I have a file containing a list of strings, for example:
abc
search1
lkj
sdfdgs
search2
kkd
#search3
search3
How can I keep all lines matching search1, search2 or search3, that is the
command search1,search2,search3
for
expected output is:
search1
search2
search3
if command search3,search1,search2
for
expected output is:
not matching or print nothing
Note that the #search3 line should be removed.
You can use the grep utility:
% grep '^\(search1\|search2\|search3\)' file.txt
search1
search2
search3
Related
I have two files. One file contains a pattern that I want to match in a second file. I want to use that pattern to print between that pattern (included) up to a specified character (not included) and then concatenate into a single output file.
For instance,
File_1:
a
c
d
and File_2:
>a
MEEL
>b
MLPK
>c
MEHL
>d
MLWL
>e
MTNH
I have been using variations of this loop:
while read $id;
do
sed -n "/>$id/,/>/{//!p;}" File_2;
done < File_1
hoping to obtain something like the following output:
>a
MEEL
>c
MEHL
>d
MLWL
But have had no such luck. I have played around with grep/fgrep awk and sed and between the three cannot seem to get the right (or any output). Would someone kindly point me in the right direction?
Try:
$ awk -F'>' 'FNR==NR{a[$1]; next} NF==2{f=$2 in a} f' file1 file2
>a
MEEL
>c
MEHL
>d
MLWL
How it works
-F'>'
This sets the field separator to >.
FNR==NR{a[$1]; next}
While reading in the first file, this creates a key in array a for every line in file file.
NF==2{f=$2 in a}
For every line in file 2 that has two fields, this sets variable f to true if the second field is a key in a or false if it is not.
f
If f is true, print the line.
A plain (GNU) sed solution. Files are read only once. It is assumed that characters in File_1 needn't to be quoted in sed expression.
pat=$(sed ':a; $!{N;ba;}; y/\n/|/' File_1)
sed -E -n ":a; /^>($pat)/{:b; p; n; /^>/ba; bb}" File_2
Explanation:
The first call to sed generates a regular expression to be used in the second call to sed and stores it in the variable pat. The aim is to avoid reading repeatedly the entire File_2 for each line of File_1. It just "slurps" the File_1 and replaces new-line characters with | characters. So the sample File_1 becomes a string with the value a|c|d. The regular expression a|c|d matches if at least one of the alternatives (a, b, c for this example) matches (this is a GNU sed extension).
The second sed expression, ":a; /^>($pat)/{:b; p; n; /^>/ba; bb}", could be converted to pseudo code like this:
begin:
read next line (from File_2) or quit on end-of-file
label_a:
if line begins with `>` followed by one of the alternatives in `pat` then
label_b:
print the line
read next line (from File_2) or quit on end-of-file
if line begins with `>` goto label_a else goto label_b
else goto begin
Let me try to explain why your approach does not work well:
You need to say while read id instead of while read $id.
The sed command />$id/,/>/{//!p;} will exclude the lines which start
with >.
Then you might want to say something like:
while read id; do
sed -n "/^>$id/{N;p}" File_2
done < File_1
Output:
>a
MEEL
>c
MEHL
>d
MLWL
But the code above is inefficient because it reads File_2 as many times as the count of the id's in File_1.
Please try the elegant solution by John1024 instead.
If ed is available, and since the shell is involve.
#!/usr/bin/env bash
mapfile -t to_match < file1.txt
ed -s file2.txt <<-EOF
g/\(^>[${to_match[*]}]\)/;/^>/-1p
q
EOF
It will only run ed once and not every line that has the pattern, that matches from file1. Like say if you have a to z from file1,ed will not run 26 times.
Requires bash4+ because of mapfile.
How it works
mapfile -t to_match < file1.txt
Saves the entry/value from file1 in an array named to_match
ed -s file2.txt point ed to file2 with the -s flag which means don't print info about the file, same info you get with wc file
<<-EOF A here document, shell syntax.
g/\(^>[${to_match[*]}]\)/;/^>/-1p
g means search the whole file aka global.
( ) capture group, it needs escaping because ed only supports BRE, basic regular expression.
^> If line starts with a > the ^ is an anchor which means the start.
[ ] is a bracket expression match whatever is inside of it, in this case the value of the array "${to_match[*]}"
; Include the next address/pattern
/^>/ Match a leading >
-1 go back one line after the pattern match.
p print whatever was matched by the pattern.
q quit ed
I have a file named file.txt
$cat file.txt
1./abc/cde/go/ftg133333.jpg
2./abc/cde/go/ftg24555.jpg
3./abc/cde/go/ftg133333.gif
4./abt/cte/come/ftg24555.jpg
5./abc/cde/go/ftg133333.jpg
6./abc/cde/go/ftg24555.pdf
MY GOAL: To get only one line from lines who's first, second and third PATH are the same and have the same file EXTENSION.
