Hi I need to help on finding the below pattern using sed
<b><a href="/home/document.do?assetkey=x-y-abcde-1&searchclause=photo">
and replace as below in the same line on the index file.
<b><a href="/abcde.html">
thx in advance.
Mari (5 Replies)
Hello All,
I have a string "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031", and I just want to extract LLSV1, but I dont get the expected result when using the sed command below.
# echo "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031" | awk '{print... (4 Replies)
myfile:
AAAaaa
BBBbbb
CCCccc
AAAeee
DDDddd
how to replace the last AAA as EEEEE using sed?
like this:
AAAaaa
BBBbbb
CCCccc
EEEEEeee
DDDddd (14 Replies)
Hi ,
My file have data like
4:ALMOST NEVER PR 1925836
5:NEVER PR W DDA 5857610
6:NEVER PR WO DDA 26770205
but i want to replace the spaces before last numric digits out put should be like this
4:ALMOST NEVER PR=1925836
5:NEVER PR W DDA=5857610
6:NEVER PR WO... (7 Replies)
I have a file with multiple lines like this:
<junk><PATTERN><junk><PATTERN><junk>
<junk><PATTERN><junk><PATTERN><junk><PATTERN><junk>
Note that
1. There might be variable number occurrences of PATTERN in a line.
2. <> are just placeholders, they do not form part of the pattern.
I need... (4 Replies)
Hi all,
I'm trying to replace a pattern/string in about 100 files with the filename using following commands but getting nowhere:
for f in *.fa; do sed "s/^>.*/>$f/g" $f > $f_v1.fa; done
for f in *.fa; do sed 's/^>.*/>`echo $f`/' > $fa_v1.fa; done
Basically I want to change any line... (5 Replies)
I've got a file like so:
...lots of lines, etc.
push "route 10.8.0.0 255.255.255.0"
push "route 192.168.1.123 255.255.255.0"
...lots of lines, etc.
I want to sed find/replace the IP address in the second line, whatever it is, with a new IP address, but I don't want to touch the first line.... (5 Replies)
Hi Guys!
Unix newbie here!
Have a requirement for which I have been scouting the forums for a solution but has been out of luck so far :(
I have a file which contains the following:-
TEST1|TEST2|"TEST3|1@!2"|TEST5
My sed command should result in either one the following output:-... (6 Replies)
Hi,
i want to replace the following lines in such a way that if the word merge exists in first column it must replace the 3rd column as M and if parse exists in first column then the last column must P, if neither it must mark it as X. I have tried the solution using awk, but it is saying... (6 Replies)
Discussion started by: charlie87
6 Replies
LEARN ABOUT PHP
datefmt_set_pattern
DATEFMT_SET_PATTERN(3) 1 DATEFMT_SET_PATTERN(3)IntlDateFormatter::setPattern - Set the pattern used for the IntlDateFormatter
Object oriented style
SYNOPSIS
public bool IntlDateFormatter::setPattern (string $pattern)
DESCRIPTION
Procedural style
bool datefmt_set_pattern (IntlDateFormatter $fmt, string $pattern)
Set the pattern used for the IntlDateFormatter.
PARAMETERS
o $fmt
- The formatter resource.
o $pattern
- New pattern string to use. Possible patterns are documented at http://userguide.icu-project.org/formatparse/datetime.
RETURN VALUES
Returns TRUE on success or FALSE on failure. Bad formatstrings are usually the cause of the failure.
EXAMPLES
Example #1
datefmt_set_pattern(3) example
<?php
$fmt = datefmt_create(
'en_US',
IntlDateFormatter::FULL,IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'MM/dd/yyyy'
);
echo 'pattern of the formatter is : ' . datefmt_get_pattern($fmt);
echo 'First Formatted output with pattern is ' . datefmt_format($fmt, 0);
datefmt_set_pattern($fmt, 'yyyymmdd hh:mm:ss z');
echo 'Now pattern of the formatter is : ' . datefmt_get_pattern($fmt);
echo 'Second Formatted output with pattern is ' . datefmt_format($fmt, 0);
?>
Example #2
OO example
<?php
$fmt = new IntlDateFormatter(
'en_US',
IntlDateFormatter::FULL,IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'MM/dd/yyyy'
);
echo 'pattern of the formatter is : ' . $fmt->getPattern();
echo 'First Formatted output is ' . $fmt->format(0);
$fmt->setPattern('yyyymmdd hh:mm:ss z');
echo 'Now pattern of the formatter is : ' . $fmt->getPattern();
echo 'Second Formatted output is ' . $fmt->format(0);
?>
The above example will output:
pattern of the formatter is : MM/dd/yyyy
First Formatted output with pattern is 12/31/1969
Now pattern of the formatter is : yyyymmdd hh:mm:ss z
Second Formatted output with pattern is 19690031 04:00:00 PST
SEE ALSO datefmt_get_pattern(3), datefmt_create(3).
PHP Documentation Group DATEFMT_SET_PATTERN(3)