SourceFile output for some files gets embedded within quotation marks (")

[U_turn]

While trying to list all images with some tag set, I get most image filenames and paths in plain, but some are embedded within quotation marks ("). And I can't figure out the pattern or a cause for this behavior.  I see that it doesn't happen random, as the same image filenames and paths are always embedded. I would guess that it's because of space char in their filename or path, but not all filenames and paths with space char are embedded...

Used command:
exiftool.exe -charset FileName=cp1257 -r . -ext JPG -m -if "$sometag" -csv -sourcefile

[StarGeek]

Do you have some examples?

How about a comma in the name?  If there's a comma in the name, then it must be quoted.   Otherwise no program would know the proper place to separate the columns.

Accent marks, special characters, or similar stuff?

I just ran a simplified form of your command on a directory with over 1,000 files and nothing came back quoted, so more info is needed.

[U_turn]

Thanks for a tip, StarGeek, and sorry for failing to notice commas in some of my filenames
Yes, it was about having commas...

In any case, as I'm trying to parse output with another command, is there a way to get all filenames quoted or not quoted?
As far as I understand, -sep option doesn't work with -csv

Thanks

[StarGeek]

In any case, as I'm trying to parse output with another command, is there a way to get all filenames quoted or not quoted?

Not with the -csv option alone, you would have to roll your own csv with the -p option.  See FAQ #12.  Near the end of that entry, it gives you a example of a FMT file that would duplicate a CSV output.  Just replace it with the tags you want to pull and put quotes around them, like this:

Code Select Expand

#[HEAD]FileName, Aperture, ISO
"$filename", "$aperture", "$iso"


Example output:

Code Select Expand

C:\>exiftool -p temp.txt y:\!temp\Test3.jpg
FileName, Aperture, ISO
"Test3.jpg", "0.33", "640"

[U_turn]

Thanks, StarGeek, but what should I put as a format string to get SourceFile? $SourceFile is not recognized by -p...

Not with the -csv option alone, you would have to roll your own csv with the -p option.  See FAQ #12.

[U_turn]

No worries, I found solution by myself. -p $directory/$filename does the trick instead of SourceFile.

Thanks, StarGeek, but what should I put as a format string to get SourceFile? $SourceFile is not recognized by -p...

Not with the -csv option alone, you would have to roll your own csv with the -p option.  See FAQ #12.

[StarGeek]

There is also the FilePath tag.