Hi,
I have a date field in my input file.
I just want to check if its in the format "DD-MM-YYYY".
Is there any command which can achieve this?
Thanks and Regards,
Abhishek (2 Replies)
I am very new to Perl. I am struggling so hard to search a date (such as 10/09/2009, 10-09-2009) from a text file and replace with a string (say DATE) using Perl. Please help me out. Thanks in advance.
Regds
Doren (4 Replies)
how to check input date format.
for example $input_date must be in format dd.mm.gg
script is execute like this:
bin/script1.sh 14.12.2009
script1.sh code:
#!/bin/sh
input_date=$1
CMD="/app/si/test/test.sh $input_date"
echo "*****"
$CMD (2 Replies)
Hi All,
I am learning PERL for one of the projects, and in one of these scripts, I read a flat text file and print in the terminal.
The problem is, the text file has a date field. The format is yyyymmdd. I need to display this as dd-mon-yyyy.
Any ideas to do this? Thanks a lot for the... (9 Replies)
My source file having one date column. The formate of the date column is yyyymm. I need to validate whether all the rows are in same format in the given file. If it is not I have captured that records in a separate file. I am very new to Unix. I don't how to achieve this. Plz help me to achieve... (2 Replies)
I have a filename,
This can be any of any format,
I want to check if the filename has hours,mins and seconds part. If it is present, i want to replace it with a " * " (star symbol)
output needed:
IMP: The time part can be in any pattern.
How can this be done?:confused:... (3 Replies)
hi there
I have file names in different format as below
triss_20111117_fxcb.csv
triss_fxcb_20111117.csv
xpnl_hypo_reu_miplvdone_11172011.csv
xpnl_hypo_reu_miplvdone_11-17-2011.csv
xpnl_hypo_reu_miplvdone_20111117.csv
xpnl_hypo_reu_miplvdone_20111117xfb.csv... (10 Replies)
Hi Experts,
I am checking how to get day in Perl.
If it is “Monday” I need to process…below is the pseudo code.
Can you please prove the code for below condition.
if (today=="Monday" )
{
while (current_time LESS THAN 9:01 AM)
... (1 Reply)
Hi! how do i know if the input is the same as the required date format? the date should be dd/mm/YYYY ex. 2/3/2012 or 15/11/2012
all the following conditions must return an error:
*input of string
*day is > 31 or < 1
*month is > 12 or < 1
*year is < 2013
suppose the date format is stored... (1 Reply)
Hello All,
I have a requirement where i need to get the EXTRACT_DATE from a file and check if the date is of valid format or not and then mail it if it is not valid. Appreciate if you can help me with this.
I did the following so far.
awk '{for(i=1;i++<=NF;)if($i~/^EXTRACT_DATE/) print $i}'... (11 Replies)
Discussion started by: Ariean
11 Replies
LEARN ABOUT PHP
datefmt_is_lenient
DATEFMT_IS_LENIENT(3) 1 DATEFMT_IS_LENIENT(3)IntlDateFormatter::isLenient - Get the lenient used for the IntlDateFormatter
Object oriented style
SYNOPSIS
public bool IntlDateFormatter::isLenient (void )
DESCRIPTION
Procedural style
bool datefmt_is_lenient (IntlDateFormatter $fmt)
Check if the parser is strict or lenient in interpreting inputs that do not match the pattern exactly.
PARAMETERS
o $fmt
- The formatter resource.
RETURN VALUES
TRUE if parser is lenient, FALSE if parser is strict. By default the parser is lenient.
EXAMPLES
Example #1
datefmt_is_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 : -2147483
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_set_lenient(3), datefmt_create(3).
PHP Documentation Group DATEFMT_IS_LENIENT(3)