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String parse question

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# 1  
Old 03-06-2007
String parse question
I have a string of data that looks like this:

[1] private.enterprises.954.1.1.1.1.1.2618 \(OctetString\): U [2] private.enterprises.954.1.1.1.1.2.2618 \(OctetString\): 2618

I am trying to parse the string to only return the values after the ":". Ex from above "U" and "2618".

Any suggestions?
# 2  
Old 03-06-2007
One way:
Code:
$ cat string
#! /usr/bin/ksh

string="[1] private.enterprises.954.1.1.1.1.1.2618 \(OctetString\): U [2] private.enterprises.954.1.1.1.1.2.2618 \(OctetString\): 2618"
echo string = $string
string=${string#*: }
first=${string%% *}
second=${string##*: }
echo first = $first
echo second = $second
exit 0
$ ./string
string = [1] private.enterprises.954.1.1.1.1.1.2618 \(OctetString\): U [2] private.enterprises.954.1.1.1.1.2.2618 \(OctetString\): 2618
first = U
second = 2618

# 3  
Old 03-06-2007
Quote:
Originally Posted by mnreferee
I have a string of data that looks like this:

[1] private.enterprises.954.1.1.1.1.1.2618 \(OctetString\): U [2] private.enterprises.954.1.1.1.1.2.2618 \(OctetString\): 2618

I am trying to parse the string to only return the values after the ":". Ex from above "U" and "2618".

Any suggestions?
In any POSIX shell (bash, ksh, ash, etc.), you can use parameter expansion to extract parts of a string:

Code:
var="abc:def ghi jkj:lmn opq"
left1=${var%%:*} ## Everything to the left of the first colon
left2=${var%:*} ## Everything to the left of the rightmost colon
right1=${var##* } ## Everything to the right of the rightmost space
right2=${var#* } ## Everything to the right of the first space

# 4  
Old 03-06-2007
Thanks for the help. I will give it a try.
# 5  
Old 03-07-2007
Assuming that the variable var contains the above data

echo $var | awk -F: '{print substr($2,1,2)" " $3;}'
# 6  
Old 03-07-2007
Code:
echo "[1] private.enterprises.954.1.1.1.1.1.2618 \(OctetString\): U [2] private.enterprises.954.1.1.1.1.2.2618 \(OctetString\): 2618" | sed 's/\(.*\): \(.*\) \[\(.*\): \(.*\)/\2\
\4/'

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