I have a variable name that appears in multiple locations of a text file. This variable will always start with the same string but not always end with the same characters. For example, it can be var_name or var_name_TEXT.
I'm looking for a way to extract the first occurrence in the text file of this string starting with var_name and ending with , (but I don't want the comma in the output).
Example1: var_name, some_other_var, another_one, ....
Output: var_name
Example2: var_name_TEXT, some_other_var, another_one, ...
Output: var_name_TEXT
grep -oPm1 '\bvar_name[^, ]*(?=,)' file | head -1
match and output only variables starting with var_name and ending with comma, do not include comma in the output, quit after the first line of match and pick the first match on that line (if there are more than one)
ps. you have to include space in the regex as well.
I suggest with GNU grep:
grep -o '\bvar_name[^,]*' file | head -n 1
All you need is (GNU awk):
$ awk 'match($0,/\<var_name[^,]*/,a){print a[0]; exit}' file
var_name_TEXT
To print the field only (i.e., var_name or var_name_TEXT only; not the line containing it) you could use awk:
awk -F, '{for (i=1;i<=NF;i++) if ($i~/^var_name/) print $i}' file
If you actually have spaces before or after the commas (as you show in your example) you can change to awk field separator:
awk -F"[, ]+" '{for (i=1;i<=NF;i++) if ($i~/^var_name/) print $i}' file
You can also use GNU grep with a word boundary assertion:
grep -o '\bvar_name[^,]*' file
Or GNU awk:
awk '/\<var_name/' file
If you want only one considered, add exit to awk or -m 1 to grep to exit after the first match.
Related
FILE1.TXT
0020220101
or
01 20220101
Need to extra date part from file where text starts from 2
Options tried:
t_FILE_DT1='awk -F"2" '{PRINT $NF}' FILE1.TXT'
t_FILE_DT2='cut -d'2' -f2- FILE1.TXT'
echo "$t_FILE_DT1"
echo "$t_FILE_DT2"
1st output : 0101
2nd output : 0220101
Expected Output: 20220101
Im new to linux scripting. Could some one help guide where Im going wrong?
Use grep like so:
echo "0020220101\n01 20220101" | grep -P -o '\d{8}\b'
20220101
20220101
Here, GNU grep uses the following options:
-P : Use Perl regexes.
-o : Print the matches only (1 match per line), not the entire lines.
SEE ALSO:
grep manual
perlre - Perl regular expressions
Using any awk:
$ awk '{print substr($0,length()-7)}' file
20220101
20220101
The above was run on this input file:
$ cat file
0020220101
01 20220101
Regarding PRINT $NF in your question - PRINT != print. Get out of the habit of using all-caps unless you're writing Cobol. See correct-bash-and-shell-script-variable-capitalization for some reasons.
The 2 in your scripts is telling awka and cut to use the character 2 as the field separator so each will carve up the input into substrings everywhere a 2 occurs.
The 's in your question are single quotes used to make strings literal, you were intending to use backticks, `cmd`, but those are deprecated in favor of $(cmd) anyway.
I would instead of looking for "after" the 2 .. (not having to worry about whether there is a space involved as well) )
Think instead about extracting the last 8 characters, which you know for fact is your date ..
input="/path/to/txt/file/FILE1.TXT"
while IFS= read -r line
do
# read in the last 8 characters of $line .. You KNOW this is the date ..
# No need to worry about exact matching at that point, or spaces ..
myDate=${line: -8}
echo "$myDate"
done < "$input"
About the cut and awk commands that you tried:
Using awk -F"2" '{PRINT $NF}' file will set the field separator to 2, and $NF is the last field, so printing the value of the last field is 0101
Using cut -d'2' -f2- file uses a delimiter of 2 as well, and then print all fields starting at the second field, which is 0220101
If you want to match the 2 followed by 7 digits until the end of the string:
awk '
match ($0, /2[0-9]{7}$/) {
print substr($0, RSTART, RLENGTH)
}
' file
Output
20220101
The accepted answer shows how to extract the first eight digits, but that's not what you asked.
grep -o '2.*' file
will extract from the first occurrence of 2, and
grep -o '2[0-9]*' file
will extract all the digits after every occurrence of 2. If you specifically want eight digits, try
grep -Eo '2[0-9]{7}'
maybe also with a -w option if you want to only accept a match between two word boundaries. If you specifically want only digits after the first occurrence of 2, maybe try
sed -n 's/[^2]*\(2[0-9]*\).*/\1/p' file
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 \|?
I have a file that, occasionally, has split lines. The split is signaled by the fact that the line starts with a space, empty line or a nonnumeric character. E.g.
40403813|7|Failed|No such file or directory|1
40403816|7|Hi,
The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
...
I'd like join the split line back with the previous line (as mentioned below):
40403813|7|Failed|No such file or directory|1
40403816|7|Hi, The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
...
using a Unix command like sed/awk. I'm not clear how to join a line with the preceeding one.
Any suggestion?
awk to the rescue!
awk -v ORS='' 'NR>1 && /^[0-9]/{print "\n"} NF' file
only print newline when the current line starts with a digit, otherwise append rows (perhaps you may want to add a space to ORS if the line break didn't preserve the space).
