Custom batch rename files
Hello, Mac OS X takes screen shot's in a very long format of filename. I would like to rename any of them that sit at path /Users/me/desktop.
Here are some examples of the filenames:
Screen Shot 2012-08-02 at 1.15.29 AM.png
Screen Shot 2012-08-02 at 1.22.12 AM.png
Screen Shot 2012-08-02 at 1.22.14 PM.png
Screen Shot 2012-08-02 at 1.22.16 PM.png
I was once told, not to do a for loop against an ls so I am trying globbing this time around. So far, this is all I can come up with, but done know how to karen wrap the expression and then get that to a file rename in the format I desire:
for i in *; do
screen_name=$(echo $i | grep --only-matching --extended-regexp '(Screen\ Shot)\ [0-9]+-[0-9]+-[0-9]+\ at\ [0-9]+\.[0-9]+.[0-9]+.[AP]M\.png');
echo $screen_name;
done
I am not sure about the hour of the time, it may be safest to assume possible 2 digits on all chunks of the time, so 1.14.29 and 01.15.29
ss.08-02-12-01.15.29-AM.png
ss.08-02-12-01.22.12-AM.png
ss.08-02-12-01.22.14-PM.png
ss.08-02-12-01.22.16-PM.png
The end goal, is a bash script that when run will rename ALL files at the above mentioned path to the new format listed.
Thank you for any help.
for i in "Screen Shot"*.png; do
new=`echo $i |awk '
{
split($3,a,"-")
split($5,b,".")
printf("ss.%s-%s-%s-%02d.%02d.%02d-%s",a[2],a[3],a[1],b[1],b[2],b[3],$6)
}
'`
mv "$i" $new
done
Before:
Screen Shot 2012-08-02 at 1.22.16 PM.png
Screen Shot 2012-09-02 at 13.42.06 PM.png
After:
ss.08-02-2012-01.22.16-PM.png
ss.09-02-2012-13.42.06-PM.png
EDIT:
as suggested by steve
printf("ss.%s-%s-%s-%02d.%02d.%02d-%s",a[2],a[3],substr(a[1]3,2),b[1],b[2],b[3],$6)
which yields
ss.08-02-12-01.22.16-PM.png
ss.09-02-12-13.42.06-PM.png
You can use stream editor sed to match and substitute using regular expressions. You would do something like this
echo $i | sed "s/PATTERN/REPLACE/"
to genereate the filename out of $i. sed will read from stdin, search (s command) for pattern and replace it with REPLACE.
In your REGEXP pattern you can mark seperate groups by surrounding them with brackets (), in most situations you will have to escape them by () and access these parts in the replace pattern by using #, where # is the number of the subgroup starting from 1. Here's a simple example:
echo "ScreenShotXYZ.png" | sed "s/ScreenShot\(.*\)\.png/\1.png/"
Here, the XYZ is matched by the expression in brackets and can be accessed using \1 in the replacment string. The whole pattern in thus replaced by XYZ.png.
So use your regexp for matching, put brackets around the relevant blocks and do something like
ss.\1.\2.(and so on)
for your replacement pattern. There's still some way to optimize the process by first using sed to replace dashes by dots, then grouping the whole time block in just one pattern but for a start it's easier to code like that.
Related
I've scraped a large amount (10GB) of PDFs and converted them to text files, but due to the format of the original PDFs, there is an issue:
Many of the words which break across lines have a dash in them that artificially breaks up the word, like this:
You can see that this happened because the original PDFs files have breaks:
What would be the cleanest and fastest way to "join" every word instance that matches this pattern inside of a .txt file?
Perhaps some sort of Regex search, like for a [a-z]\-\s \w of some kind (word character followed by dash followed by space) would work?
Or would some sort of sed replacement work better?
Currently, I'm trying to get a sed regex to work, but I'm not sure how to translate this to use capture groups to replace the selected text:
sed -n '\%\w\- [a-z]%p' Filename.txt
My input text would look like this:
The dog rolled down the st- eep hill and pl- ayed outside.
And the output would be:
The dog rolled down the steep hill and played outside.
Ideally, the expression would also work for words split up by a newline, like this:
The rule which provided for the consid-
eration of the resolution, was agreed to earlier by a
To this:
The rule which provided for the consideration
of the resolution, was agreed to earlier by a
It's straightforward in sed:
sed -e ':a' -e '/-$/{N;s/-\n//;ba
}' -e 's/- //g' filename
This translates roughly as "if the line ends with a dash, read in the next line as well (so that you have a line with a carriage return in the middle) then excise the dash and carriage return, and loop back the beginning just in case this new line also ends with a dash. Then remove any instances of - ".
