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Full Discussion: if statement problem
Top Forums Shell Programming and Scripting if statement problem Post 302276633 by methyl on Wednesday 14th of January 2009 09:02:03 AM
Old 01-14-2009
You can force the parameter to be a number not a string.

num=$(($1 + 0))
if [ ${num} -le 2279 ]
then
echo "A"
else
echo "B"
fi
 

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NUMFMT_PARSE(3) 							 1							   NUMFMT_PARSE(3)

NumberFormatter::parse - Parse a number

	Object oriented style

SYNOPSIS
public mixed NumberFormatter::parse (string $value, [int $type], [int &$position]) DESCRIPTION
Procedural style mixed numfmt_parse (NumberFormatter $fmt, string $value, [int $type], [int &$position]) Parse a string into a number using the current formatter rules. PARAMETERS
o $fmt -NumberFormatter object. o $type - The formatting type to use. By default, NumberFormatter::TYPE_DOUBLE is used. o $position - Offset in the string at which to begin parsing. On return, this value will hold the offset at which parsing ended. RETURN VALUES
The value of the parsed number or FALSE on error. EXAMPLES
Example #1 numfmt_parse(3) example <?php $fmt = numfmt_create( 'de_DE', NumberFormatter::DECIMAL ); $num = "1.234.567,891"; echo numfmt_parse($fmt, $num)." "; echo numfmt_parse($fmt, $num, NumberFormatter::TYPE_INT32)." "; ?> Example #2 OO example <?php $fmt = new NumberFormatter( 'de_DE', NumberFormatter::DECIMAL ); $num = "1.234.567,891"; echo $fmt->parse($num)." "; echo $fmt->parse($num, NumberFormatter::TYPE_INT32)." "; ?> The above example will output: 1234567.891 1234567 SEE ALSO
numfmt_get_error_code(3), numfmt_format(3), numfmt_parse_currency(3). PHP Documentation Group NUMFMT_PARSE(3)
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