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Dropdown Date Selector That Automatically Selects Todays Date


Does anyone have a PHP generated dropdown date selector that automatically selects todays date on page load that i can pinch?




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Get The Todays Date- Yesterdays Date And Tomorrows Date
I get the todays date- yesterdays date and tomorrows date, like so:

$yesterday = date ("Y-m-d", mktime (0,0,0,date("m"),(date("d")-1),date("Y")));

$tomorrow = date ("Y-m-d", mktime (0,0,0,date("m"),(date("d")+1),date("Y")));

$actualdate = date("Y-m-d");

Which returns the date in the format "200-11-13" But I need to add 13 hours to all three - how owuld I do this?

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Todays Date NOW() Function
I simply want to add todays date when inserting into MYSQL.

   $query = "insert INTO home_page (title, content, date) ";
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I keep getting the error: Call to undefined function showerror().

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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.

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Drop Down Menu With Todays Date
What is the best way to add a drop down menu to generate the below html

<br>
<select name="date">
<selected><option value="20-11-2007">20-11-2007</option></selected>
<option value="21-11-2007">21-11-2007</option>
<option value="22-11-2007">22-11-2007</option>
<option value="23-11-2007">23-11-2007</option>
<option value="24-11-2007">24-11-2007</option>
<option value="25-11-2007">25-11-2007</option>
<option value="26-11-2007">26-11-2007</option>

<option value="27-11-2007">27-11-2007</option>
<option value="28-11-2007">28-11-2007</option>
<option value="29-11-2007">29-11-2007</option>
<option value="30-11-2007">30-11-2007</option>
</select> DD-MM-YYYY<br>

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How would I go about not display items that are in the past using date? I have a page that displays events but if the date is past I dont want it to display anymore.

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Variables Constantly Echo Todays Date . . . ?
I'm trying to have the news section spit out the date of the event that it is displaying in a prettier format. For example, displaying "2007-09-27" as "September 27" . however, it won't spit out anything but todays date. Code:

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Take Todays Date, Compare It With The Dates Held In The Db
I have a db which has a field called 'date' and is set to date type. I want to be able to take todays date, compare it with the dates held in the db and print out the record with the next date in it. Say today's date is 20/09/06, the record with the next date, say 23/09/06, should be printed out. How can I do this easily. I am very new to this, and can't seem to get it right.

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Select Query To Get Only Todays Date Records
I want to use the following query to show only users who registered today with todays date.format(m/d/y)field name is: a.date_registration

select a.id, a.fname, a.sname, a.status, a.login, a.login_count, b.lang_1 as gender, a.email from pro_user a left join pro_reference_lang_spr b on b.table_key=&#3912;' and b.id_reference=a.gender where a.date_registration = (Todays Date m/d/y) order by a.login limit 0, 10 0, 10

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Auto Fill Date Dropdown List
How would I modify this code to display the years in the dropdown list like so:

current year
2006
2005
back 50 years

<select name="year" id="year">
<?PHP

for($i=date("Y");
$i<=date("Y")+2;
$i++)
if($year == $i)
echo "<option value='$i' selected>$i</option>";
else
echo "<option value='$i'>$i</option>";
?>
</select>

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Insert Date To Database Using Dropdown Boxes
I be able to insert the date into my database if i have 3 dropdown boxes wherein it contain the month,day & year. i know the basic of inserting data into my database but with this 3 dropdown box i don't have any idea. after submitting, the data selected on these 3 boxes shall be in one field in my database with fieldname Date.

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Changing Link Automatically On A Certain Date
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Dropdown List Which Selects A Post A Title (to Edit)
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How Do You Make A Date Change Automatically Every 3 Months Please?
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How do you do this with PHP please?

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I am making a form and i want to insert the current date automatically to the database, how i can do the without putting another date field on the form ??

$select="INSERT INTO dir_all(date,wbname,wbdisc,wblink) VALUES ('','$sitename','$sitedisc','$sitelink')";

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Automatically Changes The Display Of Another Dropdown Box
I currently have a dropdown box on a page. What I would like is that when a user selects something from the dropdown box it automatically changes the display of another box.

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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.

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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");.....

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<?
$date_book = date("Y-m-d H:i:s");

$date_return = $_POST['date_return']; //Returns a date like this 2007-05-25

                         // needs to be date_return - 1 day like 2007-05-24
$date_flag  = mktime(date("Y", $date_return), date("m", $date_return), date("d", $date_return)-1);
?>

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When I save the record the data field "f_date" displays "0000-00-00" which I take it the date input is not getting inserted to the table. My insert field code is PHP Code:

<input name="f_date"  type="text" class="bodytext" id="f_date" onclick='scwShow(this,this);' value="" size="15" />

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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?

