sed to manipulate output of imagemagick in bash - bash

Question

I'm trying to write a bash script to trim the scanner white space around some old photos that were scanned in ages ago. I've got hundreds of photos so I'm not doing it manually.
Fred's imagemagick scripts don't manage to select the appropriate area.
I am no programmer so please dont be too offended by my terrible attempts at scripting!
I've found a combination of commands using imagemagick that does it.
first I use a blurring filter to confuse imagemagick into correctly selecting the photo size:
convert input -virtual-pixel edge -blur 0x15 -fuzz 15% -trim info:
This spits out data as follows:
0001.jpeg JPEG 3439x2437 4960x6874+1521+115 8-bit DirectClass 0.070u 0:00.009
I then use the numbers to do a crop which has been very accurate on my scans. The following is an example using the numbers from above.
convert inputfile -crop 3439x2437+1521+115 +repage outputfile
My problem is in writing the bash file to go through a directory of pictures and automate the process.
Here's what I have so far:
#!/bin/bash
ls *.jpeg > list
cat list | while read line; do
convert $line -virtual-pixel edge -blur 0x15 -fuzz 15% -trim info: > blurtrim.txt
#need a line to manipulate the output of the above to spit out the crop coordinates for the next command
crop=$(<crop.txt)
convert $line -crop $crop +repage trim$line.jpeg
rm blurtext.txt
rm crop.txt
done
rm list
The key bit I can't do is changing the string output of the first imagemagick command.
the file goes along the lines of:
input fileformat 1111x2222 3333x4444+5555+666 and then a load of crap i dont care about
the numbers I need in my script are:
1111x2222+5555+666
the cherry on the top is that while most of the numbers are four digits long not all of them are so I cant rely on that.
any ideas on how to use sed or preferably something else less demonic to get the above numbers in my script?
an explanation of the syntax would be nice (but i understand if the explantion is the size of a book then its best left out).
thanks in advance!

Answer 1

You don't need to parse anything! ImageMagick can tell you the trim box directly itself, using the %# format:
convert image.jpg -virtual-pixel edge -blur 0x15 -fuzz 15% -format "%#" info:
1111x2222+5555+666
So, you can say:
trimbox=$(convert image.jpg -virtual-pixel edge -blur 0x15 -fuzz 15% -format "%#" info:)
convert image.jpg -crop $trimbox ...
Benefits include the fact that this approach works on Windows too, where there is no sed.
So, the full solution would be something like:
#!/bin/bash
shopt -s nullglob
for f in *.jpeg; do
trimbox=$(convert "$f" -virtual-pixel edge -blur 0x15 -fuzz 15% -format "%#" info:)
convert "$f" -crop "$trimbox" +repage "trimmed-$f"
done

Answer 2

Solution
This will parse your file line by line, extract the desired parameters, concatenate them together, and use it as the argument value to 'crop' for the convert program:
regex='([0-9]+x[0-9]+) [0-9]+x[0-9]+\+([0-9]+\+[0-9]+)'
while read line
do
if [[ $line =~ $regex ]]
then
cropParam="${BASH_REMATCH[1]}+${BASH_REMATCH[2]}"
convert inputfile -crop $cropParam +repage outputfile
else
echo "ERROR: Line was not in the expected format ($line)"
exit 1;
fi
done < blurtrim.txt
Explanation
The regex variable holds a regular expression (brief introduction to regular expressions in bash here: http://www.tldp.org/LDP/abs/html/x17129.html) which describes the format of the numbers you describe in your question. The () around parts of the pattern denotes something called a capture group. If the pattern matches, the part that is in the first () is captured in a bash variable BASH_REMATCH[1], and the second () is captured in BASH_REMATCH[2]. BASH_REMATCH[0] contains the whole match, in case you're wondering why we start at index 1.
The line [[ $line =~ $regex ]] is what actually executes the pattern matching algorithm for us. In Bash [[ is called the extended test command, and the operator =~ is called the regular expression matching operator. This article explains the operator in more detail: http://www.linuxjournal.com/content/bash-regular-expressions.

Answer 3

I would propose a similar solution to Jonathan:
re='([0-9x]+) [0-9x]+(\+[0-9+]+)'
for file in *.jpeg; do
output=$(convert "$file" -virtual-pixel edge -blur 0x15 -fuzz 15% -trim info:)
if [[ $output =~ $re ]]; then
crop="${BASH_REMATCH[1]}${BASH_REMATCH[2]}"
convert "$file" -crop "$crop" +repage "trim$file.jpeg"
fi
done
The regular expression captures any group containing characters within the range 0-9 or x and then a + followed by numbers and + characters. It is a less strict pattern as it includes the x and + inside the bracket expressions, so technically would allow things like 0x9x9x0 but I can't imagine that this would present a problem based on the output you've shown us.
The other differences between this and your original attempt are that no temporary files are created and the loop is run over the list of files, rather than using ls, the parsing of which should generally be avoided in scripts.