Date() With Leap Year
I am using this function below in a script to create a count down script.
<? date(("z"), time() - (14400 * 1)) ?> My question is; does the "z" (0 - 365) go up to a maxium of 366 when it is leap year?
View Complete Forum Thread with Replies
See Related Forum Messages: Follow the Links Below to View Complete Thread
Last Day Of Year Date Bug?
Guys, the line below just returned "Dec 07" as the date for one month back from today. Hardly life-threatening, but any thoughts? <?php print date("M `y", mktime(0, 0, 0, date("m")-1, date("d"), date("Y")));?> AS...
Need Date For Every Monday Of Year
Is there a way to get the date for every Monday for a particular year and have it populate a drop-down box? Need something to start by.
Can't Get Month Year From Variable Using Date()
I have a very simple question about date(). I am using date('l, F d, Y'); to show Tuesday, August 22, 2006. But now when I am trying to display just Month - Year i.e (August - 2006) from variable, Its not coming through and displaying January - 1970. Code: I have done date("F - Y", $entry_date) for this. Can any one just point me out or suggest where I am wrong here?
Date/Time Column, Accessing Only Year
I'm trying to generate a table that gives some statistics. I've got a date time column that registers hits. I need to calculate the number of hits for 2006 using the 'date' column, which is the date time column.
Check Week Number In End Of Year And Begiining Of New Year?
I am creating a little own calendar. Now I want to add a "previous week" and "next week" functionality to the calendar. It is easy to take $thisWeek - 1; and $thisWeek + 1;, but there has to be a check wether it is the last week of the year and it should actually be a new year and head to week number 1. HOW do I make this kind of function? I use the following string names: $thisYear $thisWeek
Starting A Year Loop With Last Year
I have the following which loops through the present year and adds two more years on: However, I want it to be exactly the same but to start from last year rather than this year. Is this possible? Code:
Unix Time And Leap Seconds
I have Red Hat Enterprise Linux 4. I was just reading up about UTC and leap seconds. Is it true on my system that the Unix time may skip up or down by one second at midnight when there is a leap second? By "Unix time" I mean the integer returned by time() and similar functions. I'm concerned about the "down" case. Some of the software I've written assumes monotonically-increasing time.
Week Of Year
Is there a function which returns the week of the year, given a specific date?
Day Of Year Count
I have a script that reports that day on which a post expires as: 2000-06-12. What i would like to do is post it instead as: Expires in: 11 Days. So basically, I'm looking for a PHP3 way to do a Day of year Subtraction. I know with MySQL its: (TO_DAYS(Expires) - TO_DAYS(Current_Date)) is there a way to do it without going back into MySQL, using PHP3? OR how to i configure my Select so that it will automatically output "Expires" date as a the day count UNTIL expires.
Calendar Displaying A Whole Year?
I'm reasonably new to php and have been looking everywhere for tutorials on creating calendars. I've found so many, but they're all relatively the same, where they only display one month at a time. If anyone could point me in the direction off a tutorial where I can create a calendar to display a whole year in one go, instead of one month that would be absolutely great. I'd quite like a tutorial instead of a simple download, because I can learn more by going through it all. I don't need any fancy stuff like events. It's litterally just a calandar with nothing on it but the dates.
Dynamicaly Display Month And Year
Hi how can i display month and years dynamically without using database in a drop down menu. <select name=month> <option value=1>January</option> <option value=2>February</option> ... </select> Also the year drop down since 2003 value will be added with every year ie in 2004 it shd show: <select name=year> <option value=2003>2003</option> <option value=2004>2004</option> </select> and so on.. Any help?
Selecting Year & Month From Timestamp(14)
What I'm trying to do is create a link to a page that displays the headings of articles from a previous month of the year. The link will be something like <a href="/prevnews.php?yearmonth=$var"> and then on the page I will query the database with a line like: SELECT * FROM table WHERE TimeEntered=$var My problem is, I don't know how to select all columns from a table where the year and month are whatever I specified in $var. My TimeEntered column is timestamp(14), so YYYYMMDDHHMMSS. If I could just get the YYYYMM part, that would be fine. How do you do that? Also, I run this query on the page. Could I somehow use the variable fDate to get the YYYYMM ? PHP Code:
Exclusive REQ: PHP Developer In Sunnyvale, CA - 1 Year
We have an exclusive requirement for PHP Developer in Sunnyvale, CA. I can get the interview with in 1 hour and PO by end of the day. There is no client interview for this position. Please forward updated resumes with contact info, expecting rate ASAP. Interview in 1 hour, PO by end of the day Location: Sunnyvale, CA Duration: 12 Months Start: Tomorrow No of needs: 5 Rate: $Open/Hr Client: Yahoo Required Skills: ·3+ years of PHP Development experience ·UNIX or LINUX, My SQL, Apache, Java Script ·HTML, DHTML, CCS ·Web Applications experience is a plus ·Candidates, who have worked with search engines like Google, Yahoo will be of great help, though it is not mandatory ·Should have excellent communication skills
Printing Full Year Calendar?