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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.

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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.

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How To Format The Date To Read Add 1 Hour To The Date Stamp.
i have a field in mysql which contains a timestamp: 2007-07-25 22:52:14 i was going to add a new variable but adding 1 hour to it so it would read: 2007-07-25 23:52:14
im not sure how to format the date to read add 1 hour to the date stamp.

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i have got an date in  string format  ddmmyyyy (eg: 15052007).i have to validate this date. i have alrady validate this date for a number. now all i have to do is check the day (dd)  against month (mm). example, 29 feb is invalid date, if not a leap year.i have tried to wrote some bit of code but still struggling with it.

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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?

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Pick From A Drop Down List That Automatically Gives Options In A Second Dropdown Box
does anyone know how to create a form that allows the user to firstly pick from a drop down list that automatically gives options in a second dropdown box. eg.

'category1'
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this then populates dropdown box 2 with the options
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or if
'category2'
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then dropdown box 2 gives the options of
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i f any one knows how I should code this please help, if not point me in the direction of a tutorial that can.

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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?

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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:

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Date Using Today's Date And A Number Of Days
I need to have a date using today's date and a number of days:
dateX = todays date - x number of days

can someone give me an example of php code .

I'm using windows, appache server.

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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?

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Date Functions That Return Date As A String
Does anyone know if there is a date function in php that would return me the date as a string, ie today would be 10 July 2001.

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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.

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Date Comparison - Checks If A Date Is In Between Two Other Dates
I'm having trouble writing a program that checks if a date is in between two other dates or not.

$today = date("d-m-y");
$a = "09-09-06"; //September 9th, 2006
$c = "09-09-09"; //September 9th, 2009

if($a <= $today && $today<$c){
       echo "Wee";}
else{
echo "Wrong";}

When I try outputting this, it outputs always outputs "Wrong" for some reason, although $today is clearly inbetween $a and $c (Sept.09'06 and Sept.09'09). Am I doing something wrong?

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Compare Date Of Today And The Date In The Database
i have a db that stores a date in a table with a specific column.

db=test
table=test_date_compare
column name=date_inserted

ok i hv a php form that allows user to input date and it enters in the above column (date) as in format eg: 21/03/2007

now i wat i would like to do is i would like to compare date of today and the date in the database if it is lesser then todays date it deletes that row of data. deleting the row of data i know how but the problem comes wen coparing the date i dunno how to ...

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The Server Date Cant Match With My Country Date.
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I use a lot of -30 days searches and stuff like that. As I insert the records into the database (which come in from a .csv file) the "SCHEDULED_DATE" looks like that above but that, as we all know, won't work with things like 'SCHEDULED_DATE > $onemonth' ($onemonth being a date that is a month ago).

So while the records are imported I have been trying to take that date and convert it into a usable MySQL date. Although my attempts have completely failed. Here is one of the newest things I have used, which didn't work and I really don't know why. Code:

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Convert Input Date To Mysql Date
Wanted to see if there was a cleaner way to do this. My form takes a user-input date in the format MM/DD/YYYY. I need to convert that to YYYY-MM-DD to query against my database. The code below works, but basically I take input dates convert them to a UNIX timestamp with strtotime() and then convert them back to the format I need using strftime()

PHP Code:

$format = '%Y-%m-%d'
$startDate =  strtotime($startDate);
$startDate = strftime($format, $startDate);

$endDate =  strtotime($endDate);
$endDate = strftime($format, $endDate);

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Convert The Date Separator Before Inserting The Date Into The DB
I need to convert the date separator before inserting the date into the DB. Right now, I have 01/01/2006 and I want 01-01-2006. The dates are coming from a CSV and so are an array element. I've tried strrpos and a few other functions trying to just replace the slashes with hypens and that's not working (I think maybe because it's an array and those functions don't work on arrays?) Could someone post example code for something like this?

I have an array called $items[3] that holds the date with the separator mentioned above. I'm sure there are several ways to do it in both php and with mysql functions but I've been stuck on it for a while and REALLY need to get through this for a project I'm working on right now.

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Return The Current Date/time Using Date('c')
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2006-05-09T18:00:00-08:00

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2006-08-16 21:03:54

and convert it into a date this:

16 August 2006 @ 11:03:54pm

Does anyone have some insight as to how I would go about doing that?

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Unix Date Stamp Convert To UK Date
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textbox1-29/08/2007
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upon entering it into the database the following dates should be converted already to the mysql date format like 2007-08-29

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Date Different - How To Get Previous Month Of Current Date.
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I use date("d/m/Y") for return the current date.

But i need to get one day less.

How can I do that?

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