This may be a bit complex, but here it goes:
Assuming I have an XML that looks as follows:
<a>
<b>000</b>
<c>111</c>
<b>222</b>
<d>333</d>
<c>444</c>
</a>
How can I, using sed on a mac, get a resulting an XML that looks as follows:
<a>
<b>111 000</b>
<b>222</b>
<d>333</d>
<c>444</c>
</a>
Basically:
Matching 2 consecutive lines that are of the form <b>...</b> followed by </c>...</c>
Taking the value between <c>...</c> and placing it (plus a space character) right after <b> on the line before it
Removing the second line <c>...</c>
Thank you.
If sed is too much for this, please advise anything else as long as I can run it from a mac shell.
Not the most beautiful solution but it seams to work :-)
$ tr '\n' # < input | sed 's#<b>\([0-9]\+\)</b>#<c>\([0-9]\+\)</c>#<b>\2 \1</b#g' | tr # '\n'
output:
<a>
<b>111 000</b
<b>222</b>
<d>333</d>
<c>444</c>
</a>
or a bit more general:
$ tr '\n' # < f1 | sed 's#<b>\([^<]*\)</b>#<c>\([^<]*\)</c>#<b>\2 \1</b#' | tr # '\n'
using [^<] to match anything between brackets
Ruby would support multi-line patterns:
ruby -e 'print gets(nil).sub(/<b>([^\n]*)<\/b>\n<c>([^\n]*)<\/c>/m,"<b>\\2 \\1</b>")' file.txt
Related
I have many files named a, b, c and so on. These files contain line like this:-
11.077-105.882
-22.134-302.321
-1.011-201.254
I want to add a space when - sign come in mid of line. I want my output file look like this:-
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
I have tried this command:-
cat a |sed 's/-/ -/g' >out.txt
But it do not give desired result
Require (and capture) a character before each - to replace:
$ sed 's/\(.\)-/\1 -/g' < tmp.txt
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
This will only match a - that is not line-initial, and will include the preceding character in the replacement text.
You could combine 2 sed commands:
$ sed 's/-/ -/g' a | sed 's/^ //'
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
Or, in a single line solution add whitespaces only before - that come after a digit:
$ sed 's,\([0-9]\)-,\1 -,' a
11.077 -105.882
-22.134 -302.321
-1.011 -201.254
How can I reverse a four length of letters with sed?
For example:
the year was 1815.
Reverse to:
the raey was 5181.
This is my attempt:
cat filename | sed's/\([a-z]*\) *\([a-z]*\)/\2, \1/'
But it does not work as I intended.
not sure it is possible to do it with GNU sed for all cases. If _ doesn't occur immediately before/after four letter words, you can use
sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
\b is word boundary, word definition being any alphabet or digit or underscore character. So \b will ensure to match only whole words not part of words
$ echo 'the year was 1815.' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
the raey was 5181.
$ echo 'two time five three six good' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
two emit evif three six doog
$ # but won't work if there are underscores around the words
$ echo '_good food' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
_good doof
tool with lookaround support would work for all cases
$ echo '_good food' | perl -pe 's/(?<![a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])(?!=[a-z0-9])/$4$3$2$1/gi'
_doog doof
(?<![a-z0-9]) and (?!=[a-z0-9]) are negative lookbehind and negative lookahead respectively
Can be shortened to
perl -pe 's/(?<![a-z0-9])[a-z0-9]{4}(?!=[a-z0-9])/reverse $&/gie'
which uses the e modifier to place Perl code in substitution section. This form is suitable to easily change length of words to be reversed
Possible shortest sed solution even if a four length of letters contains _s.
sed -r 's/\<(.)(.)(.)(.)\>/\4\3\2\1/g'
Following awk may help you in same. Tested this in GNU awk and only with provided sample Input_file
echo "the year was 1815." |
awk '
function reverse(val){
num=split(val, array,"");
i=array[num]=="."?num-1:num;
for(;i>q;i--){
var=var?var array[i]:array[i]
};
printf (array[num]=="."?var".":var);
var=""
}
{
for(j=1;j<=NF;j++){
printf("%s%s",j==NF||j==2?reverse($j):$j,j==NF?RS:FS)
}}'
This might work for you (GNU sed):
sed -r '/\<\w{4}\>/!b;s//\n&\n/g;s/^[^\n]/\n&/;:a;/\n\n/!s/(.*\n)([^\n])(.*\n)/\2\1\3/;ta;s/^([^\n]*)(.*)\n\n/\2\1/;ta;s/\n//' file
If there are no strings of the length required to reverse, bail out.
Prepend and append newlines to all required strings.
Insert a newline at the start of the pattern space (PS). The PS is divided into two parts, the first line will contain the current word being reversed. The remainder will contain the original line.
Each character of the word to be reversed is inserted at the front of the first line and removed from the original line. When all the characters in the word have been processed, the original word will have gone and only the bordering newlines will exist. These double newlines are then replaced by the word in the first line and the process is repeated until all words have been processed. Finally the newline introduced to separate the working line and the original is removed and the PS is printed.
