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Assigning variable to output gives error with expected result

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# 1  
Old 10-10-2016
Assigning variable to output gives error with expected result
Hello,
I am trying to print out the first string matching query with grep and I need your help.

My scenario:

Database
Code:
John F 
4433 Street No 88 CA
Elisabeth Taylor
7733 Street No 26 ON
Jack Nicholson
0133 Green Park No 34 AR
John F 2
9399 Southpark No 02D UT

test.sh
Code:
#!/bin/bash
A="$(cat Database | grep "$1" -A1 | tail -n 1)"
printf "$A"

When I run
Code:
./test.sh John F

, it prints out the second John F.
How can I print the first one and how come can I make it searching accurately?

The most importan question is:
What would happen if I simultaneously and continuously run this script at the background with different search queries ( ten or twenty script.sh processes will be running I mean), do you think that $A will be confused and will give wrong answers?

I'd appreciate if you could give an answer to solve this or lead me to right direction.

Thanks in advance
Boris


Moderator's Comments:
Mod Comment Please use CODE tags correctly as required by forum rules!
Last edited by RudiC; 10-10-2016 at 05:57 PM.. Reason: Added CODE tags.
# 2  
Old 10-10-2016
The most important question first: ALL (interactive as well as background) processes run in their own environment and thus there's no danger the variables although commonly named will be confused.

For the other problem: in fact, grep matches and thus lists BOTH John Fs, but the tail -n1 command removes all but the very last line of grep's output.
You could use grep's -x (--line-regexp) option (if available) to eliminate the John F 2 line, but - it will fail unless you set $1 to "John F " as the file has a trailing space in this line...
Code:
set -- "John F"
grep -x "$1" -A1 file | tail -n 1
set -- "John F "
grep -x "$1" -A1 file | tail -n 1
4433 Street No 88 CA

This User Gave Thanks to RudiC For This Post:
baris35
# 3  
Old 10-10-2016
Note also that when you invoke your script with:
Code:
./test.sh John F

$1 in your script expands to John; not John F. If you want your script to use John F as your search pattern, you need to call it with:
Code:
./test.sh "John F"

or:
Code:
./test.sh 'John F'

or:
Code:
./test.sh John\ F

# 4  
Old 10-10-2016
Hi,
In complement to say RudiC, you could use also option -m nth occurence (if available):
Code:
$ grep -A1 -m 1 "John F" file | tail -1
4433 Street No 88 CA
$ grep -A1 -m 2 "John F" file | tail -1
9399 Southpark No 02D UT

Beware: One line with 2 occurrences will count for 2 to -m option. So in this case, if you would second line with occurence you must set -m 3.

Regards.
# 5  
Old 10-10-2016
Thanks for your answers Rudic and Don,

In my understanding (--line-regexp) is not available in my computer as below ways are not giving an output:


Database:
Code:
"John F "
4433 Street No 88 CA
"Elisabeth Taylor"
7733 Street No 26 ON
Jack Nicholson
0133 Green Park No 34 AR
"John F 2"
9399 Southpark No 02D UT

test.sh

Code:
#!/bin/bash
A="$(cat Database | grep -x "$1" -A1 | tail -n 1)"
printf "$A"

Terminal:
Code:
./test.sh "John F"
./test.sh 'John F'
./test.sh John\ F

Only when I do not use -x I see output.
thanks
Boris


Moderator's Comments:
Mod Comment Please use CODE tags as required by forum rules!
Last edited by RudiC; 10-11-2016 at 04:12 AM.. Reason: Added CODE tags.
# 6  
Old 10-10-2016
I was responding to your 1st post in this thread (where there was no -x option in your grep command). It doesn't work with -x because of the trailing space in your file. If you want to exactly match a line that ends with a space you'll need to use something like:
Code:
./test.sh "John F "

to match either line,
Code:
./test.sh "John F $"

to match only the first line, or
Code:
./test.sh "John F 2"

to match only the last line.
This User Gave Thanks to Don Cragun For This Post:
baris35
# 7  
Old 10-10-2016
Thanks Don,
The one with $ worked.


Thank You All
Boris
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