How do I pull a substring from a string. For example, from the string:
'/home/auto/gift/surprise'
take only:
'/home/auto/'
Note that '/home/auto/gift/surprise' may vary, i.e., instead of having 4 directory levels, it may go to 6 or 8, yet I'm only interested in the first 2 folders.
Here's what I've tried so far, without success:
$ pwd
'/home/auto/gift/surprise'
$ pwd | sed 's,^\(.*/\)\?\([^/]*\),\1,'
'/home/auto/gift/'
I think it is better to use cut for this:
$ echo "/home/auto/gift/surpris" | cut -d/ -f1-3
/home/auto
$ echo "/home/auto/gift/surpris/bla/bla" | cut -d/ -f1-3
/home/auto
Note that cut -d/ -f1-3 means: strip the string based on the delimiter /, then print from the 1st to the 3rd parts.
Or also awk:
$ echo "/home/auto/gift/surpris" | awk -F/ 'OFS="/" {print $1,$2,$3}'
/home/auto
$ echo "/home/auto/gift/surpris/bla/bla" | awk -F/ 'OFS="/" {print $1,$2,$3}'
/home/auto
You may use parameter substitution, which is POSIX defined:
$ s="/home/auto/gift/surprise"
$ echo ${s%/*/*}
/home/auto
Related
Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 is my string and the result I want is vm-1.0.3
What is the best way to do this
Below is what I tried
$ echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F _ {'print $2'} | awk -F - {'print $1,$2'}
vm 1.0.3
I also tried
$ echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F _ {'print $2'} | awk -F - {'print $1"-",$2'}
vm- 1.0.3
Here I do not need space in between
I tried using cut and I got the expected result
$ echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F _ {'print $2'} | cut -c 1-8
vm-1.0.3
What is the best way to do the same?
Making assumptions from the 1 example you provided about what the general form of your input will be so it can handle that robustly, using any sed:
$ echo 'Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2' |
sed 's/^[^-]*-[^-]*-[^_]*_\(.*\)-[^-]*$/\1/'
vm-1.0.3
or any awk:
$ echo 'Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2' |
awk 'sub(/^[^-]+-[^-]+-[^_]+_/,"") && sub(/-[^-]+$/,"")'
vm-1.0.3
You don't need 2 calls to awk, but your syntax with the single quotes outside the curly's, including printing the hyphen:
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 |
awk -F_ '{print $2}' | awk -F- '{print $1 "-" $2}'
If your string has the same format, let the field separator be either - or _
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F"[-_]" '{print $4 "-" $5}'
Or split the second field on - and print the first 2 parts
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 | awk -F_ '{
split($2,a,"-")
print a[1] "-" a[2]
}'
Or with gnu-awk a bit more specific match with a capture group:
echo Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2 |
awk 'match($0, /^Apps-[^_]*_(vm-[0-9]+\.[0-9]+\.[0-9]+)/, a) {print a[1]}'
Output
vm-1.0.3
This is the easiest I can think of:
echo "Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2" | cut -c 25-32
Obviously you need to be sure about the location of your characters. In top of that, you seem to be have two separators: '_' and '-', while both characters also are part of the name of your entry.
echo 'Apps-10.00.00R000-B1111_vm-1.0.3-x86_64.qcow2' | sed -E 's/^.*_vm-([0-9]+).([0-9]+).([0-9]+)-.*/vm-\1.\2.\3/'
Lets say we got date and time to a log in the format of
2022-05-18-11:57:140100
I need to remove seconds from this time. It means the final output should be like
2022-05-18-11:57
I tried the following
echo "2022-05-18-11:57:140100" | rev | cut -d/ -f6- | rev`:"
this was not successful and I have no idea how this cut command works. can anybody explain. Thanks in advance
1st solution: With GNU awk you can simply do it like following. Simple explanation would be, set FS and OFS as : and then in main block of awk program decrease NF with 1 and print the line.
echo "2022-05-18-11:57:140100" | awk 'BEGIN{FS=OFS=":"} NF--'
2nd solution: In any awk using awk's match function please try following.
echo "2022-05-18-11:57:140100" | awk 'match($0,/.*:/){print substr($0,RSTART,RLENGTH-1)}'
3rd solution: Using sed's capability of capturing groups(keep values in temp buffer) please try following code.
echo "2022-05-18-11:57:140100" | sed -E 's/(^[^:]*):([^:]*):.*/\1:\2/'
4th solution(OP's efforts FIX): Fixing OP's efforts using rev + cut + rev combination here.
echo "2022-05-18-11:57:140100" | rev | cut -d':' -f2- | rev
This can be done easily in bash shell also:
s='2022-05-18-11:57:140100'
echo "${s%:*}"
2022-05-18-11:57
# or else use cut
cut -d: -f1-2 <<< "$s"
2022-05-18-11:57
# or sed
sed 's/:[^:]*$//' <<< "$s"
2022-05-18-11:57
any one of these 5 solutions should work for gawk, mawk-1, mawk-2, and macos nawk :
echo "2022-05-18-11:57:140100" |
mawk2 NF=NF FS=':[^:]*$' OFS=
or
gawk -- --NF FS=':[^:]*$'
or
gawk NF-- FS=':[^:]*$'
or
nawk NF=1 FS=':[^:]*$'
or
mawk '$!_=$!_' FS=':[^:]*$'
2022-05-18-11:57
The most compacted version would be like
mawk -F':[^:]*$' NF--
I need to write a simple bash script that takes a text line
some-pattern something-else
and erases some-pattern and returns only something-else. I wrote a script to do the opposite with grep -o, but I don't know how I could do with this case. Any help is very much appreciated.
