I am beginner in shell script .
I have one variable containing value having = character.
I want to add quote in fields after = Character.
abc="source=TDG"
echo $abc|awk -F"=" '{print $2}'
My code is printing one field only.
my expected output is
source='TDG'
$ abc='source=TDG'
$ echo "$abc" | sed 's/[^=]*$/\x27&\x27/'
source='TDG'
[^=]*$ match non = characters at end of line
\x27&\x27 add single quotes around the matched text
With awk
$ echo "$abc" | awk -F= '{print $1 FS "\047" $2 "\047"}'
source='TDG'
-F= input field separator is =
print $1 FS "\047" $2 "\047" print first field, followed by input field separator, followed by single quotes then second field and another single quotes
See how to escape single quote in awk inside printf
for more ways of handling single quotes in print
With bash parameter expansion
$ echo "${abc%=*}='${abc#*=}'"
source='TDG'
${abc%=*} will delete last occurrence of = and zero or more characters after it
${abc#*=} will delete zero or more characters and first = from start of string
Sed would be the better choice:
echo "$abc" | sed "s/[^=]*$/'&'/"
Awk can do it but needs extra bits:
echo "$abc" | awk -F= 'gsub(/(^|$)/,"\047",$2)' OFS==
What is taking place?
Using sub to surround TDG with single quotes by its octal nr to avoid quoting problems.
echo "$abc" | awk '{sub(/TDG/,"\047TDG\047")}1'
source='TDG'
Related
/home/user/views/links/user1/gitsrc/database/src/
This is my string. I want to cut it in 2 strings such as
"/home/user/views/links/user1/"
"/database/src/"
so the delim is not actally a single character but a group of characters ie "gitsrc".
You can only define a single character as delimiter in cut.
You could use awk where the field separator can be a single character, a null string or a regular expression, e.g.
$ echo '/home/user/views/links/user1/gitsrc/database/src/' |
awk -F'gitsrc' '{ print $1 " " $2 }'
/home/user/views/links/user1/ /database/src/
or
$ echo '/home/user/views/links/user1/gitsrc/database/src/' |
awk -F'gitsrc' '{ print $1 ORS $2 }'
/home/user/views/links/user1/
/database/src/
In your shell you could or use a parameter expansion to get the first and second part:
$ str=/home/user/views/links/user1/gitsrc/database/src/
$ echo "${str%%gitsrc*}" # remove longest suffix `gitsrc*`
/home/user/views/links/user1/
$ echo "${str#*gitsrc}" # remove shortest prefix `*gitsrc`
/database/src/
I Have a file name abc.lst i ahve stored that in a variable it contain 3 words string among them i want to grep second word and in that i want to cut the word from expdp to .dmp and store that into variable
example:-
REFLIST_OP=/tmp/abc.lst
cat $REFLIST_OP
34 /data/abc/GOon/expdp_TEST_P119_*_18112017.dmp 12-JAN-18 04.27.00 AM
Desired Output:-
expdp_TEST_P119_*_18112017.dmp
I Have tried below command :-
FULL_DMP_NAME=`cat $REFLIST_OP|grep /orabackup|awk '{print $2}'`
echo $FULL_DMP_NAME
/data/abc/GOon/expdp_TEST_P119_*_18112017.dmp
REFLIST_OP=/tmp/abc.lst
awk '{n=split($2,arr,/\//); print arr[n]}' "$REFLIST_OP"
Test Results:
$ REFLIST_OP=/tmp/abc.lst
$ cat "$REFLIST_OP"
34 /data/abc/GOon/expdp_TEST_P119_*_18112017.dmp 12-JAN-18 04.27.00 AM
$ awk '{n=split($2,arr,/\//); print arr[n]}' "$REFLIST_OP"
expdp_TEST_P119_*_18112017.dmp
To save in variable
myvar=$( awk '{n=split($2,arr,/\//); print arr[n]}' "$REFLIST_OP" )
Following awk may help you on same.
awk -F'/| ' '{print $6}' Input_file
OR
awk -F'/| ' '{print $6}' "$REFLIST_OP"
Explanation: Simply making space and / as a field separator(as per your shown Input_file) and then printing 6th field of the line which is required by OP.
