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PHP - Days In The Month
Hello. I want to generate the initial of the day of all the days of a chosen month. Meaning, if I choose this month it will give me:
TFSSMTWTFSSMTWTFSSMTWTFSSMTWTF Is that clear? Can anybody help me? thanks! Similar TutorialsI am trying to create a loop that will work through the dates of the current month; I have the following piece of code for the loop: Code: [Select] $monthFloor = DateTime::createFromFormat('j H:i:s', '1 0:0:0'); $monthCeiling = DateTime::createFromFormat('j H:i:s', $monthFloor->format('t').' 0:0:0'); for ( $monthDate = DateTime::createFromFormat('Y-m-d H:i:s', $monthFloor->format('Y-m-d H:i:s')); $monthDate->format('d') <= $monthCeiling->format('d'); $monthDate->modify('+ 1 day') ) { echo $monthDate->format('Y-m-d H:i:s')."<br />"; } But when I run this, it loops through the dates of the current month twice!? why is it doing this? and how can I rectify the loop so it only goes through the month once? Please check the img link and you will understand Code: [Select] for ($index = 1; $index <= $numdays; $index++) { $users="SELECT invoice_date, DAY(invoice_date) as invoiceday, sum(total_amount) as fullamount from invoices WHERE MONTH(invoice_date)='".$month."' AND YEAR(invoice_date)='".$year."' GROUP BY invoiceday"; $res=mysql_query($users); $row=mysql_num_rows($res); while($fetch=mysql_fetch_array($res)){ echo "<tr id='". ( ( $j %2 == 0 ) ? 'row3' : 'row4' ) . "'>"; $explodedate=explode("-",$fetch['invoice_date']); $days=$explodedate[2]; $month1=$explodedate[1]; $year1=$explodedate[0]; $dates=date("jS", strtotime($fetch['invoice_date'])); echo "<td style='text-align:left;padding-left:12px'>".$index."</td>"; if($index==$dates){ echo "<td style='text-align:right;padding-right:12px'>".$fetch['fullamount']."</td>"; }else{ echo "<td style='text-align:right;padding-right:12px'>--</td>"; } echo "</tr>"; } } $total="SELECT SUM(total_amount) as fulltotal from invoices where MONTH(invoice_date)='".$month1."' AND YEAR(invoice_date)='".$year1."';"; $totres=mysql_query($total); $totfetch=mysql_fetch_assoc($totres); if($totfetch['fulltotal']==""){ echo "<tr><td>NOTHING FOUND</td></tr>"; }else{ ECHO "<TR>"; echo "<td style='color:#BD0000;padding-left:12px;border:1px solid #ccc;width:20%;height:40px;text-align:left'>TOTAL AMOUNT FOR ".$displaydate."</td>"; echo "<td style='color:#BD0000;border:1px solid #ccc;width:20%;height:40px;text-align:right;padding-right:12px'>".$totfetch['fulltotal']."</TD></TR>"; } echo "</table>"; Trying to get the below to generate the month in word form from a numeric entry in a database. Code: [Select] elseif($_GET['rma']=="calander"){ $sql101010="SELECT DISTINCT rma_year_issued FROM $tbl_name4 ORDER BY rma_year_issued"; $result101010=mysql_query($sql101010); while($row101010=mysql_fetch_array($result101010)){ extract($row101010); $content.='<a href="./acp_admincp.php?rma=calander&year='.$rma_year_issued.'">'.$rma_year_issued.'</a> <br />'; } if(isset($_GET['year'])){ $logout.=' | <a href="./acp_admincp.php?rma=calander">Back to RMA Calander</a>'; $rma_year_issued=$_GET['year']; $sql111010="SELECT DISTINCT rma_month_issued FROM $tbl_name4 WHERE rma_year_issued='$rma_year_issued' ORDER BY rma_month_issued"; $result111010=mysql_query($sql111010); while($row111010=mysql_fetch_array($result111010)){ extract($row111010); $months = array('Janurary', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December'); for($i=0; $i <=11; $i++){ if($rma_month_issued=$i){ $rma_month_issued2=$months[$i]; } } $content=""; $content.='<a href="./acp_admincp.php?rma=calander&year='.$rma_year_issued.'&month='.$rma_month_issued.'">'.$rma_month_issued2.'