Note each PATH is separated by forward slash "/". Eg in the first line of the list, the first PATH is abc, second PATH is cde and third PATH is go.
File EXTENSION is .jpg, .gif,.pdf... always at the end of the line.
HERE IS WHAT I TRIED
sort -u -t '/' -k1 -k2 -k3
My thoughts
Using / as a delimiter gives me 4 fields in each line. Sorting them with "-u" will remove all but 1 line with unique First, Second and 3rd field/PATH. But obviously, I didn't take into account the EXTENSION(jpg,pdf,gif) in this case.
MY QUESTION
I need a way to grep only 1 of the lines if the first, second and third field are same and have the same EXTENSION using "/" as delimiter to divide it into fields. I want to output it to a another file, say file2.txt.
In the file2.txt, how do I add a word say "KALI" before the extension in each line, so it will look something like /abc/cde/go/ftg13333KALI.jpg using line 1 as an example in file.txt above.
Desired Output
/abc/cde/go/ftg133333KALI.jpg
/abt/cte/come/ftg24555KALI.jpg
/abc/cde/go/ftg133333KALI.gif
/abc/cde/go/ftg24555KALI.pdf
COMMENT
Line 1,2 & 5 have the same 1st,2nd and 3rd field, with same file extension
".jpg" so only line 1 should be in the output.
Line 3 is in the output even though it has same 1st,2nd and 3rd field with
1,2 and 5, because the extension is different ".gif".
Line 4 has different 1st, 2nd and 3rd field, hence it in output.
Line 6 is in the output even though it has same 1st,2nd and 3rd field with
1,2 and 5, because the extension is different ".pdf".
$ awk '{ # using awk
n=split($0,a,/\//) # split by / to get all path components
m=split(a[n],b,".") # split last by . to get the extension
}
m>1 && !seen[a[2],a[3],a[4],b[m]]++ { # if ext exists and is unique with 3 1st dirs
for(i=2;i<=n;i++) # loop component parts and print
printf "/%s%s",a[i],(i==n?ORS:"")
}' file
Output:
/abc/cde/go/ftg133333.jpg
/abc/cde/go/ftg133333.gif
/abt/cte/come/ftg24555.jpg
/abc/cde/go/ftg24555.pdf
I split by / separately from .s in case there are .s in dir names.
Missed the KALI part:
$ awk '{
n=split($0,a,/\//)
m=split(a[n],b,".")
}
m>1&&!seen[a[2],a[3],a[4],b[m]]++ {
for(i=2;i<n;i++)
printf "/%s",a[i]
for(i=1;i<=m;i++)
printf "%s%s",(i==1?"/":(i==m?"KALI.":".")),b[i]
print ""
}' file
Output:
/abc/cde/go/ftg133333KALI.jpg
/abc/cde/go/ftg133333KALI.gif
/abt/cte/come/ftg24555KALI.jpg
/abc/cde/go/ftg24555KALI.pdf
Using awk:
$ awk -F/ '{ split($5, ext, "\\.")
if (!(($2,$3,$4,ext[2]) in files)) files[$2,$3,$4,ext[2]]=$0
}
END { for (f in files) {
sub("\\.", "KALI.", files[f])
print files[f]
}}' input.txt
/abt/cte/come/ftg24555KALI.jpg
/abc/cde/go/ftg133333KALI.gif
/abc/cde/go/ftg24555KALI.pdf
/abc/cde/go/ftg133333KALI.jpg
another awk
$ awk -F'[./]' '!a[$2,$3,$4,$NF]++' file
/abc/cde/go/ftg133333.jpg
/abc/cde/go/ftg133333.gif
/abt/cte/come/ftg24555.jpg
/abc/cde/go/ftg24555.pdf
assumes . doesn't exist in directory names (not necessarily true in general).
For example, I have a file containing a line as below:
"abc":"def"
I need to insert 123 between "abc":" and def" so that the line will become: "abc":"123def".
As "abc" appears only once so I think I can just search it and do the insertion.
How to do this with bash script such as sed or awk?
