Working with large checks if elif

Question

I am writing a script that will automatically work with MySQL. I do not understand how to implement this:

MYSQL_RESULT=mysql -u LOGIN -pPASS -D "TABLE_NAME" -e "SQL request";
for i in $MYSQL_RESULT; do
    if [ $i -le 100 ]
        then r_kills = '100';

    elif [ $i -le 500 ]
        then r_kills = '100';

    elif [ $i -le 1000 ]
        then r_kills = '500';

    echo $r_kills;
    fi
done;

sh: 21: cron.rankme.sh: r_kills: not found

Hi, check out [How to create a Minimal, Complete, and Verifiable example](https://stackoverflow.com/help/mcve) and edit your question to receive more attention. :) — iksajotien, Mar 29 '18 at 09:57
@karl-richter ,in php it works, since php works slowly, I decided to rewrite it to the shell. example php code: if ( $i <= 100 ) $type = '100'; elseif ( $i <= 500) $type = '100'; — Vlad Effect, Mar 29 '18 at 10:54
@VladEffect That doesn't explain anything related to the question. — Kalle Richter, Mar 29 '18 at 10:55
@KarlRichter Cut all $i with a smaller range and give a fixed result — Vlad Effect, Mar 29 '18 at 10:58
That's probably one of expected and actual outcome, but giving only one is as useless as giving none. — Kalle Richter, Mar 29 '18 at 11:05
@VladEffect u don't need `fi` in each elif checking, just need it at last. — Kjuly, Mar 29 '18 at 11:08
Edit the question please. Code in comments is unreadable. Also it avoids new readers from having to go through all the comments to figure out WTH your are talking about... — Nic3500, Mar 29 '18 at 11:30
@VladEffect Variable assignments in Bash mustn't have spaces around `=`. — Biffen, Mar 29 '18 at 11:54
@VladEffect `MYSQL_RESULT=mysql -u …` doesn't do what you think it does. You'll want some `$(…)` around the command. — Biffen, Mar 29 '18 at 11:59
You have 2 times '= 100' (which should be =100) in your code, while the one ( -le 100) is a subcase of the other (-le 500). That doesn't make much sense. — user unknown, Mar 29 '18 at 12:04
All semicolons except the one in the `for` line are useless and should be removed. — Jens, Mar 29 '18 at 12:10

user unknown · Answer 1 · 2018-03-29T12:06:37.903

This is an invalid assignment. You may not have spaces in front of or behind the equal sign in the shell - in contrast to most programming languages I know:

if [ $i -le 100 ]
        then r_kills = '100';

if [ $i -le 100 ]
        then r_kills=100;

Note that there isn't a way to distinguish numbers and text in the shell, either. Single- and double quotes are only used to mask special characters, including whitespace.

The message

sh: 21: cron.rankme.sh: r_kills: not found

mislead me to think, you used something which is not in the path, but in reality, the tokenizer had decided, that r_kills = '100' is a programm call to r_kills with two parameters, '=' and '100'.

The first line - without knowing mysql very well,

MYSQL_RESULT=mysql -u LOGIN -pPASS -D "TABLE_NAME" -e "SQL request";

most probably has to be

MYSQL_RESULT=$(mysql -u LOGIN -pPASS -D "TABLE_NAME" -e "SQL request";)

or

MYSQL_RESULT=($(mysql -u LOGIN -pPASS -D "TABLE_NAME" -e "SQL request";))

to create an array, since you try to iterate over the variable.

root@linux:~# ***yash cron.rankme.sh*** _[: `kills' is not a valid integer_ — Vlad Effect, Mar 29 '18 at 13:04

Working with large checks if elif

1 Answers1