I have multiple lines, and i need to select one of them, i have found the required lines using grep,
but now i want only the first line from the result.
How can i do it using, grep, awk, sed etc..
This is first line.
This is second line.
This is seventh line.
Using grep i got the o/p
grep "This is s" file.txt.
This is second line.
This is seventh line.
now i need the first line from this.
How can i use '\n' as a field separator.
Print the first line that matches This is s and quit with awk:
$ awk '/This is s/{print $0; exit}'
This is second line.
However GNU grep has the -m option which stops are the given number of matches:
$ grep -Fm 1 'This is s' file
This is second line.
Note: the -F is for fixed string matching instead of regular expressions.
And for completeness with sed you could do:
$ sed '/This is s/!d;q' file
This is second line.
However the example seems slightly strange as you could just do grep 'second' file.
Related
I have a file with many lines, one is:
COMPOSER_HOME=/home/glen/.composer
I want to extract the string /home/glen/.composer from this file in my shell script. How can I?
I can get the whole line with grep but not sure how to remove the first part.
Here:
grep 'COMPOSER_HOME=' file| cut -d= -f2
cut cut's by delimiter = and the 2nd portion would be whatever is After the = e.g.: /home/glen/.composer , with -f1 you would get COMPOSER_HOME
Since you tagged linux, you have GNU grep which includes PCRE
grep -oP 'COMPOSER_HOME=\K.+' file
The \K means match what comes before, then throw it out and operate on the rest of the line.
You can also use awk
awk -F "=" '$1 == "COMPOSER_HOME" {print $2}' file
Maybe this is enough
sed -nE 's/COMPOSER_HOME=(.*)/\1/p' your_file
It does not print any line unless you explicitly request it (-n), it matches the line starting with COMPOSER_HOME= and captures what follows (.*) (using () instead of \(\), thanks to -E), and puts in the replacement only what is captured. Then requests the printing of the line with the p flag of the substitution command.
I am trying to number the lines of a txt file and hide the empty ones . I use this code :
cat -n file.txt | grep . file.txt
But it doesnt work . It ignores the cat command . I want to display all the non-empty lines and number them ( the txt file is not a static one , like a list that a user can type in ).
edit : Given the great solutions below , i would also add that grep . file.txt | cat -n also worked .
I assume you want to number the lines that remain after the empty lines are removed.
Solution #1
Use sed '/^$/d' to delete the empty lines then pipe its output to cat -n to number them:
sed '/^$/d' file.txt | cat -n
The sed program contains only one command: d (delete the line). The sed commands can be prefixed by zero, one or two addresses that tell what lines the command applies to.
In this case there is only one address /^$/. It is a regex (enclosed in /) that selects the empty lines; the lines where start of the line (^) is followed by the end of the line ($).
Solution #2
You can also use grep -v '^$' to filter out the empty lines:
grep -v '^$' file.txt | cat -n
Again, ^$ is a regular expression that matches the empty lines. -v reverses the condition and tells grep to display the lines that do not match the regex.
The commands above do not modify the file. They read the content of file.txt, process it and display the result on screen.
Update
As #robc suggests in a comment, nl is even better than cat -n to number the lines. Thank you #robc, I didn't know about nl until now (I didn't know about cat -n either). It is never too late to learn new things.
This could be easily done with awk. This will print line with line numbers and ignore empty lines.
awk 'NF{print FNR,$0}' file.txt
Explanation: Adding detailed explanation for above code.
awk ' ##Starting awk program from here.
NF{ ##Checking condition if NF(number of fields) is NOT NULL in current line then do following.
print FNR,$0 ##Printing current line number by mentioning FNR and then current line value.
}
' file.txt ##Mentioning Input_file name which we are passing to awk program here.
I need to edit a text file by adding a blank line above every line starting with a period.
Before
Corn
.Apple
Words.
.Orange
Bean
After
Corn
.Apple
Words.
.Orange
Bean
Here is what I have so far.
This adds a spaces after every period. There are more in the actual file.
cat File.txt | sed -r 's/([.]+)/\n\1/g'
This displays the lines that start with a period
while read -r line; do
if [[ "$line" == "."* ]]; then
echo "$line"
fi
done < File.txt
How do I merge them together?
This produces the output that you want:
$ sed 's/^[.]/\n./' file
Corn
.Apple
Words.
