Mysql array in shell script (Linux)

Question

I am trying to read out ids from mysql and then delete the entry with id-1.

today_time=$(date -d "now" '+%d-%b-%Y %H:%M:%S')  
query_find_id="SELECT id FROM guestlist WHERE value <= '$today_time'"

array_query=$(mysql -u $USER -p$PASSWORD $TABLE_NAME -e "$query_find_id")

array_id=($(for i in $array_query; do echo $i; done))

for j in "${array_id[@]}"; k=$j-1; mysql -u $USER -p$PASSWORD $TABLE_NAME -e "DELETE FROM guestlist WHERE id = '$k'"; done

The array is empty every time. What am I doing wrong?

learn to use set -vx and examine the output from each line. You would probably see that query_find_id is empty when you get to the 3rd line. (Agreeing about answer from @tgx.) Good luck to all. — shellter
– shellter, Commented Sep 13, 2014 at 20:49

lxg · Accepted Answer · 2014-09-13 19:12:25Z

1

If your value column is a DATETIME, your date format in today_time should be %Y-%m-%d%H:%M:%S.

answered Sep 13, 2014 at 19:12

lxg

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2 Comments

Simon Over a year ago

Actually my value in the list is something like '12 Sep 2014 15:00:00'

lxg Over a year ago

You're using WHERE value <= '$today_time' in your query. What column type is value?

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