php 不上传任何内容时，$_FILES 数组不为空
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$_FILES array is not empty upon uploading nothing
提问by user1672267
Here is the form
这是表格
form action="index.php" method="POST" enctype="multipart/form-data" >
<input type="file" name="image[]" multiple="multiple">
<input type="submit" value="upload">
</form>
I am trying to only run my code when only when if(!empty($_FILES['image'])){but for some reason the array is not empty upon submitting no files and only clicking submit.
我试图仅在仅当if(!empty($_FILES['image'])){但由于某种原因在不提交文件且仅单击提交时数组不为空时才运行我的代码。
Here is the rest of the code if that will help, thanks.
如果有帮助，这是其余的代码，谢谢。
<html>
Image Upload
图片上传
<form action="index.php" method="POST" enctype="multipart/form-data" >
<input type="file" name="image[]" multiple="multiple">
<input type="submit" value="upload">
</form>
<?php
include 'connect.php';
if(!empty($_FILES['image'])){
echo $_FILES['image']['error'];
$allowed = array('jpg', 'gif', 'png', 'jpeg');
$count = 0;
foreach($_FILES['image']['name'] as $key => $name){
$image_name = $name;
$tmp = explode('.', $image_name);
$image_extn = strtolower(end($tmp)); //can only reference file
$image_temp = $_FILES['image']['tmp_name'][$count];
$filesize = filesize($_FILES['image']['tmp_name'][$count]);
$count = $count +1;
if(count($_FILES['image']['tmp_name']) > 5){
echo "You can upload a maximum of five files.";
break;
}
else if(in_array($image_extn, $allowed) === false){
echo $name." is not an allowed file type<p></p>";
}
else if($filesize > 1024*1024*0.3){
echo $name." is too big, can be a maximum of 3MB";
}
else{
$image_path = 'images/' . substr(md5($name), 0, 10) . '.' . $image_extn;
move_uploaded_file($image_temp, $image_path);
mysql_query("INSERT INTO store VALUES ('', '$image_name', '$image_path')") or die(mysql_error());
$lastid = mysql_insert_id();
$image_link = mysql_query("SELECT * FROM store WHERE id = $lastid");
$image_link = mysql_fetch_assoc($image_link);
$image_link = $image_link['image'];
echo "Image uploaded.<p></p> Your image: <p></p><a href = $image_link>$image_path</a>";
}
}
}
else{
echo "Please select an image.";
}
?>
采纳答案by Oliver Tappin
Use the is_uploaded_filePHP function instead:
改用is_uploaded_filePHP 函数：
if(is_uploaded_file($_FILES['image']['tmp_name'])) {
//code here
}
回答by The Alpha
This is how $_FILESarray looks like when nothing uploaded
这是未$_FILES上传任何内容时数组的样子
Array ( [image] => Array ( [name] => [type] => [tmp_name] => [error] => 4 [size] => 0 ) )
So it's never empty.
所以它永远不会是空的。
The error code 4 [error] => 4indicates no file was uploaded and error code 0indicates no error and file was uploaded so you can check
错误代码 4[error] => 4表示没有上传文件，错误code 0表示没有错误并且文件已上传，因此您可以检查
if($_FILES['image']['error']==0) {
// file uploaded, process code here
}
Here is another answeron SO.
这是关于 SO 的另一个答案。
回答by hakre
You should first of all take a look into the PHP manual because - you're not the first one with that problem - the solution has been written in there:
您应该首先查看 PHP 手册，因为-您不是第一个遇到该问题的人-解决方案已写在那里：
If no file is selected for upload in your form, PHP will return
$_FILES['userfile']['size']as 0, and$_FILES['userfile']['tmp_name']as none.
如果您的表单中没有选择要上传的文件，PHP 将返回
$_FILES['userfile']['size']0 和$_FILES['userfile']['tmp_name']无。
So if you actually want to find out if any file for the imageelement has has been submitted, check for it:
因此，如果您确实想知道image是否已提交该元素的任何文件，请检查它：
$noFile = $_FILES['image']['size'][0] === 0
&& $_FILES['image']['tmp_name'][0] === '';
Yes, that simple it is.
是的，就是这么简单。
The test you used:
您使用的测试：
empty($_FILE);
will only tell you if the whole form has been submitted or not. So an example in full:
只会告诉您整个表格是否已提交。所以一个完整的例子：
$submitted = empty($_FILE);
if ($submitted) {
$noFile = $_FILES['image']['size'][0] === 0
&& $_FILES['image']['tmp_name'][0] === '';
if ($noFile) {
...
}
}
回答by Pavnish Yadav
Check value is not null:
检查值不为空：
in_array(!null,$_FILES['field_name']['name'])
回答by rarenicks
if($_FILES['image']['error'] === UPLOAD_ERR_OK) {
// Success code Goes here ..
}
UPLOAD_ERR_OKreturns value 0 if there is no error, the file was uploaded successfully.
UPLOAD_ERR_OK如果没有错误，则返回值 0，文件上传成功。
回答by Calin Rusu
I have confronted this issue with a multiple file input. What I found to be working for checking if any file has been selected is:
我在多文件输入中遇到了这个问题。我发现用于检查是否选择了任何文件的是：
<?php
$size_sum = array_sum($_FILES['img']['size']);
if ($size_sum > 0) {
// at least one file has been uploaded
} else {
// no file has been uploaded
}
?>
回答by jimp
http://php.net/manual/en/features.file-upload.post-method.php
http://php.net/manual/en/features.file-upload.post-method.php
If no file is selected for upload in your form, PHP will return $_FILES['userfile']['size'] as 0, and $_FILES['userfile']['tmp_name'] as none.
如果表单中没有选择要上传的文件，PHP 将返回 $_FILES['userfile']['size'] 为 0，而 $_FILES['userfile']['tmp_name'] 为 none。
You get an array entry per "file" upload field, even if the user didn't select a file to upload.
即使用户没有选择要上传的文件，您也会为每个“文件”上传字段获得一个数组条目。
回答by Eng Hany Atwa
if(!empty($_FILES['image']['name'][0]) || !empty($_FILES['image']['name'][1]) ||
!empty($_FILES['image']['name'][2]))
or
或者
for(=0;$i<count($_FILES['image']);$i++){
if(!empty($_FILES['image']['name'][$i])){
// do something
}
}
回答by Badshah Sahib
if(isset($_FILES['file'])){//do some thing here}
