Can Exiftool be used for server databse solution? |XAMP, PHP, PhpMyAdmin, MySQL

[Petco]

Hello, Im using Exiftool for my Master of Engineering thesis.

I used Exiftool to extract some metadata from videos into .csv document. I easily imported the .csv to the table in the MySQL database, with import button in PhpMyAdmin.

This approach is good, but I was thinking... Would it be somehow possible to automate this process?

1.Lets say if I upload the video to the database, the video also "comes through" Exiftool, which is implemented from the database side.
2.It temporarily creates the .csv document which will contain only certain metadata, that I will choose on the server side.
3.Then it will store this data into the database table.

Also feel free to tell if anyone used Exiftool with XAMP, PHP, PhpMyAdmin or MySQL. I would be interested in your solution.

 

[Phil Harvey]

I do this sort of thing by adding cron jobs to watch the uploads directory.

- Phil

[Petco]

Hello, I was thinking and maybe I found out an easier solution for what I need but im not sure if it could work.
This code below executes when URL is called and extract data from a video in a fix directory.

Code Select Expand

exec('exiftool -filename -createdate -imagesize -videoframerate -mediaduration -T -ext mp4 -csv D:\XAMP\htdocs\root\videos\try > D:\XAMP\htdocs\root\videos\ukaz.csv');
Now if I add attribute formaction to my submit input, I now upload a video and I execute metadata extraction

Code Select Expand

<input type='submit' name='import-excel' value='Upload' formaction="http://localhost/root/webhook.php">
Now, the only problem is that I dont need to extract a video in D:\XAMP\htdocs\root\videos\try but the video I just uploaded.
Can this path be changed for a variable($name), in which I store my video file?

Code Select Expand

$name = $_FILES['file']['name'];


[Phil Harvey]

I don't understand what you are asking.

- Phil

[Blaumeise00]

Can this path be changed for a variable($name), in which I store my video file?

Code Select Expand

$name = $_FILES['file']['name'];


$_FILES['file']['name'] is just a name.
You will have to use the directory of the uploaded file, then call the actual uploaded file.
Otherwise, you could analyze the temporary name which points to the actual temporary file:

$_FILES['file']['tmp_name']