php文件上傳失敗
[英]php file upload failed
問題描述
我按照互聯網上的每個教程上傳了文件。 但是，它失敗了。 它給了我這個錯誤：
Warning: move_uploaded_file(/var/www/projects/upload/TASK.txt): failed to open stream: No such file or directory in /var/www/projects/test/upload.php on line 6 Warning: move_uploaded_file(): Unable to move '/tmp/phpjr2JJA' to '/var/www/projects/upload/TASK.txt' in /var/www/projects/test/upload.php on line 6 Something went wrong
的index.html
<head>
<title></title>
</head>
<body>
<form method="POST" action="upload.php" enctype="multipart/form-data">
<input type="file" name="upload" ><br />
<input type="hidden" name="MAX_FILE_SIZE" value="1024" />
<input type="submit" name="submit" value="Submit" />
</form>
</body>
</html>
upload.php的
<?php
$target_path = $_SERVER['DOCUMENT_ROOT'] . "/upload/";
$target_path = $target_path . basename( $_FILES['upload']['name'] );
if ( move_uploaded_file($_FILES['upload']['tmp_name'], $target_path) ) {
echo "has been uploaded";
} else {
echo "Something went wrong";
}
您能幫我指出我出了什么問題嗎？ 我正在使用ubuntu 12.04，也嘗試將/upload文件夾的權限更改為755並檢查php.ini的file_upload是否為ON
任何幫助將非常感激。 謝謝！
3 個解決方案
解決方案1
1 已采納 2014-01-11 16:26:47
$ _SERVER ['DOCUMENT_ROOT']給出/ var / www / projects / upload ..作為輸出/開頭是錯誤原因
於是
嘗試相對路徑
<?php
$target_path = "upload/";
$target_path = $target_path . basename( $_FILES['upload']['name'] );
if ( move_uploaded_file($_FILES['upload']['tmp_name'], $target_path) ) {
echo "has been uploaded";
} else {
echo "Something went wrong";
}
?>
為我工作
解決方案2
0 2014-01-11 15:56:54
文件和目錄權限...
將FTP程序666添加到文件中，然后輸入需要編寫的內容。
解決方案3
0 2014-01-11 16:08:56
我認為您的文件上傳位置不正確
的index.php
<html>
<head>
<title>Upload your file</title>
</head>
<body>
<form action="upload.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="Upload" value="Submit">
</form>
</body>
upload.php的
<?php
$target="give the path you want to store the file/";
if ($_FILES["file"]["error"] > 0)
{
echo "Error: " . $_FILES["file"]["error"] . "<br>";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
$target. $_FILES["file"]["name"]);
}
?>
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