Passing Variable to sed [duplicate] - unix
This question already has answers here:
sed substitution with Bash variables
(6 answers)
Closed 2 years ago.
I'm having trouble passing the variable $APN to sed, tried lots of combinations of ' " \ \
First column of .csv files is a number, this code works:
cat *.csv > "$APN".csv #combine
sort -u "$APN".csv -o "$APN".csv #sort, remove duplicate APNs
sed -i '/^715/!d' combined2.csv #remove not 715
This code third line does nothing:
cat *.csv > "$APN".csv #combine
sort -u "$APN".csv -o "$APN".csv #sort, remove duplicate APNs
sed -i '/^$APN/!d' #remove not $APN
The variables inside single quotes are not expanded, you need double quotes.
sed -i "/^$APN/!d" "${APN}.csv"
Learn how to quote properly in shell, it's very important :
"Double quote" every literal that contains spaces/metacharacters and every expansion: "$var", "$(command "$var")", "${array[#]}", "a & b". Use 'single quotes' for code or literal $'s: 'Costs $5 US', ssh host 'echo "$HOSTNAME"'. See
http://mywiki.wooledge.org/Quotes
http://mywiki.wooledge.org/Arguments
http://wiki.bash-hackers.org/syntax/words
Related
How to read nth line and mth field of text file in unix
Suppose i have | delimeted file, Line1: 1|2|3|4 Line2: 5|6|7|8 Line3: 9|9|1|0 Now i need to read 3 field at second line which is 7 in above example how i can do that using Cut or Sed Command. I'm new to unix please help
A job for awk:
awk -F '|' 'NR==2{print $3}' file
or
awk -F '|' -v row=2 -v col=3 'NR==row{print $col}' file
Output:
7
This should work:
sed -n '2p' file |awk -F '|' '{print $3}'
This might work for you (GNU sed):
sed -rn '2s/^(([^|]*)\|?){3}.*/\2/p' file
Turn off automatic printing by setting the -n option, turn on easier regexp declaration by -r option. Use pattern matching and back references to replace the whole of the second line by the third field of the same line and print the result.
The address of the substitution command is limited to only the second line.
The regexp groups the non-delimited characters followed by a delimiter a specific number of times. The second group, only retains the non-delimited characters for the specific number. Each grouping is replaced by the next and so the last grouping is reported, the .* consumes the remainder of the line and so only the third field (contents of second group) is printed.
N.B. the delimiter would be present following the final column and is therefore optional \|?
Unix command for replacing strings which have "/" [duplicate]
This question already has answers here: Using different delimiters in sed commands and range addresses (3 answers) Closed 4 years ago. I know that for replacing a string (in a file which has matchstring string), we can use following command grep -rl matchstring somedir/ | xargs sed -i 's/string1/string2/g' How can I use/change the command if my string has special characters like "/"? For example: string1: "/home/folder1" string2: "/home/folder1/folder2"
As #jamieguinan mentioned in his command, almost any delimiter character can be used. So, I changed the command as following: grep -rl matchstring somedir/ | xargs sed -i 's,string1,string2,g' Where string1 and string2 are: /home/folder1 and /home/folder1/folder2, respectively.
passing variable to this particular sed command [duplicate]
This question already has answers here: bash sed fail in while loop (2 answers) Closed 8 years ago. I am first searching for a key word and once that key word is found in a file from that particular line i am supposed delete till end of file. #! /bin/csh -f set sa = `grep -n -m 1 "^Pattern" file` set s = `echo "$sa" | cut -d':' -f1` set m = `sed '$s,$d' file | tee see > /dev/null` so first line gives me the matching line with line number, second line i am getting the line number and third line i am trying to delete from line $s say 20 till last but it is not working. I have tried all combinations but it does not take the variable $s. Please help.
But you can do it much more easier with a single line of sed: sed -n '/SEARCHPATTERN/q;p -n tells to not print the lines /SEARCHPATTERN/q exits on search pattern ;p otherwise print the lines
You need to take $s out of the quotes so it will be expanded. set m = `sed $s',$d' file | tee see > /dev/null`
using sed -n with variables
I am having a log file a.log and i need to extract a piece of information from it. To locate the start and end line numbers of the pattern i am using the following. start=$(sed -n '/1112/=' file9 | head -1) end=$(sed -n '/true/=' file9 | head -1) i need to use the variables (start,end) in the following command: sed -n '16q;12,15p' orig-data-file > new-file so that the above command appears something like: sed -n '($end+1)q;$start,$end'p orig-data-file > new-file I am unable to replace the line numbers with the variables. Please suggest the correct syntax. Thanks, Rosy
When I realized how to do it, I was looking for anyway to get line number into a file containing the requested info, and display the file from that line to EOF.
