我编辑了这段代码很多次(im noob whit php)我的问题是使用此代码上传多个文件。我只能上传1个文件。
以下是代码:
<?php
/**
* uploadFile()
*
* @param string $file_field name of file upload field in html form
* @param bool $check_image check if uploaded file is a valid image
* @param bool $random_name generate random filename for uploaded file
* @return array
*/
function uploadFile ($file_field = null, $check_image = false, $random_name = false) {
//Config Section
//Set file upload path
$path = 'c:/xampp/htdocs/'; //with trailing slash
//Set max file size in bytes
$max_size = 1000000;
//Set default file extension whitelist
$whitelist_ext = array('jpg','png','gif');
//Set default file type whitelist
$whitelist_type = array('image/jpeg', 'image/png','image/gif');
//The Validation
// Create an array to hold any output
$out = array('error'=>null);
if (!$file_field) {
$out['error'][] = "Please specify a valid form field name";
}
if (!$path) {
$out['error'][] = "Please specify a valid upload path";
}
if (count($out['error'])>0) {
return $out;
}
//Make sure that there is a file
if((!empty($_FILES[$file_field])) && ($_FILES[$file_field]['error'] == 0)) {
// Get filename
$file_info = pathinfo($_FILES[$file_field]['name']);
$name = $file_info['filename'];
$ext = $file_info['extension'];
//Check file has the right extension
if (!in_array($ext, $whitelist_ext)) {
$out['error'][] = "Invalid file Extension";
}
//Check that the file is of the right type
if (!in_array($_FILES[$file_field]["type"], $whitelist_type)) {
$out['error'][] = "Invalid file Type";
}
//Check that the file is not too big
if ($_FILES[$file_field]["size"] > $max_size) {
$out['error'][] = "File is too big";
}
//If $check image is set as true
if ($check_image) {
if (!getimagesize($_FILES[$file_field]['tmp_name'])) {
$out['error'][] = "Uploaded file is not a valid image";
}
}
//Create full filename including path
if ($random_name) {
// Generate random filename
$tmp = str_replace(array('.',' '), array('',''), microtime());
if (!$tmp || $tmp == '') {
$out['error'][] = "File must have a name";
}
$newname = $tmp.'.'.$ext;
} else {
$newname = $name.'.'.$ext;
}
//Check if file already exists on server
if (file_exists($path.$newname)) {
$out['error'][] = "A file with this name already exists";
}
if (count($out['error'])>0) {
//The file has not correctly validated
return $out;
}
if (move_uploaded_file($_FILES[$file_field]['tmp_name'], $path.$newname)) {
//Success
$out['filepath'] = $path;
$out['filename'] = $newname;
return $out;
} else {
$out['error'][] = "Server Error!";
}
} else {
$out['error'][] = "No file uploaded";
return $out;
}
}
?>
<?php
if (isset($_POST['submit'])) {
$file = uploadFile('file', true, true);
if (is_array($file['error'])) {
$message = '';
foreach ($file['error'] as $msg) {
$message .= '<p>'.$msg.'</p>';
}
} else {
$message = "File uploaded successfully";
}
echo $message;
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" enctype="multipart/form-data">
<input name="file" type="file" size="20" multiple="multiple" />
<input name="submit" type="submit" value="Upload files" />
</form>
请帮助我理解......我知道我必须使用 foreach
答案 0 :(得分:8)
您需要将文件字段声明为file[]
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" enctype="multipart/form-data">
<input name="file[]" type="file" size="20" multiple="multiple" />
<input name="submit" type="submit" value="Upload files" />
</form>
然后尝试
答案 1 :(得分:1)
正如其他人所指出的那样,您需要将上传字段名称更改为ie。 files [](包括名称后面的sqaure括号)。这是相关的,因为它告诉PHP它应该将字段视为数组。
此外,在代码中,您可以使用foreach()来访问上传的文件,如下所示:
foreach ($_FILES['field_name'] as $file)
(显然,在这种情况下,你的html字段将是名称field_name []) 这将在其五次迭代中的每一次中返回一个数组,为您提供有关您已发送的所有文件的信息。例如,如果您发送了两个文件,它可能如下所示:
["name"]=>
array(2) {
[0]=>
string(5) "dir.c"
[1]=>
string(10) "errcodes.h"
}
["type"]=>
array(2) {
[0]=>
string(11) "text/x-csrc"
[1]=>
string(11) "text/x-chdr"
}
["tmp_name"]=>
array(2) {
[0]=>
string(14) "/tmp/phpP1iz5A"
[1]=>
string(14) "/tmp/phpf31fzn"
}
["error"]=>
array(2) {
[0]=>
int(0)
[1]=>
int(0)
}
["size"]=>
array(2) {
[0]=>
int(511)
[1]=>
int(38)
}
}
重要的是要理解PHP会将那些不归类为文件,然后给每个文件属性,而是列出所有文件的属性。
我希望现在很清楚。
答案 2 :(得分:0)
您必须使用foreach循环上传多个文件。在框架中你也提供了具有相同设施的组件(即通过使用foreach循环)。
答案 3 :(得分:0)
我的建议是
答案 4 :(得分:0)
要成功从您的浏览器发送多个文件,您需要输入将数组传递给PHP。这是通过将[]
附加到<input>
名称的末尾来完成的:
<input type="file" name="filesToUpload[]" multiple>
处理这些文件是棘手的部分。 PHP处理文件上载的方式与处理数组中提供的其他POST或GET数据的方式不同。文件上载在输入的名称和上载的文件的索引之间插入了元数据键。因此$_FILES['filesToUpload']['name'][0]
将获取第一个文件的名称,$_FILES['filesToUpload']['name'][1]
将获取第二个文件的名称......依此类推。
因为 foreach
绝对是错误的循环使用。您最终将自己处理每个元数据,没有任何上下文。 那是非常不自然的。
让我们获取每个文件的索引并一次处理一个文件。我们将使用for
循环。这是一个完全自包含的功能示例,用户将多个文件上载到服务器上的文件夹中:
<?php
/*
* sandbox.php
*/
if (isset($_POST['submit'])) {
// We need to know how many files the user actually uploaded.
$numberOfFilesUploaded = count($_FILES['filesToUpload']['name']);
for ($i = 0; $i < $numberOfFilesUploaded; $i++) {
// Each iteration of this loop contains a single file.
$fileName = $_FILES['filesToUpload']['name'][$i];
$fileTmpName = $_FILES['filesToUpload']['tmp_name'][$i];
$fileSize = $_FILES['filesToUpload']['size'][$i];
$fileError = $_FILES['filesToUpload']['error'][$i];
$fileType = $_FILES['filesToUpload']['type'][$i];
// PHP has saved the uploaded file as a temporary file which PHP will
// delete after the script has ended.
// Let's move the file to an output directory so PHP will not delete it.
move_uploaded_file($fileTmpName, './output/' . $fileName);
}
}
?>
<form method="post" enctype="multipart/form-data">
<!-- adding [] Allows us to upload multiple files -->
<input type="file" name="filesToUpload[]" multiple>
<input type="submit" name="submit"/>Submit
</form>
要运行此示例,您的文件应如下所示
您可以使用以下命令启动PHP内置Web服务器:
$ php -S localhost:8000
然后转到http://localhost:8000/sandbox.php将运行示例。
重要说明:以上示例未进行任何验证。您需要验证所有上传的文件是否安全。