <= 0) ? "No" : "Yes";
    document.getElementById("demo2").innerHTML = "Typed password: " + x;

    INput = String(username + ", " + x + "\n")
    alert(INput);
    ----------------------------------------------------------------------- */

    fetch("side2.php", {
      credentials : "same-origin",
      method      : "POST",
      headers     : { "Content-Type": "application/json" },
      body        : JSON.stringify(
                    {
                      "Username": username,
                      "Password": password
                    })
    })
    .then(response => response.json())
    .then(data=>{
      if (data.status == 'good world :)')
      {
        document.getElementById("demo2").innerHTML = data.User_Psw;
      }
      else 
      {
        document.getElementById("demo2").innerHTML = '??? it was so easy, what did you miss?';
      }
    })
  </script>
</body>
</html>

side2.php side2.php

<?php
mb_internal_encoding("UTF-8");  // old security story,
$contentType = isset($_SERVER["CONTENT_TYPE"]) ? trim($_SERVER["CONTENT_TYPE"]) : '';

$ReturnInfo = array();
$ReturnInfo['status'] = 'bad world :/';

if ($contentType === "application/json") {
  //Receive the RAW post data.
  $content = trim(file_get_contents("php://input"));

  $decoded = json_decode($content, true);

  $User = $decoded['Username'];
  $Pwd  = $decoded['Password'];

  // write data in a File...
    $fp = fopen('data.txt', 'wb');
    fwrite($fp, "last connected user info =\n");
    fwrite($fp, 'Username='.$User."\n");
    fwrite($fp, 'Password='.$Pwd."\n");
    fclose($fp);

  $ReturnInfo['status']   = 'good world :)';
  $ReturnInfo['User_Psw'] =  $User.'<=>'.$Pwd;

  //If json_decode failed, the JSON is invalid.
  //if(! is_array($decoded)) { ......
}
header("Cache-Control: no-cache, must-revalidate"); // force reset cache
header("Expires: Mon, 26 Jul 2000 05:00:00 GMT"); 

header('Content-type: application/json');
echo json_encode($ReturnInfo);
exit(0);
?>

将HTML变量转换为PHP变量

问题描述

1 个解决方案

解决方案1
0 2019-04-07 22:00:42

将HTML变量转换为PHP变量

问题描述

1 个解决方案

解决方案1 0 2019-04-07 22:00:42

解决方案1
0 2019-04-07 22:00:42