using variables inside string in shell-script

Question

My variable is not getting replaced by its value in shell-script:

#!/bin/bash
read -p "Enter uuid "  uuid
read -p "Enter date in format yyyymmdd: " date

echo $uuid
echo $date

a=`zgrep 'Queue for uuid $uuid' reader_$date.gz`    
b=`zgrep 'Queue for uuid 23fef66b-fcf0-4a71-8ca3-a0761dffc473' reader_$date.gz`

echo $a
echo $b

output:

Enter uuid 23fef66b-fcf0-4a71-8ca3-a0761dffc473                                         
Enter date in format yyyymmdd: 20180323
23fef66b-fcf0-4a71-8ca3-a0761dffc473
20180323

[2018-03-23 17:27:10,535: INFO/Worker-1 None None tasks/push_to_rabbit] Queue for uuid 23fef66b-fcf0-4a71-8ca3-a0761dffc473 is 35.154.190.22_2_k_event

Why is variable a empty?

And did fixing the typo make your code work? If not, then please create a minimal reproducible example - that means something the rest of us can replicate without needing your input files. — Toby Speight
– Toby Speight, Commented Mar 26, 2018 at 16:47
Actually, it won't have fixed the problem, because you used single quotes around $uuid where you wanted double quotes (so that the parameter could be expanded). — Toby Speight
– Toby Speight, Commented Mar 26, 2018 at 16:48
Try this. It should be working as @TobySpeight suggested. a=zgrep "Queue for uuid $uuid" reader_$date.gz b=zgrep 'Queue for uuid 23fef66b-fcf0-4a71-8ca3-a0761dffc473' reader_$date.gz — vishnu narayanan
– vishnu narayanan, Commented Mar 26, 2018 at 16:54
Just tried and working fine for me. Replace with double quotes to allow expansion of variable. — vishnu narayanan
– vishnu narayanan, Commented Mar 26, 2018 at 16:56
@vishnunarayanan thanks it is working. replacing single quote with double quote. worked.Now variable a has same value as b — gourav kumar
– gourav kumar, Commented Mar 26, 2018 at 17:01

Toby Speight · Accepted Answer · 2018-03-26 17:03:33Z

3

You need double-quotes where parameters should be expanded, not single-quotes.

Wrong quotation:

u=23; d=3; a=$(zgrep 'Mar $u 23' /var/log/syslog.$d.gz); echo $a

Right quotation:

u=23; d=3; a=$(zgrep "Mar $u 23" /var/log/syslog.$d.gz); echo $a
Mar 23 23:00:01 tux201t CRON[25808]: (stefan) CMD ....

edited Mar 26, 2018 at 17:03

Toby Speight

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answered Mar 26, 2018 at 17:00

user unknown

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Comments

Toby Speight · Accepted Answer · 2018-03-26 17:07:55Z

3

Replace single quotes with double quotes to allow for bash variable substitution.

From

a=`zgrep 'Queue for uuid $uuid' reader_$date.gz`

To

a=`zgrep "Queue for uuid $uuid" reader_$date.gz`

edited Mar 26, 2018 at 17:07

Toby Speight

32.1k58 gold badges82 silver badges117 bronze badges

answered Mar 26, 2018 at 17:05

vishnu narayanan

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