In PHP, after assigning a file pointer resource to a variable with fopen(), how can I get the file name from the variable?

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自闭症患者
自闭症患者 2021-01-21 01:42

For example:

$file = fopen(\"File.txt\", \"r\");
$filename = $file->basename;

if there was a method like basename for file objects (file poi

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  •  无人及你
    2021-01-21 02:35

    No, there is not a method to do. You should rather store the filename in a variable, like this:

    Also, a little side note: a file pointer is not an object, it is a resource, which you can see by passing it to var_dump() (it would output something like resource(3) of type (stream)). This means that you cannot use it directly, you would have to use functions from the PHP core or a PHP extension to handle it. In the case of file pointers, you would use functions like fread(), fwrite() and fclose() to do so.

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