N.B. This method may be used to reverse strings of varying string length i.e. by changing the first regexp strings of any number can be reversed. Also strings between two lengths may also be reversed e.g. /\<w{2,4}\>/ will change all words between 2 and 4 character length.
It's a recurrent problem so somebody created a bash command called "rev".
echo "$(echo the | rev) $(echo year | rev) $(echo was | rev) $(echo 1815 | rev)".
OR
echo "the year was 1815." | rev | tr ' ' '\n' | tac | tr '\n' ' '
I want to write a small script to generate the location of a file in an NGINX cache directory.
The format of the path is:
/path/to/nginx/cache/d8/40/32/13febd65d65112badd0aa90a15d84032
Note the last 6 characters: d8 40 32, are represented in the path.
As an input I give the md5 hash (13febd65d65112badd0aa90a15d84032) and I want to generate the output: d8/40/32/13febd65d65112badd0aa90a15d84032
I'm sure sed or awk will be handy, but I don't know yet how...
This awk can make it:
awk 'BEGIN{FS=""; OFS="/"}{print $(NF-5)$(NF-4), $(NF-3)$(NF-2), $(NF-1)$NF, $0}'
Explanation
BEGIN{FS=""; OFS="/"}. FS="" sets the input field separator to be "", so that every char will be a different field. OFS="/" sets the output field separator as /, for print matters.
print ... $(NF-1)$NF, $0 prints the penultimate field and the last one all together; then, the whole string. The comma is "filled" with the OFS, which is /.
Test
$ awk 'BEGIN{FS=""; OFS="/"}{print $(NF-5)$(NF-4), $(NF-3)$(NF-2), $(NF-1)$NF, $0}' <<< "13febd65d65112badd0aa90a15d84032"
d8/40/32/13febd65d65112badd0aa90a15d84032
Or with a file:
$ cat a
13febd65d65112badd0aa90a15d84032
13febd65d65112badd0aa90a15f1f2f3
$ awk 'BEGIN{FS=""; OFS="/"}{print $(NF-5)$(NF-4), $(NF-3)$(NF-2), $(NF-1)$NF, $0}' a
d8/40/32/13febd65d65112badd0aa90a15d84032
f1/f2/f3/13febd65d65112badd0aa90a15f1f2f3
With sed:
echo '13febd65d65112badd0aa90a15d84032' | \
sed -n 's/\(.*\([0-9a-f]\{2\}\)\([0-9a-f]\{2\}\)\([0-9a-f]\{2\}\)\)$/\2\/\3\/\4\/\1/p;'
Having GNU sed you can even simplify the pattern using the -r option. Now you won't need to escape {} and () any more. Using ~ as the regex delimiter allows to use the path separator / without need to escape it:
sed -nr 's~(.*([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2}))$~\2/\3/\4/\1~p;'
Output:
d8/40/32/13febd65d65112badd0aa90a15d84032
Explained simple the pattern does the following: It matches:
(all (n-5 - n-4) (n-3 - n-2) (n-1 - n-0))
and replaces it by
/$1/$2/$3/$0
You can use a regular expression to separate each of the last 3 bytes from the rest of the hash.
hash=13febd65d65112badd0aa90a15d84032
[[ $hash =~ (..)(..)(..)$ ]]
new_path="/path/to/nginx/cache/${BASH_REMATCH[1]}/${BASH_REMATCH[2]}/${BASH_REMATCH[3]}/$hash"
Base="/path/to/nginx/cache/"
echo '13febd65d65112badd0aa90a15d84032' | \
sed "s|\(.*\(..\)\(..\)\(..\)\)|${Base}\2/\3/\4/\1|"
# or
# sed sed 's|.*\(..\)\(..\)\(..\)$|${Base}\1/\2/\3/&|'
Assuming info is a correct MD5 (and only) string
First of all - thanks to all of the responders - this was extremely quick!
I also did my own scripting meantime, and came up with this solution:
Run this script with a parameter of the URL you're looking for (www.example.com/article/76232?q=hello for example)
#!/bin/bash
path=$1
md5=$(echo -n "$path" | md5sum | cut -f1 -d' ')
p3=$(echo "${md5:0-2:2}")
p2=$(echo "${md5:0-4:2}")
p1=$(echo "${md5:0-6:2}")
echo "/path/to/nginx/cache/$p1/$p2/$p3/$md5"
This assumes the NGINX cache has a key structure of 2:2:2.
