PHOP标头 - 从PHP脚本打开CSV文件

Question

I have the below code to open a CSV file, but for some reason it looks to be converting shopkeeper/89137.csv to shopkeeper_89137.csv. It keeps opening a blank csv file obviously because its looking for a file that isnt there. What is the proper way to reference a file in order to open it?

            header("Content-type: text/csv");
            header("Content-Disposition: attachment; filename=shopkeeper/89137.csv");
            header("Pragma: no-cache");
            header("Expires: 0");

Answer 1

The headers should only include the base name of the file being sent. On the receiving end of things, it thinks you're trying to pass it a file with the name "shopkeeper/89137.csv" which, on most operating systems, contains an illegal character (/) for a file name, so it gets converted to an underscore.

You should probably tweak the line to read:

header("Content-Disposition: attachment; filename=89137.csv");