.Orange
Bean
If you want to change the file in-place, use sed's -i option:
sed -i 's/^[.]/\n./' file
For Mac OSX or other BSD system, use:
sed -i '' 's/^[.]/\n./' file
We use ^ which matches only at the beginning of a line. Since we are matching a period at the beginning of the line, it is not necessary to capture a group with parentheses: we know the match is a period. All that we need to do is add a newline before that period.
with sed
sed 's/^\./\n\./'
with awk
awk '/^\./{print ""} 1'
or
awk 'sub(/^\./,"\n.") 1'
Using RegExp it could be:
cat File.txt | sed -r 's/^(\..+)/\n\1/g'
I think an awk script is going to work best.
/^\./ {print "";}
{print $0;}
Put that into a file, in this case, called "awkfile" and run it like `awk -f awkfile File.txt'
$ sed -e '/^\./i\\' pru.txt
Corn
.Apple
Words.
.Orange
Bean
this command line instructs sed(1) to search for lines beginning with a dot, and then insert a blank line before it. Look in sed(1) manual page for how to use the insert, replace and append commands.
Say for instance I'm searching a line that is like this:
Color asdf
and I use grep to find that line, like grep asdf file.txt
How would I then display Color? Learning linux is hard.
With the command line tool sed you can replace stings by using regular expressions:
echo "Color asdf" | sed 's/\([^ ]*\).*/\1/'
This part: \([^ ]*\).* is a regular expresion. The first part of the regex: [^ ]*, matches any character except a space as many times as possible and what's between the \( and \) is being captured in the variable \1. Then you also match the remaining part of the string with .* and replace all of that with only the first word which was captured by \([^ ]*\) by using \1 in the replace part of the sed command.
Here some more info about sed:
http://linux.about.com/od/commands/a/Example-Uses-Of-Sed-Cmdsedxa.htm
You could use sed:
sed -n 's/[[:space:]][[:space:]]*asdf$//p' file.txt
Details:
The -n option tells sed not to print the pattern space automatically. Basically, it doesn't output anything unless you tell it to.
The s command of sed replaces text. Here, if a line ends with asdf, preceded by at least one whitespace character, we replace all of that with nothing and then print the line (notice the p flag at the end of the s command). The printing is only done if something was actually replaced. More information about the s command can be found e. g. in the GNU sed manual.
Edit for clarity: When using single quotes, parameter expansion does not work and thus, variables won't be replaced. To use variables, use double quotes:
search=asdf
sed -n "s/[[:space:]][[:space:]]*${search}\$//p" file.txt
If you'd really like to use grep here, you could pipe the output from grep into cut:
grep -h asdf *.txt | cut -s -d -f 1
Note that there have to be two spaces after the -d option to cut - the first tells cut to use a blank as the field delimiter (I'm assuming your fields are blank-delimited rather than tab-delimited), while the second separates the -d option from the following option (-f).
But, yeah, sed or awk are probably your friends here... :-)
you can color pattern in the line using grep
grep --colour -o 'asdf' file.txt
edit: the -o option will print only the patterns
I am reading files and i am doing something like:
cat file | sed s/\ //g |awk '$0 !~ /[^0-9]/'
With this line I want to clean anything different to numbers.
But i have a problem, when the file is not sorted the command works fine, but with a sorted file the command not works, the output is empty.
Who can help me?
with grep -o '[0-9]+' not works because:
I have a file like:
311435ll3e
kk13322;.
erre433
The output is:
311435
3
13322
433
And the 3 is in the second line, the output that i need is:
3114353
13322
433
As a general rule, there is no reason to have both awk and sed appearing in the same pipe, due to a large overlap of capability, and frequently the same is true of awk/grep/sed combinations.
If you just want to suppress the non-digit characters within lines of characters, use (eg) sed -e 's/[^0-9]//g' file, or if you want to do it in place with no backup, sed -i -e 's/[^0-9]//g' file, or in place with backup to a .bak file, sed -ibak -e 's/[^0-9]//g' file.
To suppress blank lines, you can append |egrep -v '^$' after the sed, but it's more efficient to just use sed's d command to delete the pattern space and start next cycle if the pattern space is empty. For example,
sed -e 's/[^0-9]//g; /^$/d' file
does a d if the line is empty after substitution.
The form suggested in 1_CR's comment,
sed -e 's/[^0-9]//g' -e '/./!d'
is an alternative. That form tests if the line has at least one character in it, and if so does not do a d.
If you want to suppress everything in the file that's not digits, use tr -cd 0-9 < file. This suppresses line feeds also.
Note, the form tr -cd [0-9] < file or tr -cd '[0-9]' < file is not correct; it will fail to suppress ] and [ characters because tr will regard them as part of SET1.