php 如何检查codeigniter中的文件字段是否为空？
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How to check if file field is empty in codeigniter?
提问by squiter
I have a form with a file field called image, but this field is not required.
When user don't choose any file in form, the do_upload()always return a error.
How can I check if user chosen a file before perform the upload action in my controller?
我有一个名为 的文件字段的表单image，但该字段不是必需的。
当用户没有在表单中选择任何文件时，do_upload()总是返回一个错误。
在我的控制器中执行上传操作之前，如何检查用户是否选择了文件？
回答by Vinoth Babu
Please use empty()
请使用空（）
if (empty($_FILES['userfile']['name'])) {
}
回答by Matanya
In your controller, on the function that receives the submitted form:
在您的控制器中，在接收提交表单的函数上：
if (isset($_FILES['image']['name']) && !empty($_FILES['image']['name'])) {
// do_upload
}
回答by BenM
Try to check if the file is valid using is_uploaded_file(). For example:
尝试使用is_uploaded_file(). 例如：
if(is_uploaded_file($_FILES['userfile']['tmp_name']))
{
do_upload();
}
回答by shiv
Here is full script to check if file field is empty or not in php
这是检查 php 中文件字段是否为空的完整脚本
<!DOCTYPE html>
<html>
<body>
<form action="#" method="post" enctype="multipart/form-data">
Select image to upload:
<input name="my_files[]" type="file" multiple="multiple" />
<input type="submit" value="Upload Image" name="submit">
</form>
<?php
if (isset($_FILES['my_files']))
{
$myFile = $_FILES['my_files'];
$fileCount = count($myFile["name"]);
for ($i = 0; $i <$fileCount; $i++)
{
$error = $myFile["error"][$i];
if ($error == '4') // error 4 is for "no file selected"
{
echo "no file selected";
}
else
{
$name = $myFile["name"][$i];
echo $name;
echo "<br>";
$temporary_file = $myFile["tmp_name"][$i];
echo $temporary_file;
echo "<br>";
$type = $myFile["type"][$i];
echo $type;
echo "<br>";
$size = $myFile["size"][$i];
echo $size;
echo "<br>";
$target_path = "uploads/$name"; //first make a folder named "uploads" where you will upload files
if(move_uploaded_file($temporary_file,$target_path))
{
echo " uploaded";
echo "<br>";
echo "<br>";
}
else
{
echo "no upload ";
}
}
}
}
?>
</body>
</html>
But be alert. User can upload any type of file and also can hack your server or system by uploading a malicious or php file. In this script there should be some validations.
但要警惕。用户可以上传任何类型的文件，也可以通过上传恶意或 php 文件来入侵您的服务器或系统。在这个脚本中应该有一些验证。
refer http://www.techzigzag.com/how-to-check-that-user-has-upload-any-file-or-not-in-php/
参考http://www.techzigzag.com/how-to-check-that-user-has-upload-any-file-or-not-in-php/
Hope it will help you.
希望它会帮助你。
回答by jones
Just use native php code to check file upload.
只需使用本机 php 代码来检查文件上传。
if(!file_exists($_FILES['myfile']['tmp_name']) || !is_uploaded_file($_FILES['myfile']['tmp_name'])) {
echo 'No upload';
}
回答by prasanthnv
if(!empty($_FILES[$file_name]['name'])){
// TODO your logic
}else{
echo "empty";
}
回答by Rauli Rajande
As file upload error "No file selected" is number 4, correct way of doing this is:
由于文件上传错误“未选择文件”是数字 4，正确的做法是：
if ($_FILES['my_image_field_name']['error'] !== 4){
if ($this->upload->do_upload('my_image_field_name')) { ...
When checking by name or tmp_name, there might be other reasons why these fields didn't get populated, and you may miss these.
在按名称或 tmp_name 检查时，可能有其他原因导致这些字段未填充，您可能会错过这些。