You may use this gnu-awk code:
cat file
The dog rolled down the st- eep hill and pl- ayed outside.
The rule which provided for the consid-
eration of the resolution, was agreed to earlier by a
Then use awk like this:
awk 'p != "" {
w = $1
$1 = ""
sub(/^[[:blank:]]+/, ORS)
$0 = p w $0
p = ""
}
{
$0 = gensub(/([_[:alnum:]])-[[:blank:]]+([_[:alnum:]])/, "\\1\\2", "g")
}
/-$/ {
p = $0
sub(/-$/, "", p)
}
p == ""' file
The dog rolled down the steep hill and played outside.
The rule which provided for the consideration
of the resolution, was agreed to earlier by a
If you can consider perl then this may also work for you:
Then use:
perl -0777 -pe 's/(\w)-\h+(\w)/$1$2/g; s/(\w)-\R(\w+)\s+/$1$2\n/g' file
You simply add backslash-parentheses (or use the -r or -E option if available to do away with the requirement to put backslashes before capturing parentheses) and recall the matched text with \1 for the first capturing parenthesis, \2 for the second, etc.
sed 's/\(\w\)\- \([a-z]\)/\1\2/g' Filename.txt
The \w escape is not standard sed but if it works for you, feel free to use it. Otherwise, it is easy to replace with [A-Za-z0-9_#] or whatever else you want to call "word characters".
I'm guessing not all of the matches will be hyphenated words so perhaps run the result through a spelling checker or something to verify whether the result is an English word. (I would probably switch to a more capable scripting language like Python for that, though.)
I have a very large file, containing the following blocks of lines throughout:
start :234
modify 123 directory1/directory2/file.txt
delete directory3/file2.txt
modify 899 directory4/file3.txt
Each block starts with the pattern "start : #" and ends with a blank line. Within the block, every line starts with "modify # " or "delete ".
I need to modify the path in each line, specifically appending a directory to the front. I would just use a general regex to cover the entire file for "modify #" or "delete ", but due to the enormous amount of other data in that file, there will likely be other matches to this somewhat vague pattern. So I need to use multi-line matching to find the entire block, and then perform edits within that block. This will likely result in >10,000 modifications in a single pass, so I'm also trying to keep the execution down to less than 30 minutes.
My current attempt is a sed one-liner:
sed '/^start :[0-9]\+$/ { :a /^[modify|delete] .*$/ { N; ba }; s/modify [0-9]\+ /&Appended_DIR\//g; s/delete /&Appended_DIR\//g }' file_to_edit
Which is intended to find the "start" line, loop while the lines either start with a "modify" or a "delete," and then apply the sed replacements.
However, when I execute this command, no changes are made, and the output is the same as the original file.
Is there an issue with the command I have formed? Would this be easier/more efficient to do in perl? Any help would be greatly appreciated, and I will clarify where I can.
I think you would be better off with perl
Specifically because you can work 'per record' by setting $/ - if you're records are delimited by blank lines, setting it to \n\n.
Something like this:
#!/usr/bin/env perl
use strict;
use warnings;
local $/ = "\n\n";
while (<>) {
#multi-lines of text one at a time here.
if (m/^start :\d+/) {
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
Each iteration of the loop will pick out a blank line delimited chunk, check if it starts with a pattern, and if it does, apply some transforms.
It'll take data from STDIN via a pipe, or myscript.pl somefile.
Output is to STDOUT and you can redirect that in the normal way.
Your limiting factor on processing files in this way are typically:
Data transfer from disk
pattern complexity
The more complex a pattern, and especially if it has variable matching going on, the more backtracking the regex engine has to do, which can get expensive. Your transforms are simple, so packaging them doesn't make very much difference, and your limiting factor will be likely disk IO.
(If you want to do an in place edit, you can with this approach)
If - as noted - you can't rely on a record separator, then what you can use instead is perls range operator (other answers already do this, I'm just expanding it out a bit:
#!/usr/bin/env perl
use strict;
use warnings;
while (<>) {
if ( /^start :/ .. /^$/)
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
We don't change $/ any more, and so it remains on it's default of 'each line'. What we add though is a range operator that tests "am I currently within these two regular expressions" that's toggled true when you hit a "start" and false when you hit a blank line (assuming that's where you would want to stop?).
