Re: [PHP-DB] Date Formatting in PHP
This worked perfectly, is very simple for me to use, and I would never have
come upon it in 10,000 years of trying. Thanks!
Thanks also to everyone who offered ideas.
> $f_date = date('F d, Y',strtotime($db_date));
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Re: [PHP-DB] Date Formatting in PHP
> I have a simple need to reformat a variable coming in as $DatePick,
brought
> forward from a MySQL select in the format:
>
> 2003-11-18
>
> I need to convert it to the format
>
> November 18, 2003
$f_date = date('F d, Y',strtotime($db_date));
---John Holmes...
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Re: [PHP-DB] Date Formatting in PHP
At 05:23 PM 6/10/2003 -0400, David Shugarts wrote:
I have a simple need to reformat a variable coming in as $DatePick, brought
forward from a MySQL select in the format:
2003-11-18
I need to convert it to the format
November 18, 2003
try the "dice-n-slice" method. use substr and concatenation to build
$DayReport from the current value of $DatePick.
I am trying to avoid having to mess with the MySQL select statement, as the
output is being passed through several documents. So I need to do it within
PHP.
I tried making a conversion like
$DayReport = date ("F d, Y", $DatePick);
But the value of $DayReport is neither correct nor does it change when
$DatePick changes. For instance, in this example, the $DatePick value of
"2003-11-18" gets converted to "December 31, 1969."
TIA
Dave Shugarts
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Re: [PHP-DB] Date Formatting in PHP
Hi David,
Try this one:
Regards,
Frank
- Original Message -
From: "David Shugarts" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, June 10, 2003 11:23 PM
Subject: [PHP-DB] Date Formatting in PHP
>
>
>
> I have a simple need to reformat a variable coming in as $DatePick,
brought
> forward from a MySQL select in the format:
>
> 2003-11-18
>
> I need to convert it to the format
>
> November 18, 2003
>
> I am trying to avoid having to mess with the MySQL select statement, as
the
> output is being passed through several documents. So I need to do it
within
> PHP.
>
> I tried making a conversion like
>
> $DayReport = date ("F d, Y", $DatePick);
>
> But the value of $DayReport is neither correct nor does it change when
> $DatePick changes. For instance, in this example, the $DatePick value of
> "2003-11-18" gets converted to "December 31, 1969."
>
> TIA
>
> Dave Shugarts
>
>
>
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> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
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[PHP-DB] Date Formatting in PHP
I have a simple need to reformat a variable coming in as $DatePick, brought
forward from a MySQL select in the format:
2003-11-18
I need to convert it to the format
November 18, 2003
I am trying to avoid having to mess with the MySQL select statement, as the
output is being passed through several documents. So I need to do it within
PHP.
I tried making a conversion like
$DayReport = date ("F d, Y", $DatePick);
But the value of $DayReport is neither correct nor does it change when
$DatePick changes. For instance, in this example, the $DatePick value of
"2003-11-18" gets converted to "December 31, 1969."
TIA
Dave Shugarts
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