So, this was my way.
with
PATTERN="pattern"
INPUT_FILE="file1"
OUTPUT_FILE="file2"
line number of first match of $PATTERN into $INPUT_FILE can be retrieved with
LINE=`grep -n ${PATTERN} ${INPUT_FILE} | awk -F':' '{ print $1 }' | head -n 1`
and the outfile will be the text from that $LINE to EOF. This way:
sed -n ${LINE},\$p ${INPUT_FILE} > ${OUTPUT_FILE}
The point here, is the way how can variables be used with command sed -n:
first witout using variables
sed -n 'N,$p' <file name>
using variables
LINE=<N>; sed -n ${LINE},\$p <file name>
Remove the single quotes thus. Single quotes turn off the shell parsing of the string. You need shell parsing to do the variable string replacements.
sed -n '('$end'+1)q;'$start','$end''p orig-data-file > new-file
How to remove blank lines from a Unix file
I need to remove all the blank lines from an input file and write into an output file. Here is my data as below. 11216,33,1032747,64310,1,0,0,1.878,0,0,0,1,1,1.087,5,1,1,18-JAN-13,000603221321 11216,33,1033196,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,059762153003 11216,33,1033246,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,000603211032 11216,33,1033280,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,055111034001 11216,33,1033287,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000378689701 11216,33,1033358,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000093737301 11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041926 11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041954 11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049326 11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049383 11216,33,1036985,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000093415580 11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781202001 11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781261305 11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781603955 11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781615746
sed -i '/^$/d' foo This tells sed to delete every line matching the regex ^$ i.e. every empty line. The -i flag edits the file in-place, if your sed doesn't support that you can write the output to a temporary file and replace the original: sed '/^$/d' foo > foo.tmp mv foo.tmp foo If you also want to remove lines consisting only of whitespace (not just empty lines) then use: sed -i '/^[[:space:]]*$/d' foo Edit: also remove whitespace at the end of lines, because apparently you've decided you need that too: sed -i '/^[[:space:]]*$/d;s/[[:space:]]*$//' foo
awk 'NF' filename awk 'NF > 0' filename sed -i '/^$/d' filename awk '!/^$/' filename awk '/./' filename The NF also removes lines containing only blanks or tabs, the regex /^$/ does not.
Use grep to match any line that has nothing between the start anchor (^) and the end anchor ($): grep -v '^$' infile.txt > outfile.txt If you want to remove lines with only whitespace, you can still use grep. I am using Perl regular expressions in this example, but here are other ways: grep -P -v '^\s*$' infile.txt > outfile.txt or, without Perl regular expressions: grep -v '^[[:space:]]*$' infile.txt > outfile.txt
sed -e '/^ *$/d' input > output Deletes all lines which consist only of blanks (or is completely empty). You can change the blank to [ \t] where the \t is a representation for tab. Whether your shell or your sed will do the expansion varies, but you can probably type the tab character directly. And if you're using GNU or BSD sed, you can do the edit in-place, if that's what you want, with the -i option. If I execute the above command still I have blank lines in my output file. What could be the reason? There could be several reasons. It might be that you don't have blank lines but you have lots of spaces at the end of a line so it looks like you have blank lines when you cat the file to the screen. If that's the problem, then: sed -e 's/ *$//' -e '/^ *$/d' input > output The new regex removes repeated blanks at the end of the line; see previous discussion for blanks or tabs. Another possibility is that your data file came from Windows and has CRLF line endings. Unix sees the carriage return at the end of the line; it isn't a blank, so the line is not removed. There are multiple ways to deal with that. A reliable one is tr to delete (-d) character code octal 15, aka control-M or \r or carriage return: tr -d '\015' < input | sed -e 's/ *$//' -e '/^ *$/d' > output If neither of those works, then you need to show a hex dump or octal dump (od -c) of the first two lines of the file, so we can see what we're up against: head -n 2 input | od -c Judging from the comments that sed -i does not work for you, you are not working on Linux or Mac OS X or BSD — which platform are you working on? (AIX, Solaris, HP-UX spring to mind as relatively plausible possibilities, but there are plenty of other less plausible ones too.) You can try the POSIX named character classes such as sed -e '/^[[:space:]]*$/d'; it will probably work, but is not guaranteed. You can try it with: echo "Hello World" | sed 's/[[:space:]][[:space:]]*/ /' If it works, there'll be three spaces between the 'Hello' and the 'World'. If not, you'll probably get an error from sed. That might save you grief over getting tabs typed on the command line.
grep . file grep looks at your file line-by-line; the dot . matches anything except a newline character. The output from grep is therefore all the lines that consist of something other than a single newline.
with awk awk 'NF > 0' filename
To be thorough and remove lines even if they include spaces or tabs something like this in perl will do it: cat file.txt | perl -lane "print if /\S/" Of course there are the awk and sed equivalents. Best not to assume the lines are totally blank as ^$ would do. Cheers
You can sed's -i option to edit in-place without using temporary file: sed -i '/^$/d' file