Assigning result of a select query to a variable in shell script

Question

I am trying to execute select query and store the result in a variable in shell script. Below is the script

   mysql="mysql -h localhost -u root"
   db="testdb";
   echo "Select Name from Details"|${mysql} ${db};

The above statement gets executed with the column 'Name' getting printed with its value. But when i try to assign the result like below

   var = `"Select Name from Details"|${mysql} ${db};`
   echo $var

   or

   var = $("Select Name from Details"|${mysql} ${db};)
   echo $var

I get the below error 'var: not found'. Where am i going wrong in trying to assign the result of the query to the variable?

Maheswaran Ravisankar · Accepted Answer · 2014-07-14 15:53:57Z

3

   var=`echo "Select Name from Details"|${mysql} ${db};`
   echo $var

There should not be any space between the variable name and =. The shell else interprets the variable name to be a command/executable. And hence the error!

edited Jul 14, 2014 at 15:53

answered Jul 14, 2014 at 15:43

Maheswaran Ravisankar

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7 Comments

Aarish Ramesh Over a year ago

i tried removing the space and executing but it throws an error of "Select Name from Details Not Found". Why does it occur?

Maheswaran Ravisankar Over a year ago

try my updated answer please. Added echo to the String

Aarish Ramesh Over a year ago

@Maheswaren: I get the result as 'Name aarish'. How do i parse it so as to retrieve the name alone?

Maheswaran Ravisankar Over a year ago

try the -ss option to turn off the headers.

Maheswaran Ravisankar Over a year ago

Sorry @aarish , Not Sure abt turning headers off in mysql. In oracle it is simple.!

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