How to combine multiple commands output in the one varialble in shell script?

Question

I'm using the following script to get the data of one of the variables from the database file

#!/bin/bash

sqlite3 pdu.db <<'END_SQL'
.timeout 2000
SELECT Variable_Value FROM Data Where Sr_No'7';
END_SQL

Now I wanted to store the output of the above commands in one variable. How we can store multiple commands output in one variable in the shell script?

Where are the multiple commands? There's just one bash command, sqlite3. It has multiple lines of input in the where-doc, but that doesn't create multiple commands. — Barmar
– Barmar, Commented May 6, 2022 at 8:37
What data are you getting? This just inserts and deletes, there's no SELECT. — Barmar
– Barmar, Commented May 6, 2022 at 8:37
#!/bin/bash variable=$(sqlite3 pdu.db <<'END_SQL' .timeout 2000 SELECT Variable_Value FROM Data Where Sr_No'7'; END_SQL ) — Omkar Dixit
– Omkar Dixit, Commented May 6, 2022 at 8:46
You can stuff into a variable as much information as you like (since you can append to a variable). I would rather think of how the end effect should be represented: Just all the output pieces strung together, or with a separator? Or may be an array where each element contains the output of one command execution? — user1934428
– user1934428, Commented May 6, 2022 at 9:53

Barmar · Accepted Answer · 2022-05-06 08:39:08Z

1

There's no restriction against putting a multiline command inside a command substitution.

variable=$(sqlite3 /var/www/dbs/ha.db <<'END_SQL'
.timeout 2000
INSERT INTO table1 SELECT * FROM table2;
DELETE FROM table2;
END_SQL
)

answered May 6, 2022 at 8:39

Barmar

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