CODEWITHSUNDEEP
mysqlbash
xis
xis

Reputation: 24850

Different results from MySQL when run directly/run via bash

I run the code

user@host:~$ mysql -h mysql-server -uuser -ppassword scans -N -B -e 
'SELECT scan_id, count(*) FROM scan_info NATURAL JOIN found_results  
WHERE fp_id = "1669" AND timestamp >= "2016-08-31"  
ORDER BY scan_id  LIMIT 1'

And the result is

52169471    9

However when I run it via bash -c I got

user@host:~$ bash -c "mysql -h mysql-server -uuser -ppassword scans -N -B -e
'SELECT scan_id, count(*)  FROM scan_info  NATURAL JOIN found_results  
WHERE fp_id = "1669" AND timestamp >= "2016-08-31"  
ORDER BY scan_id  LIMIT 1'"

I got a different result:

51278887    23

Any ideas why?

Upvotes: 0

Views: 25

Answers (1)

redneb
redneb

Reputation: 23850

There are three levels of quoting if the above command. This will cause a problem with the inner double quotes. Here's the command that bash will actual execute:

mysql -h mysql-server -uuser -ppassword scans -N -B -e 'SELECT scan_id, count(*)  FROM scan_info  NATURAL JOIN found_results  WHERE fp_id = 1669 AND timestamp >= 2016-08-31  ORDER BY scan_id  LIMIT 1'

As you can see, the inner double quotes are gone. You can solve that, by escaping the inner double quotes:

bash -c "mysql -h mysql-server -uuser -ppassword scans -N -B -e
'SELECT scan_id, count(*)  FROM scan_info  NATURAL JOIN found_results  
WHERE fp_id = \"1669\" AND timestamp >= \"2016-08-31\"  
ORDER BY scan_id  LIMIT 1'"

Upvotes: 1

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