Don't do anything based on the values of the strings in your fields as that could go wrong. You COULD get a wrapping line that starts with a digit, for example. Instead just print after every complete record of 5 fields:
$ awk -F'|' '{rec=rec $0; nf+=NF} nf>=5{print rec; nf=0; rec=""}' file
40403813|7|Failed|No such file or directory|1
40403816|7|Hi, The Conversion System could not be reached.|No such file or directory||1
40403818|7|Failed|No such file or directory|1
Try:
awk 'NF{printf("%s",$0 ~ /^[0-9]/ && NR>1?RS $0:$0)} END{print ""}' Input_file
OR
awk 'NF{printf("%s",/^[0-9]/ && NR>1?RS $0:$0)} END{print ""}' Input_file
It will check if each line starts from a digit or not if yes and greater than line number 1 than it will insert a new line with-it else it will simply print it, also it will print a new line after reading the whole file, if we not mention it, it is not going to insert that at end of the file reading.
If you only ever have the line split into two, you can use this sed command:
sed 'N;s/\n\([^[:digit:]]\)/\1/;P;D' infile
This appends the next line to the pattern space, checks if the linebreak is followed by something other than a digit, and if so, removes the linebreak, prints the pattern space up to the first linebreak, then deletes the printed part.
If a single line can be broken across more than two lines, we have to loop over the substitution:
sed ':a;N;s/\n\([^[:digit:]]\)/\1/;ta;P;D' infile
This branches from ta to :a if a substitution took place.
To use with Mac OS sed, the label and branching command must be separate from the rest of the command:
sed -e ':a' -e 'N;s/\n\([^[:digit:]]\)/\1/;ta' -e 'P;D' infile
If the continuation lines always begin with a single space:
perl -0000 -lape 's/\n / /g' input
If the continuation lines can begin with an arbitrary amount of whitespace:
perl -0000 -lape 's/\n(\s+)/$1/g' input
It is probably more idiomatic to write:
perl -0777 -ape 's/\n / /g' input
You can use sed when you have a file without \r :
tr "\n" "\r" < inputfile | sed 's/\r\([^0-9]\)/\1/g' | tr '\r' '\n'
How to use "grep" shell command to show specific word from a line starting with a specific word.
Ex:
I want to print a string "myFTPpath/folderName/" from the line starting with searchStr in the below mentioned line.
searchStr:somestring:myFTPpath/folderName/:somestring
Something like this with awk:
awk -F: '/^searchStr/{print $3}' File
From all the lines starting with searchStr, print the 3rd field (field seperator set as :)
Sample:
AMD$ cat File
someStr:somestring:myFTPpath/folderName/:somestring
someStr:somestring:myFTPpath/folderName/:somestring
searchStr:somestring:myFTPpath/folderName/:somestring
someStr:somestring:myFTPpath/folderName/:somestring
AMD$ awk -F: '/^searchStr/{print $3}' File
myFTPpath/folderName/
Remember that grep isn't the only tool that can usefully do searches.
In this particular case, where the lines are naturally broken into fields, awk is probably the best solution, as #A.M.D's answer suggests.
For more general case edits, however, remember sed's -n option, which suppresses printing out a line after edits:
sed -n 's/searchStr:[^:]*:\([^:]*\):.*/\1/p' input-file
The -n suppresses automatic printing of the line, and the trailing /p flag explicitly prints out lines on which there is a substitution.
This matching pattern is fiddly – use awk in this fielded case – but don't forget sed -n.
You could get the desired output with grep itself but you need to enable -P and -o parameters.
$ echo 'searchStr:somestring:myFTPpath/folderName/:somestring' | grep -oP '^searchStr:[^:]*:\K[^:]*'
myFTPpath/folderName/
\K discards the characters which are matched previously from printing at the final leaving only the characters which are matched by the pattern exists next to \K. Here we used \K instead of a variable length positive lookbehind assertion.
In count (non-blank) lines-of-code in bash they explain how to count the number of non-empty lines.
But is there a way to count the number of blank lines in a file? By blank line I also mean lines that have spaces in them.
Another way is:
grep -cvP '\S' file
-P '\S'(perl regex) will match any line contains non-space
-v select non-matching lines
-c print a count of matching lines
If your grep doesn't support -P option, please use -E '[^[:space:]]'
One way using grep:
grep -c "^$" file
Or with whitespace:
grep -c "^\s*$" file
You can also use awk for this:
awk '!NF {sum += 1} END {print sum}' file
From the manual, "The variable NF is set to the total number of fields in the input record". Since the default field separator is the space, any line consisting in either nothing or some spaces will have NF=0.
Then, it is a matter of counting how many times this happens.
Test
$ cat a
aa dd
ddd
he llo
$ cat -vet a # -vet to show tabs and spaces
aa dd$
$
ddd$
$
^I$
he^Illo$
Now let's' count the number of blank lines:
$ awk '!NF {s+=1} END {print s}' a
3
grep -v '\S' | wc -l
(On OSX the Perl expressions are not available, -P option)
grep -cx '\s*' file
or
grep -cx '[[:space:]]*' file
That is faster than the code in Steve's answer.
Using Perl one-liner:
perl -lne '$count++ if /^\s*$/; END { print int $count }' input.file
To count how many useless blank lines your colleague has inserted in a project you can launch a one-line command like this:
blankLinesTotal=0; for file in $( find . -name "*.cpp" ); do blankLines=$(grep -cvE '\S' ${file}); blankLinesTotal=$[${blankLines} + ${blankLinesTotal}]; echo $file" has" ${blankLines} " empty lines." ; done; echo "Total: "${blankLinesTotal}
This prints:
<filename0>.cpp #blankLines
....
....
<filenameN>.cpp #blankLines
Total #blankLinesTotal