php 简单的文件上传脚本
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simple file upload script
提问by Ahmad
I have written a simple file upload script but it gives me the error of undefined index file1.
我写了一个简单的文件上传脚本，但它给了我未定义索引文件1的错误。
<html>
<body>
<form method="post">
<label for="file">Filename:</label>
<input type="file" name="file1" id="file1" />
<br />
<input type="submit" name="submit" value="Submit" />
</form>
</body>
</html>
<?php
if(isset($_POST['submit'])) {
if ($_FILES["file1"]["error"] > 0) {
echo "Error: " . $_FILES["file1"]["error"] . "<br />";
} else {
echo "Upload: " . $_FILES["file1"]["name"] . "<br />";
echo "Type: " . $_FILES["file1"]["type"] . "<br />";
echo "Size: " . ($_FILES["file1"]["size"] / 1024) . " Kb<br />";
echo "Stored in: " . $_FILES["file1"]["tmp_name"];
}
}
?>
What is the problem in code?
代码中有什么问题？
回答by Wesley van Opdorp
You lack enctype="multipart/form-data"in your <form>element.
你缺乏enctype="multipart/form-data"你的<form>元素。
回答by T.Todua
Another solution for simple php file upload scriptis here :
(make a yourfile.phpand insert the below code. then put that yourfile.php on your website)
另一个简单的 php 文件上传脚本的解决方案在这里：（
制作一个yourfile.php并插入下面的代码。然后把 yourfile.php 放在你的网站上）
<?php
$pass = "YOUR_PASSWORD";
session_start();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html><head><meta http-equiv="Content-Type" content="text/html; charset=windows-1256" /></head><body>
<?php
if (!empty($_GET['action']) && $_GET['action'] == "logout") {session_destroy();unset ($_SESSION['pass']);}
$path_name = pathinfo($_SERVER['PHP_SELF']);
$this_script = $path_name['basename'];
if (empty($_SESSION['pass'])) {$_SESSION['pass']='';}
if (empty($_POST['pass'])) {$_POST['pass']='';}
if ( $_SESSION['pass']!== $pass)
{
if ($_POST['pass'] == $pass) {$_SESSION['pass'] = $pass; }
else
{
echo '<form action="'.$_SERVER['PHP_SELF'].'" method="post"><input name="pass" type="password"><input type="submit"></form>';
exit;
}
}
?>
<form enctype="multipart/form-data" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
Please choose a file: <input name="file" type="file" /><br />
<input type="submit" value="Upload" /></form>
<?php
if (!empty($_FILES["file"]))
{
if ($_FILES["file"]["error"] > 0)
{echo "Error: " . $_FILES["file"]["error"] . "<br>";}
else
{echo "Stored file:".$_FILES["file"]["name"]."<br/>Size:".($_FILES["file"]["size"]/1024)." kB<br/>";
move_uploaded_file($_FILES["file"]["tmp_name"],$_FILES["file"]["name"]);
}
}
// open this directory
$myDirectory = opendir(".");
// get each entry
while($entryName = readdir($myDirectory)) {$dirArray[] = $entryName;} closedir($myDirectory);
$indexCount = count($dirArray);
echo "$indexCount files<br/>";
sort($dirArray);
echo "<TABLE border=1 cellpadding=5 cellspacing=0 class=whitelinks><TR><TH>Filename</TH><th>Filetype</th><th>Filesize</th></TR>\n";
for($index=0; $index < $indexCount; $index++)
{
if (substr("$dirArray[$index]", 0, 1) != ".")
{
echo "<TR>
<td><a href=\"$dirArray[$index]\">$dirArray[$index]</a></td>
<td>".filetype($dirArray[$index])."</td>
<td>".filesize($dirArray[$index])."</td>
</TR>";
}
}
echo "</TABLE>";
?>
回答by Ibrahim Azhar Armar
Make the following changes and try.
进行以下更改并尝试。
<form method="post" action="" enctype="multipart/form-data" >
回答by arunwebber
Html
html
<!DOCTYPE html>
<html>
<body>
<form action="upload.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
</body>
</html>
Php
php
<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "File is an image - " . $check["mime"] . ".";
$uploadOk = 1;
} else {
echo "File is not an image.";
$uploadOk = 0;
}
}
?>
回答by jafar
<html>
<body>
<form action="" method="post" ectype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"></br>
<input type="submit" name="submit" value="Submit">
</form>
<?php
if(isset($_POST['submit']))
{
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 20000000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("upload/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
}
}
}
else
{
echo "Invalid file";
}enter code here
}
?>
</body>
</html>
回答by Kiran
Primary issue is your form does not have option to send file content over http . To send binary data along with the text data from input elements you need to add an extra attribute in form tag .
主要问题是您的表单没有通过 http 发送文件内容的选项。要将二进制数据与来自输入元素的文本数据一起发送，您需要在表单标签中添加一个额外的属性。
<form method="post" enctype="multipart/form-data">
Then in php code
然后在php代码中
try this line
试试这条线
<?php
print_r($_FILES);
?>
above code will display all information regarding file uploading from your form .
上面的代码将显示有关从表单上传文件的所有信息。