I use a calendar script, Keith Deven's PHP Calendar. I've modified it to incorporate events into it and everything works great. The problem I'm having trouble with is when I print the year long calendar. For some reason it only reads the dates in the database for the current month, in this case March and it outputs the same days for every single month. Code:
Search Data Between Two Specific Year?
I have tabel called "candidate" in that i have two fields call Graduation Passing year and Post Graduation Passing year. In that table i have more than thousands number of candidate. Here i want to make a search php form. Where i can get data of specific candidate who pass between specific year. I have make a form for search with below fields: Graduation Candidate search : Year From and Year To fields and submit button to call query form and display result. Code:
Select Distinct Year From Unix Timestamp
I'm trying to select all distinct years from a unixtimestamp field in MySQL database. I have a query: SELECT DISTINCT YEAR(date_field) As theYear FROM table but it gives me an empty array. What am I doing wrong?
List Weeks Of Year Using Calendar Function
I've been asked to design a small site which contains a page with a list of dates to show a visitor if accommodation (running from Saturday - Friday) is booked or available. Something like this: 2007 2nd Jan - 8th Jan - Available 9th Jan - 15th Jan - Available 16th Jan - 22nd Jan - Booked 23rd Jan - 29th Jan - Available 5th Feb - 11th Feb - Available 12th Feb - 17th Feb - Available etc, etc I'd like to dynamically create the list of dates using the php calendar functions and a mysql database so that each row (52 in total?) could be flagged 'booked' or 'available' allowing easy updating using a back end form. Ideally the page showing the list would have a link taking the visitor to another page showing the following year's dates. Each list would always show a year to view and always run from Jan to Dec.
Convert From 4-digit To 2-digit Year
I am trying to convert from a 4-digit to a 2-digit year. Here is my code: $date = "2007-10-02"; $explode_date = explode("-", $date); echo $new_date = $explode_date[1].$explode_date[2].$explode_date[0]; The result is 10022007 but I need it to be 100207(mmddyy).
Check Date From Input Must Greater Or Equal Today Date?
I have 3 select box. Code: <select name="day"> <option value="1">1</option> <option value="2">2</option> ....... <option value="31">31</option> </select> <select name="month"> <option value="01"> Jan </option> <option value="02"> Feb </option> ..... </select> <select name="year"> <option value="2002">2002 </option> <option value="2003">2003</option> </select> When I select and click submit in form. I will have 3 variable : $day, $month and $year How can I check the day that I choose in form must equal or greater than today date? If it less than today date. I must print error message to user.
Date Is Not Being Pulled From MYSQL Database, Instead Current Date Is Displayed!
I am having trouble pulling a date from a database using PHP at the <a href="http://www.mytuneslive.com/ameshkin69/"> following page.</a> Here is the code. As you can see, it is just making the date the current time. The values in the database are UNIX timestamp, and the DATE() function is used to convert from UNIX to readable date. Can <td width="55%" align="left" valign="top"><?php $row_comment['timestamp'] = date("n d Y g:i A");.....
Read Today's Date Print Tomorrow's Date
I've got a small script that writes a random number and the date to a text file. Then the script opens this text file and if date = date from this text file something must happen. If something happened this script must open this text file again and replace it with a new random number and tomorrow's date. <?php $date = date("d-m-Y"); mt_srand((double)microtime()*10000000000); $num=mt_rand(0, 20); if ($num == $num && $date == "$tomorrowdate") { echo"$num -- $date";} else echo " ";?> So if the script gets activated on the 05-05-2003 it must write tomorrow's date into the text file (06-05-2003). How will I get tomorrow's date today?
Help: Date Function Not Returning Correct Date & Time
I have just notices that the date() function is not returning the correct date/time on my "server". I am running apache2 on my winxp pro laptop. My system clock is set to the correct date, time and timezone, get the results returned by date() are 11 hours behind.
PHP Date Question: How To Determine The Date Each Of The Past 3 Sundays
How can I determine what the date was (in YYYY-MM-DD format) last Sunday, and the Sunday before that? For example: Today is Thursday, August 18, 2004 (2004-08-18). I would like to have the following variables: $this_past_sunday = 2004-08-14 $two_sundays_ago = 2004-08-07 $three_sundays_ago = 2004-07-31 etc. etc.
Save Date To Table Field Date Type
I have a calendar picker on a form that fills a text box with XX/XX/XXXX date format. If I run a insert statement the date field in the database shows 0000-00-00. If i change the field type to varchar the correct format displays. What is the correct way to save a date to a mysql database? Furthermore to query on the date how about a select * from db_name where servicedate >= $startdate and servicedate >= $enddate I supose this will not work Any pointers?
Date Dropdown From Mysql Date Entries
For the PHP gurus out there, here is what I want to do: create a dropdown list of available dates from a mysql database date entries. But the dropdown/s should have 3 separate fields in month, day and year which in effect shows only those months, days and years that have corresponding entries from the database. Planning to use this in the archive section of an online news publication so people can select issues to view via dropdown list that have corresponding entries only.
