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Top Forums Shell Programming and Scripting Simple Variable substitution in ksh not working Post 302960997 by RTY on Monday 23rd of November 2015 02:00:58 AM
Old 11-23-2015
Hi,

Just do

echo $db_name$i

It would print fine.
 

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ECHO(1) 						    BSD General Commands Manual 						   ECHO(1)

NAME
echo -- write arguments to the standard output SYNOPSIS
echo [-n] [string ...] DESCRIPTION
The echo utility writes any specified operands, separated by single blank (' ') characters and followed by a newline (' ') character, to the standard output. The following option is available: -n Do not print the trailing newline character. The end-of-options marker -- is not recognized and written literally. The newline may also be suppressed by appending 'c' to the end of the string, as is done by iBCS2 compatible systems. Note that the -n option as well as the effect of 'c' are implementation-defined in IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002. For porta- bility, echo should only be used if the first argument does not start with a hyphen ('-') and does not contain any backslashes (''). If this is not sufficient, printf(1) should be used. Most shells provide a builtin echo command which tends to differ from this utility in the treatment of options and backslashes. Consult the builtin(1) manual page. EXIT STATUS
The echo utility exits 0 on success, and >0 if an error occurs. SEE ALSO
builtin(1), csh(1), printf(1), sh(1) STANDARDS
The echo utility conforms to IEEE Std 1003.1-2001 (``POSIX.1'') as amended by Cor. 1-2002. BSD
November 12, 2010 BSD
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