To see the field number and field's value you could use following command too:
awk -F'/| ' '{for(i=1;i<=NF;i++){print i,$i}}' "$REFLIST_OP"
Using sed with one of these regex
sed -e 's/.*\/\([^[:space:]]*\).*/\1/' abc.lst capture non space characters after /, printing only the captured part.
sed -re 's|.*/([^[:space:]]*).*|\1|' abc.lst Same as above, but using different separator, thus avoiding to escape the /. -r to use unescaped (
sed -e 's|.*/||' -e 's|[[:space:]].*||' abc.lst in two steps, remove up to last /, remove from space to end. (May be easiest to read/understand)
myvar=$(<abc.lst); myvar=${myvar##*/}; myvar=${myvar%% *}; echo $myvar
If you want to avoid external command (sed)
When I try to run the code below, the shell is replacing (because they are not defined as a bash variable) $4 and $2 with blanks. My question is, how do I keep bash from trying to evaluate the positional parameters for awk as its variables?
I've tried putting double and single quotes around the positional parameters, however, that did not suppress bash from interpreting them as local variables instead of strings.
This is what is returned when I echo "$x$i$y"
date -r /root/capture/capture11.mp4 | awk '{print }' | awk -F":" '/1/ {print }'
Code:
#!/bin/sh
i=$(cat /etc/hour.conf)
x="date -r /root/capture/capture"
y=".mp4 | awk '{print $4}' | awk -F\":\" '/1/ {print $2}'"
$x$i$y
Any help would be greatly appreciated!
Variables are interpolated inside double quotes. Use single quotes, or escape them like \$2.
However, the way you're trying to split up the command into separate variables won't work. Instead, you should use a function. Then you don't need to deal with quotes and escaping at all. For instance:
do_thing() {
date -r "/root/capture/capture$1.mp4" | awk '{print $4}' | awk -F':' '/1/ {print $2}'
}
do_thing "$(cat /etc/hour.conf)"
$4 is doubled quoted. Though there are single quotes, it is included in double quotes. So the single quotes are just part of the string and it won't keep the literal meaning of $. So you can escape the $:
y=".mp4 | awk '{print \$4}' | awk -F\":\" '/1/ {print \$2}'"
Or, use single quotes around the whole part:
y='.mp4 | awk "{print \$4}" | awk -F':' "/1/ {print \$2}"'
Concatenating variables like that to build a command line sort of works, but quotes within the variables don't quote anything, they'll just be taken as literal quotes.
This sort of works (but is horrible):
$ x=prin; y="tf %f\\n"; z=" 123456789"
$ $x$y$z
123456789.000000
This doesn't do what you want:
$ z='"foo bar"'; printf $y ; echo
"foo
Instead of one argument foo bar, printf gets the two arguments "foo and bar".
I have string like this
my/path/to/home/file.txt
Now I want to get the number of parts in this string on the basis of delimeter (/). So for the above string the answer would be 5. I need this in my linux shell script. How to get that without using a for loop.
$ awk -F'/' '{print NF}' <<< "my/path/to/home/file.txt"
5
-F'/' : This will tell awk that fields are separate by / .
NF : This is the last field number. In this case "my" is field 1,path is 2nd..... and file.txt is 5th field.
{print NF}: This will print the last field number.
It greps the delimiter, counts the occurences and adds 1 :
echo $(($(echo "my/path/to/home/file.txt" | grep -o "/" | wc -l)+1))
#=> 5
You can use awk:
string="my/path/to/home/file.txt"
count="$(awk -F/ '{print $NF}' <<< "$string")"
-F/ splits the string into fields based on / as the delimiter. $NF contains the number of those fields.
only pure bash (fastest way):
#!/bin/bash
a=my/path/to/home/file.txt
b=${a//[^\/]}
echo $[ ${#b} +1 ]
I have a command output from which I want to remove the double quotes ".
Command:
strings -a libAddressDoctor5.so |\
grep EngineVersion |\
awk '{if(NR==2)print}' |\
awk '{print$2}'
Output:
EngineVersion="5.2.5.624"
I'd like to know how to remove unwanted characters with awk or sed.
Use sed's substitution: sed 's/"//g'
s/X/Y/ replaces X with Y.
g means all occurrences should be replaced, not just the first one.
Using just awk you could do (I also shortened some of your piping):
strings -a libAddressDoctor5.so | awk '/EngineVersion/ { if(NR==2) { gsub("\"",""); print $2 } }'
I can't verify it for you because I don't know your exact input, but the following works:
echo "Blah EngineVersion=\"123\"" | awk '/EngineVersion/ { gsub("\"",""); print $2 }'
See also this question on removing single quotes.
tr can be more concise for removing characters than sed or awk, especially when you want to remove multiple different characters from a string.
Removing double quotes:
echo '"Hi"' | tr -d \"
# Prints Hi without quotes
Removing different kinds of brackets:
echo '[{Hi}]' | tr -d {}[]
# Prints Hi without brackets
-d stands for "delete".