I am trying to validate user input, at the moment what i have is as below :-
echo "\n\tPlease, enter package name to import: \c"
read PACKAGE
So far this works perfect , But if the user does not input any thing it stalls :(
What I need is, If the user does not enter the name of the... (9 Replies)
Hi guys!
Im tryin to set up a login script that allows a user to login with only the following "forename.surname" .
The username all contain lowercase letters so thats not a problem to sort out, but im having problems making sure that they can only enter 1 " . " . the code i have so far is this... (1 Reply)
I'm kinda new in shell scripting. How do i validate an input from a user to conform to requirement. For example,
echo "Enter First Name: "
read FName
echo "Enter Date of Employment (dd/mm/yyyy): "
read DoE
If the user enters data that is alphanumeric, it accepts it. I hope i've... (1 Reply)
I am validating a form using php script and I want to "echo" the error message near to the text box itself & not below all the controls....Can I Position the display messages ?..Pls help me... (2 Replies)
Hi,
I am new to Unix shell scripting and need help to add some validation to an existing script.
I've made a script that takes two argument (input) but I want the script to display an error message when nothing (null) is entered. So far I managed to validate the fist argument but fail to... (2 Replies)
Hi All,
I need to write a small piece of code to check the following.
name should contain (A-Z), spaces, hyphens & apostrophes
I need to generate regular expressions for the same.
Please help me out as i am not familiar with regular expressions. (1 Reply)
Hey people, I got small little error which says I entered too many argument in the if else list.
The stock contains a list of information.
The 4th field consist of the quantity available, so I'm trying to alert the user that the stock is less than 3 and needed to restock.
Is this the correct way... (7 Replies)
Hi there,
As a part of file validation, I needed to check for delimiter count in the file.
My aim is to find, how many records have failed to have predefined numbers of delimiters in the file.
My code looks like below
i=`awk -F '|' 'NF != 2 {print NR, $0} ' ${pinb_fldr}/${pfile}DAT |... (3 Replies)
Hi All
I am trying to validate a value using if condition
requirement is need to check whether its a valid numeric value
the input contains ( space, #N/A and negative and positive decimal values and Zeros)
if it contains the space, I need to display the error message as space
... (15 Replies)
I am learning Shell scripting on own. I am trying to do an assignment to get details from the user like username their individual marks ,DOB and send a report in mail with the Details calculated like total and average.
validate_marks() {
local Value=$1
if &&
then
return 0
else... (1 Reply)
Discussion started by: JayashreeRobin
1 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)
03-06-2012
