我有一个表单,我可以提交文本文件,我希望它将这些文本文件添加到另一个现有的表单

时间:2017-04-19 01:30:41

标签: php html forms

我有一个用于上传文本文件的html表单:

      <div class="form">
        <h3>Upload File:</h3>
        <form action="networkSelector.php" method="post" enctype="multipart/form-data" name="FileUploadForm" id="FileUploadForm">
          <label for="UploadFileField"></label>
          <input type="file" name="UploadFileField" id="UploadFileField" />
          <input type="submit" name="UploadButton" id="UploadButton" value="Upload" />
        </form>
      </div>

表单代码的php部分:

   <?php
require('db.php');
include("auth.php");

if(isset($_FILES['UploadFileField'])){
    // Creates the Variables needed to upload the file
    $UploadName = $_FILES['UploadFileField']['name'];
    $UploadName = mt_rand(100000, 999999).$UploadName;
    $UploadTmp = $_FILES['UploadFileField']['tmp_name'];
    $UploadType = $_FILES['UploadFileField']['type'];
    $FileSize = $_FILES['UploadFileField']['size'];

    // Removes Unwanted Spaces and characters from the files names of the files being uploaded
    $UploadName = preg_replace("#[^a-z0-9.]#i", "", $UploadName);
    // Upload File Size Limit
    if(($FileSize > 125000)){

        die("Error - File too Big");

    }
    // Checks a File has been Selected and Uploads them into a Directory on your Server
    if(!$UploadTmp){
        die("No File Selected, Please Upload Again");
    }else{
        move_uploaded_file($UploadTmp, "C:/xampp/htdocs/meg/$UploadName");
    }

}

?>

它运行良好,如'move_upload_file命令所示,它将它们直接放入该目录。

然而,我想要实现的是使用此表单上传这些文件,然后将其添加到同一页面上的另一个表单。

以下是我的另一种形式的示例:

 <form action="networkCompiler.php" method="POST">
            <h3>Choose Network/Function:</h3>
            <select id ="network" name="network" />
             <option value="networkA">A</option>
             <option value="networkB">B</option>
            </select>

理想情况下,如果我在第一个表单上上传networkC,我希望它然后显示在第二个表单上。我主要在这个项目上使用PHP,并试图找到该语言的解决方案。到目前为止,我已尝试将文件上传保存为变量,然后将其添加到表单的底部。

    <?php 
       if (isset($_POST['UploadButton'])) {
          if (is_uploaded_file($_FILES['UploadFileField']['tmp_name'])) {
             $trying = $_POST['FileUploadForm'];
          }
       } 
     ?>

任何输入都将不胜感激。谢谢。

1 个答案:

答案 0 :(得分:0)

使用file_get_contents()

<?php
    if(your_condition)
        echo '<option value="the_value_you_want">' . file_get_contents('path_of_uploaded_file') . '</option>';
?>