Question

所以基本上我的主html页面有3个按钮，每个按钮都有链接的不同背景图像。我试图通过ajax发布将该图片的源传递到我的Php文件中，以写入txt文件。

我的JS：

// MAIN
$(function() {

$(".artButton").click(function(){
    var artId = $(this).attr("id");
    var source = $("#"+artId).css("background-image").replace(/^url\(['"](.+)['"]\)/, '$1');


    // DEBUG
    console.log(source) //example: http://localhost/crowd.jpg
    console.log(typeof source)//example: string


$.ajax({  
            type: 'POST',  
            url: 'writeToTxt.php', 
            data: source,
            success: function(response) {
            console.log("Succes");
    }
});
});
});

我的Php文件：

<?php


$my_file = 'toExpose.txt';
$handle = fopen($my_file, 'w') or die('Cannot open file:  '.$my_file);
$data = $_POST['data'];
fwrite($handle, $data);

?>

Chrome控制台导致以下错误： 405 Ajax Post error

请注意，我不是编程专家...更多关于“我需要一些东西，所以我想自己编写”的更多信息。就编码而言，很多方法可能不是最佳实践，但我只需要它即可按预期工作！非常感谢您的帮助！

注意：一切都在localhost上运行

更新：这是页面的HTML：

<html>

<link href="style.css" rel="stylesheet" type="text/css"/>
 <!-- Required meta tags -->
    <meta charset="utf-8">
    <meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">

    <!-- Bootstrap CSS -->
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css" integrity="sha384-Gn5384xqQ1aoWXA+058RXPxPg6fy4IWvTNh0E263XmFcJlSAwiGgFAW/dAiS6JXm" crossorigin="anonymous">


<body>
<header class="navbar">
<div class="container">

<h1>Welcome to oneFRAME</h1>

<nav class="navbar navbar-expand-lg navbar-dark bg-dark">

  <button class="navbar-toggler" type="button" data-toggle="collapse" data-target="#navbarSupportedContent" aria-controls="navbarSupportedContent" aria-expanded="false" aria-label="Toggle navigation">
    <span class="navbar-toggler-icon"></span>
  </button>

  <div class="collapse navbar-collapse" id="navbarSupportedContent">
    <ul class="navbar-nav mr-auto">
      <li class="nav-item">
        <a class="nav-link" href="Home.html">Home <span class="sr-only">(current)</span></a>
      </li>
      <li class="nav-item active">
        <a class="nav-link" href="#">Library</a>
      </li>

      <li class="nav-item">
        <a class="nav-link" href="LibraryV2.html">Library v2</a>
      </li>

      <li class="nav-item dropdown">
        <a class="nav-link dropdown-toggle" href="#" id="navbarDropdown" role="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false">
          Dropdown
        </a>
        <div class="dropdown-menu" aria-labelledby="navbarDropdown">
          <a class="dropdown-item" href="#">About</a>
          <div class="dropdown-divider"></div>
          <a class="dropdown-item" href="#">Contact</a>
        </div>
      </li>

    </ul>
    <form class="form-inline my-2 my-lg-0">
      <input class="form-control mr-sm-2" type="search" placeholder="Search" aria-label="Search">
      <button class="btn btn-outline-success my-2 my-sm-0" type="submit">Search</button>
    </form>
  </div>
</nav>
</div>
</header>



    <button id="choice1" class="artButton" type="button">Nature</button>
    <button id="choice2" class="artButton" type="button">Art</button>
    <button id="choice3" class="artButton" type="button">Crowd</button>


<script src="jquery-3.3.1.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.12.9/umd/popper.min.js" integrity="sha384-ApNbgh9B+Y1QKtv3Rn7W3mgPxhU9K/ScQsAP7hUibX39j7fakFPskvXusvfa0b4Q" crossorigin="anonymous"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/js/bootstrap.min.js" integrity="sha384-JZR6Spejh4U02d8jOt6vLEHfe/JQGiRRSQQxSfFWpi1MquVdAyjUar5+76PVCmYl" crossorigin="anonymous"></script>
<script src="main_temp.js"></script>

</body>

</html>

Answer 1

HTTP 405错误表示您的脚本和/或Web服务器存在问题。该错误表示您正在尝试回传一个在服务器上没有必需权限的文件，以从另一个脚本接收Post信息。

Answer 2

尝试在

之类的对象中传递数据

$.ajax({  
            type: 'POST',  
            url: 'writeToTxt.php', 
            data: {data: source},
            success: function(response) {
            console.log("Succes");
    }

Answer 3

由于在htdocs或www目录中没有写许可权，因此出现错误。

使用sudo chmod -R 777 /directory
将ajax方法POST更改为GET（有时无法使用）。

不确定，但我怀疑您的路径一定是http://localhost/example/index.php

更新的代码：

<!DOCTYPE html>
<html>
<head>
    <title></title>
    <script type="text/javascript" src="jquery.min.js"></script>

</head>
<body>

    <button id="choice1" class="artButton" type="button" style="background-image:url('http://localhost/Veridocglobalportal2/veridocedu/public/img/edu-logo.png');">Nature</button>
    <button id="choice2" class="artButton" type="button">Art</button>
    <button id="choice3" class="artButton" type="button">Crowd</button>
    <script type="text/javascript">
        // MAIN
$(function() {

$(".artButton").click(function(){
    var artId = $(this).attr("id");
     var source = $("#"+artId);
switch (artId) {
    case "choice1":
        var source = "http://localhost/crowd.jpg";
        break;
    case "choice2":
        var source = "http://localhost/people.jpg";
        break;
    case "choice3":
        var source = "http://localhost/alone.jpg";
        break;
 }


// DEBUG
console.log(source) //example: http://localhost/crowd.jpg
console.log(typeof source)//example: string

$.ajax({  
            type: 'POST',  
            url: 'writeToTxt.php', 
            data: {"data":source},
            success: function(response) {
            console.log("Succes");
    }
});
});
});
    </script>

</body>
</html>

writeTotxt.php

   <?php
$cwd=getcwd();
$data = $_POST['data'];
$myfile = file_put_contents("$cwd/logs.txt", $data.PHP_EOL,FILE_APPEND | LOCK_EX );

Answer 4

尝试一下

$(function() {

$(".artButton").click(function(){
  var artId = $(this).attr("id");
  var source = $("#"+artId).css("background-image").replace(/^url\(['"](.+)['"]\)/, '$1');


// DEBUG
console.log(source) //example: http://localhost/crowd.jpg
console.log(typeof source)//example: string


$.ajax({  
        type: 'POST',  
        url: 'writeToTxt.php', 
        data: {source:source},
        success: function(response) {
          console.log("Succes");
        }
     });
  });
});

Answer 5

您遇到客户端错误。 4.x.x都是关于客户端错误的。

这样更改您的ajax请求-

$.ajax({  
                type: 'POST',  
                url: 'writeToTxt.php', 
                data: 'name=xxx&address=xxxx',
                success: function(response) {
                console.log("Succes");
        }
    });

用您的帖子数据替换'name = xxx＆address = xxxx'。

，在PHP中使用如下方式- $ name = $ _POST ['name']; $ adress = $ _POST ['address'];

405 Ajax POST错误

5 个答案: