convert date like this: 2453748 to 2005-Jan-14

[djbeta]

Please forgive my newbidity, I've tried to search the php forums and manual but I'm not even sure what sort of date I'm dealing with.

I have a MySQL database with a date field (startDate) that has data in it of this form:

2453748

My first question is, what format is that ?

My second question is, how can my php script put it into a form that looks like this: 2005-Jan-06

Many thanks,
beta

 

[sleipnir214]

What is the exact type of the column in MySQL? That value looks like a Unix-epoch time integer -- if it is, you can use PHP's date() function (

http://www.php.net/date)

to format it anyway you like.

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TANSTAAFL!!

 

[vacunita]

Have you tried the [blue]date()[/blue] function.

Supposing it is a Unix Timestamp, it will convert it to a readable format. here is the link:

http://www.php.net/manual/en/function.date.php

Hope this helps.

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.

 

[jpadie]

the date looks like a unix timestamp. it corresponds to a date in 1970 by the looks of it.

to convert a timestamp to humanreadable form:

date ("Y-M-d",$timestamp);

 

[djbeta]

Ug... the above code is returning 1970-Jan-29 for all the records.. and I know they're all different dates.

I checked the MySQL table, and it's an Int(7) field.


I'm basically trying to write a little interface that will allow a user to quickly edit records in an existing php mySQL application: Easy PHP Calendar.

Could this mean that the author of the application is using his own calculation to render dates into and from a 7 digit integer ? (possibly for some sort of copy protection)

 

[sleipnir214]

Have you verified that the correct values are getting from the database to the variable you use as a parameter in date()?

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TANSTAAFL!!

 

[djbeta]

Yes, i have verified that..

I just found on the developer's forum that he uses Julian Date format

is Gregorian the format I want to be in for the above code to work ?

I see there is a jdtogregorian function

to use it it says jdtogregorian(int date)
what is the "int" in there.. do i just leave the
letters "int" in there ??

am i going about this correctly ???

should I declare a variable to change the date to the right format ?

 

[djbeta]

i think there's also the issue of whether it is a day count versus a date ??? oh god i'm so confused.. :-/

here is a link to the developer's forum:

http://www.easyphpcalendar.com/forums/showthread.php?t=3742&highlight=date+mysql

 

[Itshim]

If you run the integer you posted through jdtojulian() you will get the date in this format: '12/30/2005'.

If you then run a date through: juliantojd() you will get the Julian Day Count (the integer you posted)

Examples:

<?php
//this line prints '2453748' to the screen
echo juliantojd(12, 30, 2005);

//this line prints '12/30/2005' to the screen
echo jdtojulian(2453748);
?>

After retrieving the date through jdtojulian(), you can now format it any way you like using the date() function, along with explode(), or any other number of options.

 

[Itshim]

One thing I forgot to mention, (and you probably already realized) when a user submits a new date you will need to covert that to mm, dd, yy. Then send that through juliantojd() to get an integer before inserting/updating the db. This way the rest of the application will know what to do with the data.

 

[jpadie]

don't you need to make the field bigger? the current number of seconds since 1/1/1970 has at 10 figures so an int(7) is not going to contain the date properly.

i can't think of a date format (other than "yymmdd" a non-y2k compliant form of year) that takes only 7 figures.

 

[Itshim]

jpadie:
The date is in stored in the Julian Date format.

 

[jpadie]

aha. duh...