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Top Forums Shell Programming and Scripting Substring check in IF condition in shell script Post 302795201 by kshji on Wednesday 17th of April 2013 11:56:37 AM
Old 04-17-2013
Builtin solution is to use case in every shell, if [[ ... regexp is not usable.

Code:
case "$field" in
     reserverd*|fillspaces* )  some 
                   ;;
esac

ksh93 and bash you can use [[ ]] testing with patterns. Condition Test.

It's not same as [ / test cmd.

Code:
a=teststring

if [[ $a = tes* ]] ; then
        echo y1
fi
if [[ $a = tes*r* ]] ; then
        echo y2
fi
if [[ $a = *g ]] ; then
        echo y3
fi
if [[ $a = *g*1 ]] ; then   # false, no echo
        echo y4
fi

 

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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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