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Full Discussion: Variables in shell script
Top Forums Shell Programming and Scripting Variables in shell script Post 302350301 by shantanuo on Thursday 3rd of September 2009 11:20:11 AM
Old 09-03-2009
Variables in shell script

Code:
mysqldump --compact --add-drop-table -h192.168.150.80 -uroot -p somePass $combined | sed '/$combined/$table/g' | mysql $database

The sed part is not working from the above statement.
The variables combined and table are already defined and instead of showing the actual variable, it is executing the $/combined/$table in the sed part.
How do I replace the variables and then use the sed command?
 

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DROP 
TABLE(7) SQL Commands DROP TABLE(7) NAME
DROP TABLE - remove a table SYNOPSIS
DROP TABLE name [, ...] [ CASCADE | RESTRICT ] INPUTS name The name (optionally schema-qualified) of an existing table to drop. CASCADE Automatically drop objects that depend on the table (such as views). RESTRICT Refuse to drop the table if there are any dependent objects. This is the default. OUTPUTS DROP TABLE The message returned if the command completes successfully. ERROR: table "name" does not exist If the specified table does not exist in the database. DESCRIPTION
DROP TABLE removes tables from the database. Only its owner may destroy a table. A table may be emptied of rows, but not destroyed, by using DELETE. DROP TABLE always removes any indexes, rules, triggers, and constraints that exist for the target table. However, to drop a table that is referenced by a foreign-key constraint of another table, CASCADE must be specified. (CASCADE will remove the foreign-key constraint, not the other table itself.) NOTES Refer to CREATE TABLE and ALTER TABLE for information on how to create or modify tables. USAGE
To destroy two tables, films and distributors: DROP TABLE films, distributors; COMPATIBILITY
SQL92 SQL - Language Statements 2002-11-22 DROP TABLE(7)
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