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Top Forums UNIX for Dummies Questions & Answers optimize if block : shell script Post 302690107 by jim mcnamara on Wednesday 22nd of August 2012 12:53:51 PM
Old 08-22-2012
Your logic has problems. I preserved the bad logic in this code
Code:
#!/bin/ksh
num=$1
mod=`expr $num % 2` 
result=0                                            
[ $mod -eq 0 ] && result=1 
[[ $result -eq 0 &&  $num -gt 4  ]] && result=2 
[[ $result -eq 0 &&  $num -lt 8  ]] && result=3     
if [ $result -gt 0 ] 
then                                     
      echo "No. is either even or greater than 4 or less than 8"
      echo "result is $result"
fi

If you test for > 4 and set result to 2 you never get a result for less than 8.
Did you mean > 4 and <= 8? rather than just > 4?

You can change the code to fit your requirements.
 

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MYSQL_NUM_FIELDS(3)							 1						       MYSQL_NUM_FIELDS(3)

mysql_num_fields - Get number of fields in result

SYNOPSIS
Warning This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include: omysqli_field_count(3) o PDOStatement::columnCount int mysql_num_fields (resource $result) DESCRIPTION
Retrieves the number of fields from a query. o $ result -The result resource that is being evaluated. This result comes from a call to mysql_query(3). Returns the number of fields in the result set resource on success or FALSE on failure. Example #1 A mysql_num_fields(3) example <?php $result = mysql_query("SELECT id,email FROM people WHERE id = '42'"); if (!$result) { echo 'Could not run query: ' . mysql_error(); exit; } /* returns 2 because id,email === two fields */ echo mysql_num_fields($result); ?> Note For backward compatibility, the following deprecated alias may be used: mysql_numfields(3) mysql_select_db(3), mysql_query(3), mysql_fetch_field(3), mysql_num_rows(3). PHP Documentation Group MYSQL_NUM_FIELDS(3)
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