Start Date - End Date = Number Of Days
I have a starting date for var 1 and today’s date for var 2. I need to calculate how many days are between those two dates. 02/24/03(today) - 01/14/2002(start date) = number of days. In theory this is easy but there are many instances were I could get an inaccurate answer. For example one consideration is that some months have 30 days others have 31. exc. My question is, is there an easier way to solve this problem other than using endless if statments to error handle all possible instances? Does PHP4 have any built in functions that could calculate this formula correctly?
UNIX Mktime To MySQL Date Or Date
A form that sends a starting date and an ending date. I have figured out how to tell the difference between the two and even rip the difference into an array because I need to compare the individual dates against a mySQL database to pull results from a particular date in a series. This is no problem. My problem is that teh mySQL database uses DATE as a field definition and this can not be changed. I managed to pull the date differences using some cleaver mktime stuff but no I do not know how to get this 1072242000 into this 2003-12-24... Is there a function that reverses the mktime result into a usable date format? I would take anything at this point and I can explode and then array it to get what I want - but there has to be something. Code:
Formatting Date From MYSQL Date Field
i've got a date field in my table, but the default format is YYYY-MM-DD... is there any way i can pull the date out of the database and format it differently... like DD.MM.YYYY or like April 13, 2003?
Counting Days From $date To Current Date
how to count the time between two variable dates? Such as someone inputting 6/8/1991 in a form and it comparing that to todays date... then getting the number of seconds elapsed... I know I have to use the mktime() function to get the seconds.. but not quite sure how.
Date() Gives Me Yesterday's Date And Not Today's - Why?
$result[$i]->record_entered contains the current date as set up by MySQL (entered as "now()"). I verified that in the database the date is "now()", which in this case will be "2/9/2006". However, when viewing the page using the display: Code: ( php )
Begin Date And End Date, But No Weekends!
I am making a system to plan my work days. If i begin on a project on 01-10-2007, and the project is 100 hour and i work 10 hours a day, it will cost 10 days for me too work. The answer is simple, project is done @ 10-10-2007, but i dont work in the weekend. So it must be 12-10-2007, i got the following script;, but it aint working right. Because when the begin date and end date is more then 2 weeks, and more than 2 weekends, it doenst count the weekends good. Can someone help me, or is there somebody with a better solution? I am working two days on it, but i cant get it right <?php $begin_date = strtotime( date( '2007-10-01' ) ); $hours = 100; $workdays = 10; $total_days = $hours / $workdays; $total_days *= (60 * 60 * 24); $weekend = $total_days / 7 * 2; $enddate = $begin_date + $total_days + $weekend; if( date('D', $enddate ) == 'Sat' ) { echo 'End date <b>' . date('d-m-Y', $enddate + (2*86400)) . '</b> 2 days<br /><br />'; } elseif( date('D', $enddate ) == 'Sun' ) { echo 'End date <b>' . date('d-m-Y', $enddate + (1*86400)) . '</b> 1 day<br /><br />'; } else { echo 'End date <b>' . date('d-m-Y', $enddate) . '</b><br /><br />'; } ?>
Date Conversions ( Date, Mktime )
I have a date in three pieces dd mm yyyy. I'm just wondering the best way to convert this for inserting into a DATETIME field and then back again when displaying it. I have tried date, strtotime mktime but it keeps inserting a blank date. i.e. 00:00:00 00:00... I've been looking through the php manual but I can't seem to get this working. Also would I be better off using the UNIX_TIMESTAMP function?
Check Date Is Not Before Current Date
I need to validate a date entered to make sure it isnt before the current date. I have managed to this in Javascript but not sure how to do this in PHP when the value is submitted.
Date Time Not Accurate For Date();
echo "date=".date('Y-m-d h:i:s'); date=2007-11-15 15:48:14 the clock is the accurate at here..showing only 3pm ..while now my pc showing 11.51pm(localtime)
Put Today's Date Into A MySQL "date" Formated Field
My MySQL table has a field that is set as type "date." I need to get today's date, and insert it into that field. The default for that MySQL field is 2006-00-00. I know about the date() function. I have tried date("Y-m-d") and date("U"), neither worked. The date field was just filed with all zeros. Can anybody point me in the right direction?
Date And System Date
I am currently looking for a way to compare the current system time date to a date that is in my database. PHP Code:
Change $row['date'] To $date
When writing the code to display the mysql_fetch_assoc results I'd like to remove the $row[''] portion and just leave it as a variable like this: $date, $numid, $text and on and on. For the life of me I can't remember how to do it... even though I know I looked at this last week sometime.
Start Date End Date
I am creating a simple auction site. There will be a start date which I think will be a timestamp and an end date. How can I do code the end date so that it is $days until it closes? I am not asking for code (although that may be helpful) I am asking for an explanation.
Date From And Date To Fields
I have a table which contains special offers starting from a certain date and ending at another date. Therefore would I just do $date = date("Y-m-d H:i:s "); SELECT * from Tablename WHERE start_date <= '$date' && end_date >= '$date' Is this correct?
Date() Help
Here's my question. I made an litle calendar that displays the days of the month. if in a certain day i have something booked it displays a link to the day information.
|