Question

我正在用一个简单的HTML格式的html工作。我正在添加一些字段，包括文件上传。

但我面临一个奇怪的问题。当我上传图片并提交表格时。它提交但是当我不上传图像并提交表格时。它说“不支持的文件格式”

当我不上传文件并提交表单时，我检查了一下。它甚至没有发布表格。只有“不支持的文件格式”行进入页面并且整页空白。

这是我的代码：

<form action="" method="post" enctype="multipart/form-data">
    <table class="form-table">
      <tr>
        <th>Title<font color="#ff0000">*</font></th>
        <td><input name="title" type="text" value="<?=$_POST['title']?>" size="40" /></td>
      </tr>
      <tr>
        <th>Image<font color="#ff0000">*</font></th>
        <td><input type="file" name="file_name" /></td>
      </tr>
      <tr>
        <th>&nbsp;</th>
        <td> Dimensions: <?=$imgwidth?> x <?=$imgheight?> (Max: 2MB)&nbsp;<br />
          JPG format is the one recommended.</td>
      </tr>
 <tr>
        <th></th>
        <td><input type="submit" name="btnAdd_cat" class="button" value="Add" /></td>
      </tr>
    </table>
  </form>

Php代码：

<?php

if(isset($_POST['btnAdd_cat'])){


$error = "";

    $title      =   addslashes($_POST['title']);
    if(empty($title)) $error .= "Please enter title.<br/>";

    if(empty($error)){
    $sql        =   "INSERT INTO ".CATEGORIES." (`title`, `status`) VALUES ('$title', '1')";
    mysql_query($sql) or die(__LINE__.mysql_error());
    $id = $insert_id = mysql_insert_id();
    $success    = "Successfuly added.<br/>";

     $filename = $_FILES['file_name']['name'];

        if(!empty($filename)){
        $imgext = strtolower(substr($filename, -4));

        $img = ereg_replace("[^a-z0-9._]", "",str_replace(" ", "-",str_replace("%20", "-", strtolower($title))));


        $filename = "category-".$insert_id."-".$img.$imgext;
        $savefile = "../pictures/".$filename;
        //upload  
        if(copy($_FILES['file_name']['tmp_name'], $savefile)){
            //echo "....Image uploaded ";
        }else{$warning = "Failed to upload image!<br/>";}
        chmod("$savefile",0777);

        if(resize_picture("$savefile","$savefile","$imgwidth","$imgheight")){
        //echo "....Image resized ";
        }else{$warning = "Failed to resize image!<br/>";}
            $image = $filename;
        }

     if(mysql_query("UPDATE ".CATEGORIES." SET image='".$image."' WHERE id='".$id."'")){
        $success .= "Image added.<br/>";
        unset($_GET);
     } else {die(__LINE__.mysql_error());}
}

}

?>

我提交的页面没有上传文件：

http://prntscr.com/706ght

请帮助我。

由于

Answer 1

将文件上传代码块放在if(isset($_FILES['file_name'])){}中，即检查文件是否已过帐。因为正如您所说的那样，当您没有选择任何文件时发生错误，因此最好在运行上传代码之前检查文件是否已发布。

希望这有助于解决您的问题。

Answer 2

正如您所说，当您没有选择任何文件时，它会向您显示错误，那么您需要更新if条件

$filename = $_FILES['file_name']['name'];

if(!empty($filename))

到

$filename = $_FILES['file_name']['error'];

if($filename != 4) // Check no file is uploaded

php文档中有一节关于file handling。您会发现可以从file-upload-errors

检查各种错误

UPLOAD_ERR_OK
    Value: 0; There is no error, the file uploaded with success.
<...>
UPLOAD_ERR_NO_FILE
    Value: 4; No file was uploaded.

表格中的奇怪问题

2 个答案: