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Top Forums Shell Programming and Scripting any possible to run sql result and output to file Post 302108927 by happyv on Thursday 1st of March 2007 10:42:03 PM
Old 03-01-2007
Quote:
Originally Posted by ahmedwaseem2000
I am not sure which sql you are trying to connect. just for an example i am using mysql

Code:
  echo "SELECT *  FROM employee WHERE \
     emp_name="XXXXX" \
	   |$MYSQL_PATH/mysql dbname --user='whoever' --password='whatever'  -v -v -v >$DIRNAME/$outputfile 2>&1


thank..how can i check which sql I used?
 

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CUBRID_IS_INSTANCE(3)							 1						     CUBRID_IS_INSTANCE(3)

cubrid_is_instance - Check whether the instance pointed by OID exists

SYNOPSIS
int cubrid_is_instance (resource $conn_identifier, string $oid) DESCRIPTION
The cubrid_is_instance(3) function is used to check whether the instance pointed by the given $oid exists or not. PARAMETERS
o $conn_identifier -Connection identifier. o $oid -OID of the instance that you want to check the existence. RETURN VALUES
1, if such instance exists; 0, if such instance does not exist; -1, in case of error EXAMPLES
Example #1 cubrid_is_instance(3) example <?php $conn = cubrid_connect("localhost", 33000, "demodb"); $sql = <<<EOD SELECT host_year, medal, game_date FROM game WHERE athlete_code IN (SELECT code FROM athlete WHERE name='Thorpe Ian'); EOD; $req = cubrid_execute($conn, $sql, CUBRID_INCLUDE_OID); $oid = cubrid_current_oid($req); $res = cubrid_is_instance ($conn, $oid); if ($res == 1) { echo "Instance pointed by $oid exists. "; } else if ($res == 0){ echo "Instance pointed by $oid doesn't exist. "; } else { echo "error "; } cubrid_disconnect($conn); ?> The above example will output: Instance pointed by @0|0|0 doesn't exist. SEE ALSO
cubrid_drop(3), cubrid_get_class_name(3). PHP Documentation Group CUBRID_IS_INSTANCE(3)
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