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PHP - Date() Get Wday From Day/month/year
Hi Guys,
Been a while since I've been on here, but I'm currently bashing my head against a wall trying to figure this one out. Basically I have a calendar that outputs Monday through Sunday, and the user clicks on a date and outputs it to the URL.... ie. ?day=$day&month=$month&year=$year. Now the next part I need to find out what day (0-6) say the 25th Feb 2011 will be wday 5... now how can I do this in code? And before anyone says why not output it to url, I cannot as the first week and last weeks use different code Cheers James Similar TutorialsHello. I'm new to pHp and I would like to know how to get my $date_posted to read as March 12, 2012, instead of 2012-12-03. Here is the code: Code: [Select] <?php $sql = " SELECT id, title, date_posted, summary FROM blog_posts ORDER BY date_posted ASC LIMIT 10 "; $result = mysql_query($sql); while($row = mysql_fetch_assoc($result)) { $id = $row['id']; $title = $row['title']; $date_posted = $row['date_posted']; $summary = $row['summary']; echo "<h3>$title</h3>\n"; echo "<p>$date_posted</p>\n"; echo "<p>$summary</p>\n"; echo "<p><a href=\"post.php?id=$id\" title=\"Read More\">Read More...</a></p>\n"; } ?> I have tried the date() function but it always updates with the current time & date so I'm a little confused on how I get this to work. Hi, I've got a date picker on a form which puts data into a database in the YYYY-MM-DD format. Just wondering how I could also put the name of the month (extracted from that) as well as just the year into separate columns. Ie: To use the field race_date from the form to also fill the 'race_month' and 'race_year' columns in the database. This code obviously only fills the 'race_date' column so far: Code: [Select] global $_POST; $race_date = $_POST["race_date"] ; .... $query = "INSERT INTO 10k_races (race_date, race_month, race_year)" . "VALUES ( '$race_date', '$race_month', '$race_year')"; Hello this code is supposed to sort by year and month, but it doesn't. It sort by year and months are random. Any suggestions? Many thanks,
$sortdata =[];
$namecolumn = array_column($data, '0');
foreach($data as $datecolumn){
$dateexplode = explode('/', $datecolumn[3]);
$sortdata[] = (isset($dateexplode[1])) ? $dateexplode[1] : ((isset($dateexplode[0])) ? $dateexplode[0] : '');
}
array_multisort($namecolumn, (($sorttype=='DESC') ? SORT_DESC : SORT_ASC), $sortdata, (($sorttype=='DESC') ? SORT_DESC : SORT_ASC), $data);
Edited May 13 by Barand code tags added Thanks to aleX_hill and RussellReal for the help so far. I'm trying to utilise the code before to find the next month. However, it needs to be used with the current year. So I altered the below <?php $nextMonth = date("m") + 1; $daysInNextMonth = date("t",mktime(0,0,0,$nextMonth)); echo $daysInNextMonth;?> to #NEXT MONTH $nextMonth = date("m") + 1; echo $nextMonth; $daysInNextMonth = date("t",mktime(0,0,0,$nextMonth)); echo $daysInNextMonth ; $year = date("o"); echo $year; I can then pass it into my url query like so: <a href="<?php echo $url_path; ?>after=<?php echo $year; ?>-<?php echo $nextMonth; ?>-01&before=<?php echo $year; ?>-<?php echo $nextMonth; ?>-<?php echo $daysInNextMonth; ?>">Next Month</a> Brilliant I thought, until it comes to December 2010, the next month needs to be January 2011. Now I'm bamboozled. How do I get around this one? Many thanks, your help is always appreciated! right now i have my date system to say "X amount of seconds ago, X hours ago, X days ago, X years ago, How could i turn mktime() (1290874161) into a mm/dd/yyyy format? so it would say like, "Nov 27, 2010 at Hour:Minute" (whatever the hour/minute is) Hello, not sure if lack of sleep or just staring at code too long. This is what I have: Code: [Select] while($rowy = mysql_fetch_array( $resulty )) { mpdate = YYYY-MM-DD $date = $rowy['mpdate']; $year = explode('-', $date); $month = date("F", mktime(0, 0, 0, $year[1])); ?> <?php echo $year[0]; ?> <br> <?php echo $month; ?> <br> <?php echo $rowy['mptitle']; } ?> I figure if I use: if ($year[0] == date('Y"){ echo month and entry stuff } but I keep getting errors. What I am trying to do is sort everything out into year month and day ie: 2012 March Entry 1 Entry 2 February Entry 1 Entry 2 Entry 3 etc... This is what I am getting 2012 March Entry 1 2012 March Entry 2 2012 February Entry 1 etc... Is there a way to fix this? How to acheive it friends? i want my current codes to show results accouding to current month followed by year, the precious month goes to the page two.
