0

我制作了一个动态更改的HTML表单，我可以在文件上传或从远程数据库填充的下拉框之间更改表单的输入类型。调用要显示的输入类型的函数在AJAX. 当我使用下拉框提交表单时，表单提交没有任何问题，并对我编写的数据库执行更新。但是，当我在尝试上传文件时尝试提交表单时，我的错误检查脚本（很简单if... statements) 告诉我它都没有检测到任何正在上传的文件，并且该文件已经存在于数据库中。但是，当出现“数据库上已存在”错误时，它不会像我编程的那样返回我尝试上传的文件的名称，所以我怀疑我的文件没有正确提交。

有人可以告诉我我做错了什么吗？

这是我到目前为止的脚本：

文件 1：测试.php

    <html>
    <body>
    <head>
    <script>
    function getInput(value)
    {
    var xmlhttp;
    if (value=="")
      {
      document.getElementById("display").innerHTML="";
      return;
      } 
    if (window.XMLHttpRequest)
      {// code for IE7+, Firefox, Chrome, Opera, Safari
      xmlhttp=new XMLHttpRequest();
      }
    else
      {// code for IE6, IE5
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
    xmlhttp.onreadystatechange=function()
      {
      if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
        document.getElementById("display").innerHTML=xmlhttp.responseText;
        }
      }
    xmlhttp.open("GET","grabtest.php?q="+value,true);
    xmlhttp.send();
    }
  </script>
    <?php
    // connect to database on server
    $con=mysqli_connect("localhost","username","password","database name");
    // if there was an error in connecting to the database, display the error
    if (mysqli_connect_errno($con))
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    ?>
</head>
<form enctype="multipart/form-data" action="test2.php" method="POST">
   <select id='SelectInput' onchange='getInput(this.value)'>
      <option selected value=''>Select</option>
      <option value='N'>New File</option>
      <option value='E'>Existing File</option>
   </select>
<div id="display"></div><br>
<input type="submit" value="Submit">
</form>

<?php
if (!empty($_POST)){
   if ($_FILES["file"]["error"] == 0){
      $target = 'PathName/TargetFolder/';
      $target = $target . basename($FILES['file']['name']);
      if (file_exists('PathName/TargetFolder/' . $_FILES["file"]["name"])){
         echo $_FILES["file"]["name"] . " already exists on server. ";
      }
      else{
        $upload = $_FILES["file"]["name"];
        $select = mysqli_query($con, "Select Files from DB_Table where Files = '$upload'");
        if (mysqli_num_rows($select) > 0){
        echo $_FILES["file"]["name"] . " already exists in database. ";
        }
        else{
        //script for moving the uploaded file to the proper storage location on the server and adding it to the database
            move_uploaded_file($_FILES["file"]["tmp_name"], $target . $_FILES["file"]["name"]);
        echo "Stored in: " . $target . $_FILES["file"]["name"] . "<br>";
            $insert="INSERT INTO DB_Table (Files) VALUES ('$upload')";
            if (!mysqli_query($con,$insert)){
            die('Error: ' . mysqli_error($con));
        }
        echo "Your data has been added to the database";
        }
    }
    if ($_POST['recipe']=="" and !file_exists($_FILES['file']['tmp_name']) and !is_uploaded_file($_FILES['file']['tmp_name'])){
      exit("Please select or add the file.");
    }
}
mysqli_close($con);
?>

文件 2： grabtest.php

<?php
//connect to database on server
$con=mysqli_connect("localhost","username","password","database name");
//if there was an error in connecting to the database, display the error
if (mysqli_connect_errno($con))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$q=$_GET["q"];

if ($q==""){
    echo "";
}

elseif ($q=="N"){
    echo "Select recipe to upload: <input type='file' name='newfile'>";
}

elseif ($q=="E"){
    //creates a dropdown box where you can select desired field
    $list = mysqli_query($con, "select * from DB_Table");
    echo 'Files: <select name = "Files">';
    while ($row = mysqli_fetch_array($list))
    {
        echo '<option value = "' . $row["ID"] . '">' . $row["Files"] . '</option>';
    }
    echo '</select><br>';
    echo '</form>';
}



//after script is executed, close connection to database
//this improves security by ensuring the connection to the database does not remain open when there is no activity being done to change the data
mysqli_close($con);
?>
4

0 回答 0