exiftool will not work using php exec($cmd, $output) when dir is on CD

Started by pizzipie, July 13, 2023, 10:36:31 PM

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pizzipie

Hi,


When using using PHP and the source dir is on media(DVD) I can't get exiftool to work.


$cmd='exiftool -s -@ ./myargs.txt -json /media/rick/PollyPics/Pictures-Vista/March8_2005/*';

exec($cmd, $output);

echo "out ". gettype($output);

myprint($output);

out array

Array
(                <============= NOTHING HERE
)

I have to send the output of exiftool  run on the browser to a json file on my hard drive (pic.txt) first and then I can  manipulate the exiftool output with PHP.


 exiftool -s -json -@ /home/rick/DBases/Dbmysql/experiment/exiftools_php/myargs.txt /media/rick/PollyPics/Pictures-Vista/March8_2005/ > /home/rick/Desktop/PollyPics/pic.txt

    1 directories scanned
    4 image files read
[/tt]

Help with this is greatly appreciated. I am very in-experienced using exiftool so please be detailed (ie: basic details are needed.)

Thanks, R

StarGeek

What is the contents of myargs.txt?

Exiftool should at least output a json with "SourceFile" entry unless it is unable to execute properly.

What is the result if you add result_code variable?

exec($cmd, $output,$rc);
echo $rc;
* Did you read FAQ #3 and use the command listed there?
* Please use the Code button for exiftool code/output.
 
* Please include your OS, Exiftool version, and type of file you're processing (MP4, JPG, etc).

Phil Harvey

You should make sure your command works at the command line and figure out any problems there before trying to run it from PHP.

- Phil
...where DIR is the name of a directory/folder containing the images.  On Mac/Linux/PowerShell, use single quotes (') instead of double quotes (") around arguments containing a dollar sign ($).