It applies the pattern transformation if this condition is true, and it ... ignores and carries on printing if it is not.
sed's pattern ranges are your friend here:
sed -r '/^start :[0-9]+$/,/^$/ s/^(delete |modify [0-9]+ )/&prepended_dir\//' filename
The core of this trick is /^start :[0-9]+$/,/^$/, which is to be read as a condition under which the s command that follows it is executed. The condition is true if sed currently finds itself in a range of lines of which the first matches the opening pattern ^start:[0-9]+$ and the last matches the closing pattern ^$ (an empty line). -r is for extended regex syntax (-E for old BSD seds), which makes the regex more pleasant to write.
I would also suggest using perl. Although I would try to keep it in one-liner form:
perl -i -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit
Or you can use redirection of stdout:
perl -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit > new_file
with gnu sed (with BRE syntax):
sed '/^start :[0-9][0-9]*$/{:a;n;/./{s/^\(modify [0-9][0-9]* \|delete \)/\1NewDir\//;ba}}' file.txt
The approach here is not to store the whole block and to proceed to the replacements. Here, when the start of the block is found the next line is loaded in pattern space, if the line is not empty, replacements are performed and the next line is loaded, etc. until the end of the block.
Note: gnu sed has the alternation feature | available, it may not be the case for some other sed versions.
a way with awk:
awk '/^start :[0-9]+$/,/^$/{if ($1=="modify"){$3="newdirMod/"$3;} else if ($1=="delete"){$2="newdirDel/"$2};}{print}' file.txt
This is very simple in Perl, and probably much faster than the sed equivalent
This one-line program inserts Appended_DIR/ after any occurrence of modify 999 or delete at the start of a line. It uses the range operator to restrict those changes to blocks of text starting with start :999 and ending with a line containing no printable characters
perl -pe"s<^(?:modify\s+\d+|delete)\s+\K><Appended_DIR/> if /^start\s+:\d+$/ .. not /\S/" file_to_edit
Good grief. sed is for simple substitutions on individual lines, that is all. Once you start using constructs other than s, g, and p (with -n) you are using the wrong tool. Just use awk:
awk '
/^start :[0-9]+$/ { inBlock=1 }
inBlock { sub(/^(modify [0-9]+|delete) /,"&Appended_DIR/") }
/^$/ { inBlock=0 }
{ print }
' file
start :234
modify 123 Appended_DIR/directory1/directory2/file.txt
delete Appended_DIR/directory3/file2.txt
modify 899 Appended_DIR/directory4/file3.txt
There's various ways you can do the above in awk but I wrote it in the above style for clarity over brevity since I assume you aren't familiar with awk but should have no trouble following that since it reuses your own sed scripts regexps and replacement text.
I have a problem that because of PHP version, I need to change my code from $array[stringindex] to $array['stringindex'];
So I want to find all the text using regex, and replace them all. How to find all strings that look like this? $array[stringindex].
Here's a solution in PHP:
$re = "/(\\$[[:alpha:]][[:alnum:]]+\\[)([[:alpha:]][[:alnum:]]+)(\\])/";
$str = "here is \$array[stringindex] but not \$array['stringindex'] nor \$3array[stringindex] nor \$array[4stringindex]";
$subst = "$1'$2'$3";
$result = preg_replace($re, $subst, $str);
You can try it out interactively here. I search for variables beginning with a letter, otherwise things like $foo[42] would be converted to $foo['42'], which might not be desirable.
Note that all the solutions here will not handle every case correctly.
Looking at the Sublime Text regex help, it would seem you could just paste (\\$[[:alpha:]][[:alnum:]]+\\[)([[:alpha:]][[:alnum:]]+)(\\]) into the Search box and $1'$2'$3 into the Replace field.
It depends of the tool you want to use to do the replacement.
with sed for exemple, it would be something like that:
sed "s/\(\$array\)\[\([^]]*\)\]/\1['\2']/g"
If sed is allowed you could simply do:
sed -i "s/(\$[^[]*[)([^]]*)]/\1'\2']/g" file
Explanation:
sed "s/pattern/replace/g" is a sed command which searches for pattern and replaces it with replace. The g options means replace multiple times per line.
(\$[^[]*[)([^]]*)] this pattern consists of two groups (in between brackets). The first is a dollar followed by a series of non [ chars. Then an opening square bracket follows, followed by a series of non closing brackets which is then followed by a closing square bracket.
\1'\2'] the replacement string: \1 means insert the first captured group (analogous for \2. Basically we wrap \2 in quotes (which is what you wanted).
the -i options means that the changes should be applied to the original file, which is supplied at the end.
For more information, see man sed.
