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  #1  
Old November 22nd, 2012, 12:24 PM
ttremain ttremain is offline
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Without backticks?
Is there a way to do this without using backticks?

Code:
show databases WHERE `Database` LIKE "central\_%"


I need to include this in a linux command line like:

Code:
for database in `mysql -Bse 'show databases WHERE `Database` LIKE "central\_%" OR `Database` LIKE "facebook\_%"`
do
  Something
done


However backticks within backticks don't work too well...
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  #2  
Old November 23rd, 2012, 04:54 AM
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Quote:
Originally Posted by ttremain
Is there a way to do this without using backticks?
yes, there is

don't use backticks
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  #3  
Old November 23rd, 2012, 07:17 AM
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Quote:
Originally Posted by r937
yes, there is

don't use backticks


I've tried the same query without them and it doesn't work.
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Old November 23rd, 2012, 07:48 AM
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well, i guess i should have tested that myself first

sorry

yes, you need the backticks around the `Database` column name produced by the SHOW command

as for the backticks around the linux command, i'm sorry, i don't do linux, but i'd be ~very~ suprised if the backticks were legal there
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Old November 23rd, 2012, 08:31 AM
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Sinec there isn't a way to do it by modifying the mySQL command itself, I've just posted the same question in the Linux forum, to see if there is a way on that part of the puzzle.

I'm hoping there is some fancy workaround that I have not considered (or had no idea could be done)

Thanks
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  #6  
Old November 26th, 2012, 06:03 AM
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Have you tried quoting them?

Code:
show databases WHERE \`Database\` LIKE "central\_%"
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  #7  
Old November 26th, 2012, 06:25 AM
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Quote:
Originally Posted by SimonJM
Have you tried quoting them?

Code:
show databases WHERE \`Database\` LIKE "central\_%"


Yes, I tried that.

I found my solution (actually two solutions) by modifying the Bash script itself, instead of altering the mySQL query.

There is an additional thread, with solutions, in the Linux forum here:
http://forums.devshed.com/linux-help-33/backticks-934843.html

Thank you!
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  #8  
Old November 27th, 2012, 06:48 AM
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It worked for me ...

Code:
ehb@phalanx:~/Desktop$ mysql -uroot -pXXXXXXX -Be "show databases where \`database\` like 'm%';"
Database
mysql
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Old November 27th, 2012, 08:49 AM
ttremain ttremain is offline
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Quote:
Originally Posted by SimonJM
It worked for me ...


It does, until you wrap THAT with the required Linux code that ALSO includes backticks.

Remember, the actual problem consisted of having to wrap that in

Code:
for database in `mysql -uroot -pXXXXXXX -Be "show databases where \`database\` like 'm%';"`
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Old November 27th, 2012, 11:22 AM
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Backticks are deprecated. Use $( command ) instead.
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  #11  
Old November 28th, 2012, 07:56 PM
SimonJM SimonJM is offline
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I seem to recall having used nested backticks before, but it was a mess, and it was for nested commands, not delimiting a name - though I do not think that should matter.
As b49P23TIvg says, however, you should really replace the command quotes of `...` with $(...). Apart from backticks being deprecated I, personally, find the $(...) syntax far easier to read.
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Old November 28th, 2012, 08:09 PM
ttremain ttremain is offline
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I hope people don't find this too confusing...

This original thread was posted in the mySQL forum, then a second thread in the linux forum to tackle it from another direction.

It looks like this thread was later moved to the linux forum when the direction changed...

I hope that clears some things up a bit.

One of the last solutions was this:

Code:
function doit {
    mysql -Bse 'show databases WHERE `Database` LIKE "central\_%" OR `Database` LIKE "facebook\_%"'
}

for database in `doit` ; do 
    echo $database
done


I also can do it this way by changing the requirements just a bit:

Code:
for database in `mysql -Bse 'show databases'`
do
  if [[ $database =~ ^(facebook|central)\_ ]]
  then
    do something 
  fi
done




I assume the later can be modified using the $(...) syntax???
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b49P23TIvg agrees: Functions, what a thought. Great.
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  #13  
Old November 29th, 2012, 08:02 AM
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It sure can ...
Code:
ehb@phalanx:~$ for db in $(mysql -uroot -pXXXXXX -Bse "show databases where \`database\` like '%';")
do
  echo "Database is: " $db
done

Produces output like:
Code:
Database is:  information_schema
Database is:  Test
Database is:  VampireHalls
Database is:  mysql
Last edited by SimonJM : November 29th, 2012 at 08:07 AM.
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