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Top Forums Shell Programming and Scripting Help with sed to replace entire line Post 302901453 by vinodhin4 on Wednesday 14th of May 2014 12:19:04 AM
Old 05-14-2014
Please see my code:

Code:
sed -e 's/\"`\date\`"\/\"`\date\`"\ \`\cat \$\SRC\/\mail_list\`\/g' sap_prod_test_accts.ksh1 > sap_prod_test_accts.ksh2
sed: command garbled: s/\"`\date\`"\/\"`\date\`"\ \`\cat \$\SRC\/\mail_list\`\/g


Last edited by Franklin52; 05-14-2014 at 04:58 AM.. Reason: fixed code tags
 

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DATE_PARSE_FROM_FORMAT(3)						 1						 DATE_PARSE_FROM_FORMAT(3)

date_parse_from_format - Get info about given date formatted according to the specified format

SYNOPSIS
array date_parse_from_format (string $format, string $date) DESCRIPTION
Returns associative array with detailed info about given date. PARAMETERS
o $format - Format accepted by DateTime.createFromFormat(3). o $date - String representing the date. RETURN VALUES
Returns associative array with detailed info about given date. EXAMPLES
Example #1 date_parse_from_format(3) example <?php $date = "6.1.2009 13:00+01:00"; print_r(date_parse_from_format("j.n.Y H:iP", $date)); ?> The above example will output: Array ( [year] => 2009 [month] => 1 [day] => 6 [hour] => 13 [minute] => 0 [second] => 0 [fraction] => [warning_count] => 0 [warnings] => Array ( ) [error_count] => 0 [errors] => Array ( ) [is_localtime] => 1 [zone_type] => 1 [zone] => -60 [is_dst] => ) SEE ALSO
DateTime.createFromFormat(3), checkdate(3). PHP Documentation Group DATE_PARSE_FROM_FORMAT(3)
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