This can be combined with the find command, as follows:
find . -name '*.php' -exec sed -i "s/(\$[^[]*[)([^]]*)]/\1'\2']/g" '{}' \;
This will apply the sed command to all php files found.
I read some threads about batch renames in Mac OS X Terminal and understood you can do something like this:
for file in *.pdf
do
mv "$file" "<some magic regex>"
done
As you can guess, my issue is with the regex for this particular purpose:
Old file
Author-Year-Title.pdf
New file
Year-Author-Title.pdf
I did try some regex code but got stuck. I "just" want to flip Author and Year, but cannot figure out how. Any help would be appreciated.
This will do, or come quite close to doing, what you are asking for. The technique I am using is called Bash Parameter Substitution and it is documented and described very well here.
#!/bin/bash
for file in *.pdf
do
echo DEBUG: Processing file $file
f=${file%.*} # strip extension from right end
author=${f%%-*} # shortest str at start that ends with dash
title=${f##*-} # shortest str at end that starts with dash
authoryear=${f%-*} # longest string at start that ends in dash
year=${authoryear#*-}
echo DEBUG: author:$author, year:$year, title:$title
echo mv "$file" "$year-$author-$title.pdf"
done
Basically, I am extracting the Author, Year and Title into variables for you and then you can put them together in whatever order you like with whatever separators you like at the end and do the actual renaming. Note that the script actually does nothing until you remove the echo statement in front of the mv command so you can test it out and see what it would do.
Please practice on a COPY of your data in a spare, temporary directory.
Sample Output
DEBUG: Processing file Banks-2012-Something.pdf
DEBUG: author:Banks, year:2012, title:Something
mv Banks-2012-Something.pdf 2012-Banks-Something.pdf
DEBUG: Processing file Shakey-2013-SomethingElse.pdf
DEBUG: author:Shakey, year:2013, title:SomethingElse
mv Shakey-2013-SomethingElse.pdf 2013-Shakey-SomethingElse.pdf
If you like ugly sed commands you can do it more succinctly like this:
#!/bin/bash
for file in *.pdf
do
echo DEBUG: Processing file $file
new=$(sed -E 's/(.*)-([0-9]{4})-(.*)\.*/\2-\1-\3.pdf/' <<< $file)
echo $new
done
The s/xxx/yyy/ means substitute or replace xxx with yyy. Anything inside parentheses must be captured as capture groups and then the first capture group becomes available as \1 in the replacement and the second capture group becomes available as \2 and so on. So it says... save anything up to the first dash as \1, exactly 4 digits between the next pair of dashes as \2 and the other stuff as \3, and then it prints the captured groups out in a different order.
Consider the following:
var="text more text and yet more text"
echo $var | egrep "yet more (text)"
It should be possible to get the result of the regex as the string: text
However, I don't see any way to do this in bash with grep or its siblings at the moment.
In perl, php or similar regex engines:
$output = preg_match('/yet more (text)/', 'text more text yet more text');
$output[1] == "text";
Edit: To elaborate why I can't just multiple-regex, in the end I will have a regex with multiple of these (Pictured below) so I need to be able to get all of them. This also eliminates the option of using lookahead/lookbehind (As they are all variable length)
egrep -i "([0-9]+) +$USER +([0-9]+).+?(/tmp/Flash[0-9a-z]+) "
Example input as requested, straight from lsof (Replace $USER with "j" for this input data):
npviewer. 17875 j 11u REG 8,8 59737848 524264 /tmp/FlashXXu8pvMg (deleted)
npviewer. 17875 j 17u REG 8,8 16037387 524273 /tmp/FlashXXIBH29F (deleted)
The end goal is to cp /proc/$var1/fd/$var2 ~/$var3 for every line, which ends up "Downloading" flash files (Flash used to store in /tmp but they drm'd it up)
So far I've got:
#!/bin/bash
regex="([0-9]+) +j +([0-9]+).+?/tmp/(Flash[0-9a-zA-Z]+)"
echo "npviewer. 17875 j 11u REG 8,8 59737848 524264 /tmp/FlashXXYOvS8S (deleted)" |
sed -r -n -e " s%^.*?$regex.*?\$%\1 \2 \3%p " |
while read -a array
do
echo /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
It cuts off the first digits of the first value to return, and I'm not familiar enough with sed to see what's wrong.
End result for downloading flash 10.2+ videos (Including, perhaps, encrypted ones):
#!/bin/bash
lsof | grep "/tmp/Flash" | sed -r -n -e " s%^.+? ([0-9]+) +$USER +([0-9]+).+?/tmp/(Flash[0-9a-zA-Z]+).*?\$%\1 \2 \3%p " |
while read -a array
do
cp /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
Edit: look at my other answer for a simpler bash-only solution.
