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Top Forums Shell Programming and Scripting How to Check given string is number in shell script? Post 302324920 by balajiora on Friday 12th of June 2009 08:24:00 AM
Old 06-12-2009
How to assign this output into variable

I have tried the below option

X=`echo "$REQUEST_ID" | awk '$0 ~/[^0-9]/ { print TRUE }'`

echo $X

its printing empty string.

Regards
BS
 

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DATEFMT_SET_LENIENT(3)							 1						    DATEFMT_SET_LENIENT(3)

IntlDateFormatter::setLenient - Set the leniency of the parser

	Object oriented style

SYNOPSIS
public bool IntlDateFormatter::setLenient (bool $lenient) DESCRIPTION
Procedural style bool datefmt_set_lenient (IntlDateFormatter $fmt, bool $lenient) Define if the parser is strict or lenient in interpreting inputs that do not match the pattern exactly. Enabling lenient parsing allows the parser to accept otherwise flawed date or time patterns, parsing as much as possible to obtain a value. Extra space, unrecognized tokens, or invalid values ("February 30th") are not accepted. PARAMETERS
o $fmt - The formatter resource o $lenient - Sets whether the parser is lenient or not, default is TRUE (lenient). RETURN VALUES
Returns TRUE on success or FALSE on failure. EXAMPLES
Example #1 datefmt_set_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 : 66038400 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_is_lenient(3), datefmt_create(3). PHP Documentation Group DATEFMT_SET_LENIENT(3)
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