would really appreciate if someone could help me acheive it
$tableName="records"; Edited by lovephp, 02 December 2014 - 11:50 AM. Hi,
I want to display my data in this format
format.JPG 54.52KB
0 downloads
Expense data i am getting like this
<td><?php
$in = "SELECT sales_invoice.invoice_id as INID, DATE_FORMAT(sales_invoice.date_invoiced,'%M') AS month, sales_invoice_line_items.invoice_id, SUM(sales_invoice_line_items.sub_total) AS Total, SUM(sales_invoice_line_items.tax_amount) AS total_tax FROM sales_invoice INNER JOIN sales_invoice_line_items ON sales_invoice.invoice_id=sales_invoice_line_items.invoice_id GROUP BY sales_invoice.invoice_id, DATE_FORMAT(sales_invoice.date_invoiced, '%Y-%m')";
$in1 = mysql_query($in) or die (mysql_error());
$total=0;
$tax =0;
while($income = mysql_fetch_array($in1))
{
$total +=$income['Total'];
$tax +=$income['total_tax'];
echo $total - $tax; }
?>
</td>
And Profit i am getting like this
<td><?php
$in = "SELECT purchase_invoice.invoice_id as INID, DATE_FORMAT(purchase_invoice.date_invoiced,'%M') AS month, purchase_invoice_line_items.invoice_id, SUM(purchase_invoice_line_items.sub_total) AS Total, SUM(purchase_invoice_line_items.tax_amount) AS total_tax FROM purchase_invoice INNER JOIN purchase_invoice_line_items ON purchase_invoice.invoice_id=purchase_invoice_line_items.invoice_id GROUP BY purchase_invoice.invoice_id, DATE_FORMAT(purchase_invoice.date_invoiced, '%Y-%m')";
$in1 = mysql_query($in) or die (mysql_error());
$total=0;
$tax =0;
while($income = mysql_fetch_array($in1))
{
$total +=$income['Total'];
$tax +=$income['total_tax'];
echo $total - $tax; }
?>
</td>
I dont know how to display data like this. Do i need to change the the table structure in my database? Please suggest
Hello. I have a mysql database with an 'entrytime' field which contains the unix timestamp the record was added. How do I create a mysql query to select all records for a given month/year? For example, if I wanted to create a query to display all records for October 2010 how do I go about it? I am at a loss and can't find my answer via google. Thanks in advance for any help! I have a file upload script that will eventually process a ton of files. I would like to upload them into sub-directories according to what year, month, and day they are uploaded.
A typical tree should look like this:
attachments/
--/2014
-----/January
--------/01
--------/02
--------/03 , etc.
-----/February
--------/01
--------/04
--------/09
--------/18
--------/20, etc
-----/March, etc
--/2015, etc.
So a file called image.jpg uploaded on 10/31/2014 would have a URL of attachments/2014/October/31/image.jpg. I understand that every time a file is uploaded, the script would have to detect through FTP whether or not folders for the year, month, and day exist, and if they don't create them. My problem is that I have no idea what the logic of this script would be. What order should I do things in? Is there a way to use maybe foreach to detect/create the folders? Any input would be appreciated.
Hi, I'm wanting to find rows whose date is within the next week of the current month of the current year. The format of the date is, for example: 2010-10-28 Any ideas guys? Thanks lots! I am looking for a way to have my 'start_date', echo ('start_date' + 1 month). I am looking at the line that reads .... date("F Y",strtotime($row['start_date'])). Does anyone have a suggestion. The date format read, July 2014 or alphabetical full month, numerical full year. Thank you in advance.
<?php
$query = sqlsrv_query($conn, $sql);if ($query === false){ exit("<pre>".print_r(sqlsrv_errors(), true));}while ($row = sqlsrv_fetch_array($query)){ echo "<tr>
<td><a href='view_invoice.php?meter_id=$meter_id&subaccount=$subaccount&start_date=$row[start_date]'>$row[meter_id]</a></td>
<td>" . date("F Y",strtotime($row['start_date'])) . "<td>$row[invoice_no]</td><td>" . date_format($row['invoice_date'],'Y-m-d') . "<td>$row[amount_paid]</td>" . "<td>$row[payment_date]</td>" . "<td>$row[check_number]</td>"
;}sqlsrv_free_stmt($query);
?>
Edited by Butterbean, 11 January 2015 - 08:19 PM. I have two text fields in a form. Both accepts dates. I'd like to check whether the second date provided is exactly one year in the future from the first provided date.
I can easily do this in PHP but since I need to do the check once the user leaves the second field, I have to resort to JavaScript, using onblur or something.
There are two issues I have.
