Script to extract a part of string

Question

ns/APAC_BankTransfers_Publish/CMB/services/svcPublishBankTransfers/flow.xml

In the above line and similar lines like this I want to extract whatever is present in between services and flow.xml and save it to a variable DIST.

The output should be svcPublishBankTransfers.

Charles Duffy · Accepted Answer · 2014-02-19 15:23:28Z

5

Using parameter expansion mechanisms available in POSIX sh:

s=ns/APAC_BankTransfers_Publish/CMB/services/svcPublishBankTransfers/flow.xml
s=${s%/flow.xml}    # remove "/flow.xml"
s=${s##*/services/} # remove everything before "services"
echo "$s"

This has the advantages of being purely in-process (so faster than approaches that require piping through an external tool), and compatible with all POSIX shells (ash, dash, ksh, etc).

References:

edited Feb 19, 2014 at 15:23

answered Feb 19, 2014 at 15:16

Charles Duffy

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anubhava · Accepted Answer · 2014-02-19 15:17:28Z

1

Using BASH regex:

s='ns/APAC_BankTransfers_Publish/CMB/services/svcPublishBankTransfers/flow.xml'
[[ "$s" =~ /([^/]+)/[^/]*$ ]] && echo "${BASH_REMATCH[1]}"
svcPublishBankTransfers

OR else:

[[ "$s" =~ /services/([^/]+)/flow\.xml ]] && echo "${BASH_REMATCH[1]}"
svcPublishBankTransfers

answered Feb 19, 2014 at 15:17

anubhava

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C.B. · Accepted Answer · 2014-02-19 15:17:43Z

0

bash$ echo "ns/APAC_BankTransfers_Publish/CMB/services/svcPublishBankTransfers/flow.xml" | cut -d '/' -f 5
svcPublishBankTransfers
bash$

answered Feb 19, 2014 at 15:17

C.B.

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