sample input:
"SNMPv2::sysLocation.0 = STRING: someLocation"
Desired Output:
"someLocation"
Considering " are NOT in your sample Input_file and expected output, could you please try following with GNU grep.
grep -oP '.*STRING: \K(.*)' Input_file
someLocation
For \K explanation:
\K is a PCRE extension to regex syntax discarding content prior to
that point from being included in match output
You can use sed to delete the part in front of what you want to keep.
Given:
$ echo "$s"
"SNMPv2::sysLocation.0 = STRING: someLocation"
You can do:
$ echo "$s" | sed -nE 's/^.*(someLocation)/\1/p'
someLocation
And if you want to add quotes:
$ echo "$s" | sed -nE 's/^.*(someLocation)/"\1"/p'
"someLocation"
If the portion after STRING: is variable, not fixed, you can use STRING: and the capture anchor:
$ echo "$s" | sed -nE 's/^.*STRING:[[:space:]]*(.*)/"\1"/p'
"someLocation"
Or, sed to capture and print the last word after the last space:
$ echo "$s" | sed -nE 's/([^[:space:]]*$)/\1/p'
You can also use awk if the last word is space separated from the other fields:
$ echo "$s" | awk '{print $NF}'
Or a pipeline with cut and rev works too:
$ echo "$s" | rev | cut -d' ' -f 1 | rev
You can use: echo ${STRING} | awk -F" " '/someLocation/ { print $NF }'
-F will use space (represented by double-quotes with space between them) as separator; /someLocation/ will search for your location; { print $NF } will show the last part of your string (which, I believe, is the place where location is.
I have some file xxx.conf in text format. I have some text "disablelog = 1" in this file.
When I use
grep -r "disablelog" oscam.conf
output is
disablelog = 1
But i need only value 1.
Do you have some idea please?
one way is to use awk to print just the value
grep -r "disablelog" oscam.conf | awk '{print $3}'
you could also use sed to replace diablelog = with empty
grep -r 'disablelog' oscam.conf | sed -e 's/disablelog = //'
If you also want to get the lines with or without space before and after = use
grep -r 'disablelog' oscam.conf | sed 's/disablelog\s*=\s*//'
above command will also match
disablelog=1
Assuming you need it as a var in a script:
#!/bin/bash
DISABLELOG=$(awk -F= '/^.*disablelog/{gsub(/ /,"",$2);print $2}' /path/to/oscam.conf)
echo $DISABLELOG
When calling this script, the output should be 1.
Edit: No matter wether there is whitespace or not between the equals sign and the value, the above will handle that. The regex should be anchored in either way to improve performance.
Try:
grep -r "disablelog" oscam.conf | awk -F= '{print $2}'
Just for fun a solution without awk
grep -r disablelog | cut -d= -f2 | xargs
xargs is used here to trim the whitespace
Can any one advise how to search on linux for some data between a tilde character. I need to get IP data however its been formed like the below.
Details:
20110906000418~118.221.246.17~DATA~DATA~DATA
One more:
echo '20110906000418~118.221.246.17~DATA~DATA~DATA' | sed -r 's/[^~]*~([^~]+)~.*/\1/'
echo "20110906000418~118.221.246.17~DATA~DATA~DATA" | cut -d'~' -f2
This uses the cut command with the delimiter set to ~. The -f2 switch then outputs just the 2nd field.
If the text you give is in a file (called filename), try:
grep "[0-9]*~" filename | cut -d'~' -f2
With cut:
echo "20110906000418~118.221.246.17~DATA~DATA~DATA" | cut -d~ -f2
With awk:
echo "20110906000418~118.221.246.17~DATA~DATA~DATA"
| awk -F~ '{ print $2 }'
In awk:
echo '20110906000418~118.221.246.17~DATA~DATA~DATA' | awk -F~ '{print $2}'
Just use bash
$ string="20110906000418~118.221.246.17~DATA~DATA~DATA"
$ echo ${string#*~}
118.221.246.17~DATA~DATA~DATA
$ string=${string#*~}
$ echo ${string%%~*}
118.221.246.17
one more, using perl:
$ perl -F~ -lane 'print $F[1]' <<< '20110906000418~118.221.246.17~DATA~DATA~DATA'
118.221.246.17
bash:
#!/bin/bash
IFS='~'
while read -a array;
do
echo ${array[1]}
done < ip
If string is constant, the following parameter expansion performs substring extraction:
$ a=20110906000418~118.221.246.17~DATA~DATA~DATA
$ echo ${a:15:14}
118.221.246.17
or using regular expressions in bash:
$ echo $(expr "$a" : '[^~]*~\([^~]*\)~.*')
118.221.246.17
last one, again using pure bash methods:
$ tmp=${a#*~}
$ echo $tmp
118.221.246.17~DATA~DATA~DATA
$ echo ${tmp%%~*}
118.221.246.17