1. Make sure that the strings provided are indeed dates (I would use strtodate() in PHP for this)
2. Regardless of month provided, making sure that the second date is exactly a year in the future of the already provided year
I've only captured the values and passed them to Date(). However, it can only handle european style, and if I enter 1981-06-16, the value returned is Jun 15 1981 17:00:00 which is obviously wrong.
Here is the simple broken code I have
var text1 = document.getElementById("text1");
var start_date = new Date(text1.value);
var text2 = document.getElementById("text2");
var end_date = new Date(text2.value);
alert(start_date);
I need some guidance.
Start with: Saturday August 13 1978. What is the next year that August 13th falls on a Saturday? Couldn't find an answer to that anywhere! I must admit that dates and timestamps really confuse me! I have a query that gets a field named "date_purchased" and returns it like this "2010-09-05 09:58:12" What I need to do is add one year to the date, and display it like this "September 5, 2011" Basically, it's displaying the end date of a one year subscription. The database captures the purchase date with the order, so I want to grab that, add a year, and display it to the customer. Any help would be most appreciated! Hi all,
I have a search page where I'd like members to be able to search for races that they and their friends have entered into a database. When entering the data, users put the race date into the race_date field (Which is a DATETIME field) in the table.
I've set up the search and display pages, so users can search by a number of other criteria (first name, last name, race distance, time, etc), but am having trouble with searching by year. On my search form, is a field called race_year.
So, obviously I need some change in the code below, specifically in relation to the line $where .= " AND race_date='$race_year'"; because I want to extract the year part of the race_date column from the table and see if it matches $race_year .
Any advice would be appreciated.
Cheers,
Dave
$race_year = $_POST['race_year'];
if ($race_year != '')
{ // A year is selected
$where .= " AND race_date='$race_year'";
}
Hi i have a drop down menu for date which is meant to insert all 3 values into a date column bt is only sending the year how can i fix it <select name="date_of_birth"> <option value="1">January <option value="2">February <option value="3">March <option value="4">April <option value="5">May <option value="6">June <option value="7">July <option value="8">August <option value="9">September <option value="10">October <option value="11">November <option value="12">December </select> <select name="date_of_birth"> <option value="1">1 <option value="2">2 <option value="3">3 <option value="4">4 <option value="5">5 <option value="6">6 <option value="7">7 <option value="8">8 <option value="9">9 <option value="10">10 <option value="11">11 <option value="12">12 <option value="13">13 <option value="14">14 <option value="15">15 <option value="16">16 <option value="17">17 <option value="18">18 <option value="19">19 <option value="20">20 <option value="21">21 <option value="22">22 <option value="23">23 <option value="24">24 <option value="25">25 <option value="26">26 <option value="27">27 <option value="28">28 <option value="29">29 <option value="30">30 <option value="31">31 </select> <select name="date_of_birth" id="year"> <?php $year = date("Y"); for($i=$year;$i>$year-50;$i--) { if($year == $i) echo "<option value='$i' selected>Current Year</option>"; else echo "<option value='$i'>$i</option>"; } ?> Hi guys, i was having an issue with dates and although it leans a bit more into MYSQL I was hoping i could get some help ive searched all over the net for this so I dont know if im not thinking on how to ask this correctly or can't find it, but ill be as detailed as possible and hope that helps.
Ok I have a table that shows when payment was received on past invoices dating back to say 1995; The company processes 'ALL' invoices every month on the 9th, so if anything comes in from the first to the ninth it will be processed that month on the ninth. If it comes in from the tenth to the thirtieth or thirty-first it will be processed the following month.
The table already has a "Received" date filed and now I want to create a column on the table called "Processed Date" (adding the column is no problem don't need help with that part) that shows me the date we processed the invoice.
I know i have to use the "Received" field (because it shows what day the invoice was put into the system) but im not sure how to format the "Processed Date" field, so say the "Received" field shows an invoice came in on '1995-05-11', I want the "Processed Date" field to display it was payed on '1995-06-09' (notice that is the following month).
I thought i would put in a visual of what I was trying to do below:
So if I pull the two columns in MYSQL from that table I would see something like this:
Received Processed Date
1996-02-03 1996-02-09
1996-03-03 1996-03-09
1996-04-11 1996-05-09
1996-05-20 1996-06-09
1996-07-13 1996-08-09
If i could get some guidance/help on this i would really appreciate it.
thanks in advance!!
JL
Hello, I've been struggling for days on this problem now. And I'm getting frustrated! I know that the answer is simple, and I've done it before but I cannot figure out what I'm doing wrong. What do I want? Well, I get a date from the database in this format: YYYY-MM-DD. I first want to distract one month from that and then set in into this form: YYYY, MM, DD. I've got this but it's not working (I've tried a lot different values). $transformdate = date("Y-m-d", strtotime("-1 month", $d_datevalue)); $startpoint = str_replace("-", ", ", $transformdate); Thanks, NLCJ |
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