问题描述

谁能告诉我如何在 PHP 中使用 html <input type="file"> 获取文件路径?

Can somebody pls tell me how to get the filepath using html <input type="file"> in PHP?

这是我的代码:

index.php

<form action="csv_to_database.php" method="get" >
 <input type="file" name="csv_file" />
 <input type="submit" name="upload" value="Upload" />
</form>

和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>
";
    $row++;
    for ($c=0; $c < $num; $c++) {
        echo $data[$c] . "<br />
";
    }
   }
   fclose($handle);
  }

 } 
?>

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.

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.

推荐答案

你不应该只使用你现在拥有的 $_GET.您的文件基于 $_FILES["csv_file"]["tmp_name"].

You shouldn't just use the $_GET you've got now. Your file is based in $_FILES["csv_file"]["tmp_name"].

最好阅读本教程，基本上说你需要做这样的事情:

Best you review this tutorial, that basically says you need to do something like this:

<?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"];
  }
?>

你可以从那里去.如果您想从临时位置移动文件，请使用 move_uploaded_file，教程中也有说明:)

And you can go from there. Use move_uploaded_file if you want to move the file from the temp location, also explained in the tutorial :)

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