php 如何在PHP中获取html <input type="file">中的文件路径？
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How to get the file path in html <input type="file"> in PHP?
提问by Newbie Coder
Can somebody pls tell me how to get the filepath using html <input type="file">in PHP?
有人可以告诉我如何<input type="file">在 PHP 中使用 html 获取文件路径吗？
Here are my codes:
这是我的代码：
index.php
index.php
<form action="csv_to_database.php" method="get" >
<input type="file" name="csv_file" />
<input type="submit" name="upload" value="Upload" />
</form>
and in csv_to_database.php
并在 csv_to_database.php
<?php
if (isset($_GET['csv_file'])) {
$row = 1;
if (($handle = fopen($_GET['csv_file'], "r")) !== FALSE) {
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$num = count($data);
echo "<p> $num fields in line $row: <br /></p>\n";
$row++;
for ($c=0; $c < $num; $c++) {
echo $data[$c] . "<br />\n";
}
}
fclose($handle);
}
}
?>
My problem is, it only works when the csv data is in the same directory as my php files. I think I need to get the file path but I don't know how to do it.
我的问题是，它仅在 csv 数据与我的 php 文件位于同一目录中时才有效。我想我需要获取文件路径，但我不知道该怎么做。
回答by Nanne
You shouldn't just use the $_GETyou've got now. Your file is based in $_FILES["csv_file"]["tmp_name"].
你不应该只使用$_GET你现在拥有的。您的文件基于$_FILES["csv_file"]["tmp_name"].
Best you review this tutorial, that basically says you need to do something like this:
最好阅读本教程，基本上说你需要做这样的事情：
<?php
if ($_FILES["csv_file"]["error"] > 0)
{
echo "Error: " . $_FILES["csv_file"]["error"] . "<br />";
}
else
{
echo "Upload: " . $_FILES["csv_file"]["name"] . "<br />";
echo "Type: " . $_FILES["csv_file"]["type"] . "<br />";
echo "Size: " . ($_FILES["csv_file"]["size"] / 1024) . " Kb<br />";
echo "Stored in: " . $_FILES["csv_file"]["tmp_name"];
}
?>
And you can go from there. Use move_uploaded_fileif you want to move the file from the temp location, also explained in the tutorial :)
你可以从那里去。move_uploaded_file如果您想从临时位置移动文件，请使用，教程中也有说明:)
回答by Kokos
I think you would gain a lot from taking a look at the following link: POST method uploads.
我认为您会从以下链接中获益良多：POST 方法上传。
First of all, you should change your form method to post, and add enctype="multipart/form-data".
首先，您应该将表单方法更改为post，并添加enctype="multipart/form-data".
Then you can get the temporary file path from $_FILES['csv_file']['tmp_name'].
然后你可以从$_FILES['csv_file']['tmp_name'].
回答by David Fells
In your call to fopen, use $_GET['csv_file']['tmp_name']- this points to the file on the server immediately after upload.
在调用 fopen 时，使用$_GET['csv_file']['tmp_name']- 这会在上传后立即指向服务器上的文件。
