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Full Discussion: Problem for novice
Top Forums Shell Programming and Scripting Problem for novice Post 76670 by vgersh99 on Wednesday 29th of June 2005 04:06:48 PM
Old 06-29-2005
Quote:
Originally Posted by deepaksamuel
Even I am getting the same error cheche
Even you? Smilie
Code:
#!/bin/ksh

cnt=1
maxcnt=2

if (( cnt < maxcnt )) then
   echo less
else
   echo more
fi

 

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CUBRID_DB_NAME(3)							 1							 CUBRID_DB_NAME(3)

cubrid_db_name - Get db name from results of cubrid_list_dbs

SYNOPSIS
string cubrid_db_name (array $result, int $index) DESCRIPTION
Retrieve the database name from a call to cubrid_list_dbs(3). PARAMETERS
o $result - The result pointer from a call to cubrid_list_dbs(3). o $index - The index into the result set. RETURN VALUES
Returns the database name on success, and FALSE on failure. If FALSE is returned, use cubrid_error(3) to determine the nature of the error. EXAMPLES
Example #1 cubrid_db_name(3) example <?php error_reporting(E_ALL); $conn = cubrid_connect('localhost', 33000, 'demodb', 'dba', ''); $db_list = cubrid_list_dbs($conn); $i = 0; $cnt = count($db_list); while ($i < $cnt) { echo cubrid_db_name($db_list, $i) . " "; $i++; } ?> The above example will output: demodb SEE ALSO
cubrid_list_dbs(3). PHP Documentation Group CUBRID_DB_NAME(3)
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