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Full Discussion: shell script query
Top Forums Shell Programming and Scripting shell script query Post 302554952 by newaix on Tuesday 13th of September 2011 12:42:51 PM
Old 09-13-2011
shell script query

Hi Admins,

I was trying to list the failed logins as part of my daily checklist.
Here is my script
Code:
for i in `who -s /etc/security/failedlogins|tail -100|grep "$dt"|awk '{print $1"  "$6}'`
do
a=`echo $i|wc -l`
if [ $a -gt 0 ]
then
echo $i
else
echo "There are no failed logins"
fi
done

but m getting output in multiple lines.If there are two entries output will be scattered in multiple lines.Please tell me how to print the output in 2 lines(for 2 entries for example). I have used \n with echo $i.. but no use..

Regards
new aix

Moderator's Comments:
Mod Comment Use code tags, see PM, thanks.

Last edited by zaxxon; 09-13-2011 at 02:59 PM.. Reason: code tags, see PM
 

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DATEFMT_IS_LENIENT(3)							 1						     DATEFMT_IS_LENIENT(3)

IntlDateFormatter::isLenient - Get the lenient used for the IntlDateFormatter

	Object oriented style

SYNOPSIS
public bool IntlDateFormatter::isLenient (void ) DESCRIPTION
Procedural style bool datefmt_is_lenient (IntlDateFormatter $fmt) Check if the parser is strict or lenient in interpreting inputs that do not match the pattern exactly. PARAMETERS
o $fmt - The formatter resource. RETURN VALUES
TRUE if parser is lenient, FALSE if parser is strict. By default the parser is lenient. EXAMPLES
Example #1 datefmt_is_lenient(3) example <?php $fmt = datefmt_create( 'en_US', IntlDateFormatter::FULL, IntlDateFormatter::FULL, 'America/Los_Angeles', IntlDateFormatter::GREGORIAN, 'dd/mm/yyyy' ); echo 'lenient of the formatter is : '; if ($fmt->isLenient()) { echo 'TRUE'; } else { echo 'FALSE'; } datefmt_parse($fmt, '35/13/1971'); echo " Trying to do parse('35/13/1971'). Result is : " . datefmt_parse($fmt, '35/13/1971'); if (intl_get_error_code() != 0) { echo " Error_msg is : " . intl_get_error_message(); echo " Error_code is : " . intl_get_error_code(); } datefmt_set_lenient($fmt,false); echo 'Now lenient of the formatter is : '; if ($fmt->isLenient()) { echo 'TRUE'; } else { echo 'FALSE'; } datefmt_parse($fmt, '35/13/1971'); echo " Trying to do parse('35/13/1971').Result is : " . datefmt_parse($fmt, '35/13/1971'); if (intl_get_error_code() != 0) { echo " Error_msg is : " . intl_get_error_message(); echo " Error_code is : " . intl_get_error_code(); } ?> Example #2 OO example <?php $fmt = new IntlDateFormatter( 'en_US', IntlDateFormatter::FULL, IntlDateFormatter::FULL, 'America/Los_Angeles', IntlDateFormatter::GREGORIAN, "dd/mm/yyyy" ); echo "lenient of the formatter is : "; if ($fmt->isLenient()) { echo 'TRUE'; } else { echo 'FALSE'; } $fmt->parse('35/13/1971'); echo " Trying to do parse('35/13/1971'). Result is : " . $fmt->parse('35/13/1971'); if (intl_get_error_code() != 0){ echo " Error_msg is : " . intl_get_error_message(); echo " Error_code is : " . intl_get_error_code(); } $fmt->setLenient(FALSE); echo 'Now lenient of the formatter is : '; if ($fmt->isLenient()) { echo 'TRUE'; } else { echo 'FALSE'; } $fmt->parse('35/13/1971'); echo " Trying to do parse('35/13/1971'). Result is : " . $fmt->parse('35/13/1971'); if (intl_get_error_code() != 0) { echo " Error_msg is : " . intl_get_error_message(); echo " Error_code is : " . intl_get_error_code(); } ?> The above example will output: lenient of the formatter is : TRUE Trying to do parse('35/13/1971'). Result is : -2147483 Now lenient of the formatter is : FALSE Trying to do parse('35/13/1971'). Result is : Error_msg is : Date parsing failed: U_PARSE_ERROR Error_code is : 9 SEE ALSO
datefmt_set_lenient(3), datefmt_create(3). PHP Documentation Group DATEFMT_IS_LENIENT(3)
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