AMD$ sed 's/"abc":"/&123/' File
"abc":"123def"
Match "abc":", then append this match with 123 (& will contain the matched string "abc":")
If you want to take care of space before and after :, you can use:
sed 's/"abc" *: *"/&123/'
For replacing all such patterns, use g with sed.
sed 's/"abc" *: *"/&123/g' File
sed:
$ sed -E 's/(:")(.*)/\1123\2/' <<<'"abc":"def"'
"abc":"123def"
(:") gets :" and put in captured group 1
(.*) gets the remaining portion and put in captured group 2
in the replacement, \1123\2 puts 123 between the groups
awk:
$ awk -F: 'sub(".", "&123", $2)' <<<'"abc":"def"'
"abc" "123def"
In the sub() function, the second ($2) field is being operated on, pattern is used as . (which would match "), and in the replacement the matched portion (&) is followed by 123.
echo '"abc":"def"'| awk '{sub(/def/,"123def")}1'
"abc":"123def"
XYZNA0000778800Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
XYZNA0000778900Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
I have above file format from which I want to find a matching record. For example, match a number(7789) on line starting with XYZ and once matched look for a matching number (7345) in lines below starting with 1 until it reaches to line starting with 9. retrieve the entire line record. How can I accomplish this using shell script, awk, sed or any combination.
Expected Output:
XYZNA0000778900Z
17345000012300324000000004000000000000000
With sed one can do:
$ sed -n '/^XYZ.*7789/,/^9$/{/^1.*7345/p}' file
17345000012300324000000004000000000000000
Breakdown:
sed -n ' ' # -n disabled automatic printing
/^XYZ.*7789/, # Match line starting with XYZ, and
# containing 7789
/^1.*7345/p # Print line starting with 1 and
# containing 7345, which is coming
# after the previous match
/^9$/ { } # Match line that is 9
range { stuff } will execute stuff when it's inside range, in this case the range is starting at /^XYZ.*7789/ and ending with /^9$/.
.* will match anything but newlines zero or more times.
If you want to print the whole block matching the conditions, one can use:
$ sed -n '/^XYZ.*7789/{:s;N;/\n9$/!bs;/\n1.*7345/p}' file
XYZNA0000778900Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
This works by reading lines between ^XYZ.*7779 and ^9$ into the pattern
space. And then printing the whole thing if ^1.*7345 can be matches:
sed -n ' ' # -n disables printing
/^XYZ.*7789/{ } # Match line starting
# with XYZ that also contains 7789
:s; # Define label s
N; # Append next line to pattern space
/\n9$/!bs; # Goto s unless \n9$ matches
/\n1.*7345/p # Print whole pattern space
# if \n1.*7345 matches
I'd use awk:
awk -v rid=7789 -v fid=7345 -v RS='\n9\n' -F '\n' 'index($1, rid) { for(i = 2; i < $NF; ++i) { if(index($i, fid)) { print $i; next } } }' filename
This works as follows:
-v RS='\n9\n' is the meat of the whole thing. Awk separates its input into records (by default lines). This sets the record separator to \n9\n, which means that records are separated by lines with a single 9 on them. These records are further separated into fields, and
-F '\n' tells awk that fields in a record are separated by newlines, so that each line in a record becomes a field.
-v rid=7789 -v fid=7345 sets two awk variables rid and fid (meant by me as record identifier and field identifier, respectively. The names are arbitrary.) to your search strings. You could encode these in the awk script directly, but this way makes it easier and safer to replace the values with those of a shell variables (which I expect you'll want to do).
Then the code:
index($1, rid) { # In records whose first field contains rid
for(i = 2; i < $NF; ++i) { # Walk through the fields from the second
if(index($i, fid)) { # When you find one that contains fid
print $i # Print it,
next # and continue with the next record.
} # Remove the "next" line if you want all matching
} # fields.
}
Note that multi-character record separators are not strictly required by POSIX awk, and I'm not certain if BSD awk accepts it. Both GNU awk and mawk do, though.
EDIT: Misread question the first time around.
an extendable awk script can be
$ awk '/^9$/{s=0} s&&/7345/; /^XYZ/&&/7789/{s=1} ' file
set flag s when line starts with XYZ and contains 7789; reset when line is just 9, and print when flag is set and contains pattern 7345.
This might work for you (GNU sed):
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^XYZ[^\n]*7789/!b;/7345/p' file
Use the option -n for the grep-like nature of sed. Gather up records beginning with XYZ and ending in 9. Reject any records which do not have 7789 in the header. Print any remaining records that contain 7345.
If the 7345 will always follow the header,this could be shortened to:
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^XYZ[^\n]*7789.*7345/p' file
If all records are well-formed (begin XYZ and end in 9) then use:
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^[^\n]*7789.*7345/p' file
I am trying to print all the lines from a file before the first match. I have the same entries again in the file, but I don't need that lines. Tried
awk "{print} /${pattern}/ {exit}" and sed "/$pattern/q" (my serach is based on a variable). But both these commands are printing all the line before the last match
ex: my file is like
abc
bcd
def
xyz
def
lmno
def
xvd
when my pattern is 'def', i just need abc and bcd . but the above commands are printing, all the lines before the last 'def'. could you please provide some idea
This should work:
awk '!'"/${pattern}/{print} /${pattern}/ {exit}" input_file.txt