So, here the solution using sed to fetch the right groups and split them up. You later still have to use bash to read them. (And in this way it only works if the groups themselves do not contain any spaces - otherwise we had to use another divider character and patch read by setting $IFS to this value.)
#!/bin/bash
USER=j
regex=" ([0-9]+) +$USER +([0-9]+).+(/tmp/Flash[0-9a-zA-Z]+) "
sed -r -n -e " s%^.*$regex.*\$%\1 \2 \3%p " |
while read -a array
do
cp /proc/${array[0]}/fd/${array[1]} ~/${array[2]}
done
Note that I had to adapt your last regex group to allow uppercase letters, and added a space at the beginning to be sure to capture the whole block of numbers. Alternatively here a \b (word limit) would have worked, too.
Ah, I forget mentioning that you should pipe the text to this script, like this:
./grep-result.sh < grep-result-test.txt
(provided your files are named like this). Instead you can add a < grep-result-test after the sed call (before the |), or prepend the line with cat grep-result-test.txt |.
How does it work?
sed -r -n calls sed in extended-regexp-mode, and without printing anything automatically.
-e " s%^.*$regex.*\$%\1 \2 \3%p " gives the sed program, which consists of a single s command.
I'm using % instead of the normal / as parameter separator, since / appears inside the regex and I don't want to escape it.
The regex to search is prefixed by ^.* and suffixed by .*$ to grab the whole line (and avoid printing parts of the rest of the line).
Note that this .* grabs greedy, so we have to insert a space into our regexp to avoid it grabbing the start of the first digit group too.
The replacement text contains of the three parenthesed groups, separated by spaces.
the p flag at the end of the command says to print out the pattern space after replacement. Since we grabbed the whole line, the pattern space consists of only the replacement text.
So, the output of sed for your example input is this:
5 11 /tmp/FlashXXu8pvMg
5 17 /tmp/FlashXXIBH29F
This is much more friendly for reuse, obviously.
Now we pipe this output as input to the while loop.
read -a array reads a line from standard input (which is the output from sed, due to our pipe), splits it into words (at spaces, tabs and newlines), and puts the words into an array variable.
We could also have written read var1 var2 var3 instead (preferably using better variable names), then the first two words would be put to $var1 and $var2, with $var3 getting the rest.
If read succeeded reading a line (i.e. not end-of-file), the body of the loop is executed:
${array[0]} is expanded to the first element of the array and similarly.
When the input ends, the loop ends, too.
This isn't possible using grep or another tool called from a shell prompt/script because a child process can't modify the environment of its parent process. If you're using bash 3.0 or better, then you can use in-process regular expressions. The syntax is perl-ish (=~) and the match groups are available via $BASH_REMATCH[x], where x is the match group.
After creating my sed-solution, I also wanted to try the pure-bash approach suggested by Mark. It works quite fine, for me.
#!/bin/bash
USER=j
regex=" ([0-9]+) +$USER +([0-9]+).+(/tmp/Flash[0-9a-zA-Z]+) "
while read
do
if [[ $REPLY =~ $regex ]]
then
echo cp /proc/${BASH_REMATCH[1]}/fd/${BASH_REMATCH[2]} ~/${BASH_REMATCH[3]}
fi
done
(If you upvote this, you should think about also upvoting Marks answer, since it is essentially his idea.)
The same as before: pipe the text to be filtered to this script.
How does it work?
As said by Mark, the [[ ... ]] special conditional construct supports the binary operator =~, which interprets his right operand (after parameter expansion) as a extended regular expression (just as we want), and matches the left operand against this. (We have again added a space at front to avoid matching only the last digit.)
When the regex matches, the [[ ... ]] returns 0 (= true), and also puts the parts matched by the individual groups (and the whole expression) into the array variable BASH_REMATCH.
Thus, when the regex matches, we enter the then block, and execute the commands there.
Here again ${BASH_REMATCH[1]} is an array-access to an element of the array, which corresponds to the first matched group. ([0] would be the whole string.)
Another note: Both my scripts accept multi-line input and work on every line which matches. Non-matching lines are simply ignored. If you are inputting only one line, you don't need the loop, a simple if read ; then ... or even read && [[ $REPLY =~ $regex ]] && ... would be enough.
echo "$var" | pcregrep -o "(?<=yet more )text"
Well, for your simple example, you can do this:
var="text more text and yet more text"
echo $var | grep -e "yet more text" | grep -o "text"