Bash - extract url using find

Question

When I run the following command in a directory of an extracted (inflated) XLSX, I get the all URLs found within the sub-folder files.

find . -exec exiftool {} \; | grep http

Now I want to run this as part of a bash script I have developed but it simply doesn't show anything upon completion.

Here is the code in the script:

if [[ $fileName == *.xlsx ]]; then 
    #file is XML based, create new directory with file name, copy file there and extract
    echo "Creating directory\n"
    DATE=`date +%Y-%m-%d`
    mkdir "Files/$DATE-$fileName-$USER"
    cp $filePath "Files/$DATE-$fileName-$USER"
    unzip "Files/$DATE-$fileName-$USER/$fileName" -d "Files/$DATE-$fileName-$USER/"
    result="find 'Files/$DATE-$fileName-$USER/$fileName' -exec exiftool {} \; | grep http"
    printf $result
fi

Your support is appreciated in figuring out what's wrong here.

is it just me, or are you not executing the find command, at all? — Michael Speer
– Michael Speer, Commented Mar 3, 2018 at 8:23
also the == *.xlsx ... is this a real construct? wouldn't be better served by something like if echo $filename | grep -E '*[.]xlsx$' ; then — Michael Speer
– Michael Speer, Commented Mar 3, 2018 at 8:26
@MichaelSpeer There are multiple ways of performing this check , mine and yours are just two of them. Nonetheless, if the result is the same in this use case, it becomes a matter of preference. — ksa_coder
– ksa_coder, Commented Mar 3, 2018 at 8:47
@MichaelSpeer, that construct has the benefit of working in a POSIX shell, but is unnecessarily heavy in bash. Also, the regular expression in your comment is incorrect, as the * is supposed to represent "zero or more of the preceding atom". — ghoti
– ghoti, Commented Mar 4, 2018 at 14:27
@ghoti thank you for the correction. I prefixed my other comment wondering if it was a real construct, as opposed to a misunderstanding. I later looked through the extensive bash man-page and found it. I was simply unfamiliar with it previously, and therefore questioned it alongside the apparently non-functional find statement. Thank you both for educating me in the matter — Michael Speer
– Michael Speer, Commented Mar 5, 2018 at 1:29

Socowi · Accepted Answer · 2018-03-03 13:53:26Z

3

You don't call the command, there is the assignment

result="find 'Files/$DATE-$fileName-$USER/$fileName' -exec exiftool {} \; | grep http"

But the command is treated as a string and never executed. Also the variables in single quotes won't expand. Instead, write

result="$(find "Files/$DATE-$fileName-$USER/$fileName" -exec exiftool {} \; | grep http)"

Other improvements

echo does not recognize \n by default. Replace echo "Creating directory\n" with echo "Creating directory" and add another empty echo for an additional newline.

printf $result may swallow % escape sequences and the parts after the first space. Quote the variable and use %s: printf %s "$result".

Don't pass directories and the files inside them to exiftool. Either use exiftool's recursive option and drop the find part or add -type f to the find command, such that only files are listed.

edited Mar 3, 2018 at 13:53

answered Mar 3, 2018 at 8:52

Socowi

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2 Comments

ksa_coder Over a year ago

Ok....So I replaced by bit with yours but I still see no output of the find command...

Socowi Over a year ago

@ksa_coder Did you assign any value to $fileName, does that value end with .xlsx? Can you confirm that there are any files inside Files/$DATE-$fileName-$USER/$fileName and at least one of them has a meta-data field containing the search string http?

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