I want to get the string between <sometag param=' and '>
I tried to use the method from Get any string between 2 string and assign a variable in bash to get the "x":
echo "<sometag param='x'><irrelevant stuff='nonsense'>" | tr "'" _ | sed -n 's/.*<sometag param=_\(.*\)_>.*/\1/p'
The problem (apart from low efficiency because I just cannot manage to escape the apostrophe correctly for sed) is that sed matches the maximum, i.e. the output is:
x_><irrelevant stuff=_nonsense
but the correct output would be the minimum-match, in this example just "x"
Thanks for your help
You are probably looking for something like this:
sed -n "s/.*<sometag param='\([^']*\)'>.*/\1/p"
Test:
echo "<sometag param='x'><irrelevant stuff='nonsense'>" | sed -n "s/.*<sometag param='\([^']*\)'>.*/\1/p"
Results:
x
Explanation:
Instead of a greedy capture, use a non-greedy capture like: [^']* which means match anything except ' any number of times. To make the pattern stick, this is followed by: '>.
You can also use double quotes so that you don't need to escape the single quotes. If you wanted to escape the single quotes, you'd do this:
-
... | sed -n 's/.*<sometag param='\''\([^'\'']*\)'\''>.*/\1/p'
Notice how that the single quotes aren't really escaped. The sed expression is stopped, an escaped single quote is inserted and the sed expression is re-opened. Think of it like a four character escape sequence.
Personally, I'd use GNU grep. It would make for a slightly shorter solution. Run like:
... | grep -oP "(?<=<sometag param=').*?(?='>)"
Test:
echo "<sometag param='x'><irrelevant stuff='nonsense'>" | grep -oP "(?<=<sometag param=').*?(?='>)"
Results:
x
You don't have to assemble regexes in those cases, you can just use ' as the field separator
in="<sometag param='x'><irrelevant stuff='nonsense'>"
IFS="'" read x whatiwant y <<< "$in" # bash
echo "$whatiwant"
awk -F\' '{print $2}' <<< "$in" # awk
I having problems in extracting the word from a line. What i want is that it picks the first word before the symbol # but after the /. Which is the only delimiter that stand out.
A line looks like this:
,["https://picasaweb.google.com/111560558537332305125/Programming#5743548966953176786",1,["https://lh6.googleusercontent.com/-Is8rb8G1sb8/T7UvWtVOTtI/AAAAAAAAG68/Cht3FzfHXNc/s0-d/Geek.jpg",1920,1200]
I want the word Programming.
To get that line i am using this which narrows it down.
sed -n '/.*picasa.*.jpg/p' 5743548866439293105
So i want it to pretty much find # and then go backward until it hit the first /. Then print it out. In this case the word should be Programming but could be anything.
I want it to be as short as possible and have experimented with
sed -n '/.*picasa.*.jpg/p' 5743548866439293105 | awk '$0=$2' FS="/" RS="[$#]"
You can do that with sed (slightly shortened for formatting but works on your original string as well):
pax> echo ',["https://p.g.com/111/Prog#574' | sed 's/^[^#]*\/\([^#]*\)#.*$/\1/'
Prog
pax>
Explaining in more detail:
/---+------------------> greedy capture up to '/'.
/ |
| | /------+---------> capture the stuff between '/' and '#'.
| |/ |
| || | /-+-----> everything from '#' to end of line.
| || |/ |
| || || |
's/^[^#]*\/\([^#]*\)#.*$/\1/'
||
\+---> replace with captured group.
It basically searches for an entire line that has the pattern you want (first # following a /), whilst capturing (with the \( and \) brackets) just the stuff between / and #.
The substitution then replaces the entire line with just that captured text you're interested in (via \1).
Using grep with some Perl regex extensions:
echo $string | grep -P -o "(?<=/)[^/]+(?=#)"
-P tells grep to use Perl extensions. -o tells grep to display only the matched text. To understand what gets matched, break the regex into three parts: (?<=/), [^/]+?, and (?=#). The first part says that the matched text must follow a '/', without including the '/' in the match. The second parts matches a string of non-'/' characters. The last part says that the matched text must be immediately followed by a '#', without including the '#' in the match.
Another grep, using the "\K" feature to "throw away" the match up to the last '/' before the '#':
# Match as much as possible up to a '/', but throw it away, then match as much as you can
# up to the first #
echo $string | grep -oP ".*/\K.+(?=#)"
Using cut and awk to get the first field (splitting on #) followed by the last field (splitting on /):
echo $string | cut -d# -f1 | awk -F/ '{print $NF}'
Using some temporary variables and bash's parameter expansion facilities:
$ FOO=["https://picasaweb.google.com/111560558537332305125/Programming#5743548966953176786",1,["https://lh6.googleusercontent.com/-Is8rb8G1sb8/T7UvWtVOTtI/AAAAAAAAG68/Cht3FzfHXNc/s0-d/Geek.jpg",1920,1200]
$ BAR=${FOO%#*} # Strip the last # and everything after
$ echo $BAR
[https://picasaweb.google.com/111560558537332305125/Programming
$ BAZ=${BAR##*/} # Strip everything up to and including the last /
$ echo $BAZ
Programming
This might work for you:
sed '/.*\/\([^#]*\)#.*/{s//\1/;q};d' file