cutting a particular part of a line without using grep command [closed] - shell

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RPO.RE4_SND_MSG_RECO_Q_n_message
i have an line like this. I would like to cut till RPO.RE4_SND_MSG_RECO_Q . when i used
cut -d '_' _f4
it gave oly RECO as output .please help me Newbie to unix

You can use:
s='RPO.RE4_SND_MSG_RECO_Q_n_message'
cut -d_ -f1-5 <<< "$s"
RPO.RE4_SND_MSG_RECO_Q

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tracing multi-line commands in bash [closed]

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Using the uprobe example for hooking bash's readline gives one line at a time. What/How should I hook to get multi-line commands?
I tried hooking add_history but that gives me only one line.
Example input:
cat EOF
x
<< EOF
ls \
| grep

grep for a string from URLs of different lengths [closed]

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I have a text file (foo.txt) containing a number of URLs. Each URL contains a file ID:
www.url.com/etc/one/two/72f66_59875c1ffb57b5992-b18/something/maybe/
www.url.com/etc/8823cd1-ab9532a5dc74cc904cc6bd3e2/perhaps/
www.url.com/etc/something/8407fb_80bbb9c0d/1/2/6/
My expected output is just the file IDs:
72f66_59875c1ffb57b5992-b18
8823cd1-ab9532a5dc74cc904cc6bd3e2
8407fb_80bbb9c0d
I don't yet completely understand how to leverage grep to make this happen. I have been humbled.
Those look like all hex digits, so
grep -oE '/[[:xdigit:]_-]{15,}/' foo.txt | tr -d /

grep/sed/awk a specific word from a table [closed]

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The list looks like something this:
9 vm_mail running
11 vm_webserver running
13 vm_proxy running
15 vm_wsus running
Now i only need the vm name in the middle while only using "vm" as searchterm
any ideas?
grep -Po '(?<=vm_)\S*'
gives you:
mail
webserver
proxy
wsus

Reuse of the same variable wit using echo and cut [closed]

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I'm try to make a script for checking some ethx.
If the eth is up and connected my script is working
When its not working I have a problem.
RESD=$(ssh -q vmx#$1 cat /sys/class/net/$3/dormant)
I get in RESD the following result:
cat: /sys/class/net/eth3/dormant: Invalid argument
I want to put in RESD now the first letter.
How can I do this?
Thanks
As per the comment above, this what you want to do: RESD=${RESD:0:1}? This is pure bash equivalent of RESD=$(echo $RESD | cut -c1).

Unix: Trying to increment a calender year and display [closed]

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I have a unix assignment and What I have isnt working right. It doesnt format like the normal "cal" function.
#!/bin/bash
d=`date '+%Y'`;
$((++d));
calstr=`cal $d`;
echo $calstr;
You don't need the $ in line 2, and you need to wrap the $calstr in double quotes:
#!/bin/bash
d=`date '+%Y'`;
((++d));
calstr=`cal $d`;
echo "$calstr";

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