</a> <br />'; } } } Basically in the data base I may have 2,5,7 under the year 2011. I want to get the numeric months into words for use in the text of the link. The above is returning December even though the entry in the database is 5. Hi.. I need help in getting the 3 Months Name from my table field FromMonth and ToMonth. Here is the scenario. First, I select FromMonth and ToMonth. Ex. FromMonth = 5 ToMonth = 7 So it means FromMonth is May ToMonth is July. Now I save the MonthNumber to my database: table- so_month FromMonth = 5 ToMonth = 7 Now I need to get the between Months Name from FromMonth to ToMonth which are (May, June, July). How does it possible? Thank you so much. So I am creating a list of events. There will be a drop down menu to change between months. I only want the months to be listed if an event during that month exists. So if I have events listed for September and November, list only September and November, skipping October. Any idea on how to approach this or if there are any functions for such a thing? I'm trying to figure out if today's current date and time falls between Mon & Fri, but so far, I can't get it to work. Can anybody look at my code and see what I'm doing wrong? Code: [Select] <?php $day_start = date('D h:i a', strtotime("Mon 05:30")); $day_end = date('D h:i a', strtotime("Fri 10:00")); $day_current = date('D h:i a', strtotime("+1 hours")); if (($day_current > $day_start) && ($day_current < $day_end)) { echo "yup"; } else { echo "nope"; } ?> Thanks in advance This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=320501.0 Hi I am trying to add a field to a database that is 4 days from the date the record is added, but it is not adding a value Code: [Select] $end_date=strtotime("+ 4 days"); $add_vehicle_sql=mysql_query("INSERT INTO `tbl_auction_lot`(`cust_id`,`reserve`,`make`,`model`,`spec`,`fuel`,`doors`,`mot_date`,`fns`,`fos`,`rns`,`ros`,`condition`,`reg_no`,`service_history`,`sale_type`,`status`,`keepers`,`gearbox`,`emissions`,`colour`,`date_first_reg`,`date_manufacture`,`bhp`,`engine_size`,`end_date`) VALUES ('$seller_id','$reserve','$make','$model','$body_style','$fuel_type','$no_of_doors','$mot','$fns','$fos','$rns','$ros','$vehicle_condition','$vrm','$service_history','auction','$status','$prev_keepers','$gearbox','$emissions','$colour','$date_reg','$date_man','$bhp','$engine_size','$end_date')") or die(mysql_error()); What am I doing wrong and what is there a better way to achieve the desired result. Hey there, Thanks for taking the time to read my thread. My issue is if I'm given a time stamp in PHP how could I calculate the number of days until that time stamp?. Thanks for your time. Hello, I am trying to put together a mysql query that will return the number of visitors for four days ago, what i am trying to do is plot the last seven days visitors on a graph in the format of day seven, day six, etc and need to find away to get a count for each of those days. At the moment i am thinking about running various queries with each returning the results for a specific day. The code below is supposed to get the count of visitors four days ago. Not between now and four days ago but just for the 24 hour period which covers day 4. The current code just returns a value of 0. Any help would be appreciated. Code: [Select] $result = mysql_query("SELECT ip FROM ip_stats WHERE date= date_sub(NOW(), interval 4 DAY)"); $num_rows = mysql_num_rows($result); echo "$num_rows"; Hi there, for every case we have "gooo" as commen in all how this could be one code using anything like elseif or switch whatever... if($refere && (stripos($r, $refere) === false)){ echo "x1" }else{ echo"goooo" } And if($limited && ($line[hits] >= $limited)){ echo "x2" }else{ echo"goooo" } And if($pword && (isset($_POST['password']) && $_POST['password'] == $pword) ){ echo"goooo" }else{ echo"x3" } And if ($line[capt] == 1 && ($_SESSION["security_code"]) ){ echo"goooo" }else{ echo"x4" } thanks in advance When I try to add 30 days: Code: [Select] $date = date("Y-m-d"); $date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days"); echo $date; and I echo date I get 1330664400 How do I get it to echo out 3/1/2012? I know the answer lies in the strtotime but I can't figure it out. I know it's a simple problem for most of you... I have a SQL row that has a date field: ex: 2010-11-01. When a car is sold there either is a 30 day warranty, a 60 day warranty, or 0 day warranty. What I'm trying to do is display when the vehicles warranty expires, based on the date it was sold, or when did it expire based on the same sold date pulled from the database. Example using last months date: 2010-10-01 60 day: "Expires 11-30-10" 30 day: "Expired 11-01-10" I can not seem to use the date function properly... Any help would be greatly appreciated. Hi, I have db table that records the days (Sunday, Monday...) when the employee login to the system. Say that the employee logged in on Monday then Logged in on Wednesday so this means he was absent on Tuesday. I calculated the number of days to get the answer 2 (between Wednesday and Monday before logging in on Wednesday) how to get the name of the missing day "Tuesday"? code:
$curdate = date("Y-m-d");
$currday = date('l');
$lastdate = $row['indate'];
$lastdayin = $row['lastdayin'];
if ($lastdate !=Null)
{
$misseddates = strtotime($curdate) - strtotime($lastdate);
$misseddates = $misseddates / 86400;
echo $misseddates;
$misseddays = strtotime($lastdayin) - strtotime($currday);
$misseddays = $misseddays / 86400;
echo $misseddays;
}
I tried to get the name of day in the last step but it only calculates 5 in numbers how to get the name. I want the answer to be "Tuesday" as of the example. Thanks. Hi guys, I've hit a brick wall here and am in need of your help. I'm pretty new to PHP and have limited knowledge to say the least. I'll explain what it is I'm trying to do. Set start date as 01/01/2004 (dmY) $oFour Set how many days has it been since then? $today Set how many days it was from $ofour 30 days ago. $today -30 = $thirtyDaysAgo But the problem is I don't know how to make date('z'); work from 2004 and not 01/01/2010. So $today will be how many days it has been since the start of 2004 and $thirtyDaysAgo will be $today -30. I can set up $thirtyDaysAgo no problem but it's just finding out how to get the $today number... Hope anyone can offer a little light to my situation :/ Mav Hello all, The exact thing that i need is to calculate how much days there is in between two dates. The only problem is that every thing that i found dont care about leap year Anyone have a function to do that? Hi fellas, this is really kicking my arse and i know its so simple! I retrieve a date from the database, done! I am manipulating it to display as i want, done! How the hell do i add 365 days to this date? $date= ($row['date']); $subscription = strtotime($date); echo "<p>Subscription renewal date: ". date('l jS F Y', $subscription) . "</p>"; please could some one help me i need to limit the query by timestamp for the last 7 days here is my code on the php side Code: [Select] <?php $db = @mysql_connect("localhost", "8conv", "*****") or die("Connection Error: " . mysql_error()); mysql_select_db("tempmonitor") or die("Error connecting to db."); $sql = "SELECT *, UNIX_TIMESTAMP(datetime) AS datetime FROM 8conv"; $result = mysql_query($sql); $data = array(); while ($row = mysql_fetch_array($result)) { $temp1[] = array($row['datetime'] * 1000, (int)$row['temp1']);; } $temp1 = json_encode($temp1); Hi, I need to calculate absent percentage but not sure how. working days from Sunday to Thursday every week so 5 working days a week. i will calculate the absent days from joining date until current date. Example if joining date is 24-3-2020 and today's date is 7-4-2020 i should get the number of absent days to 11 days since both Friday and Saturday are excluded. How to do that? here is what I have: $workingdays = $curdate - $joindate; $workingdays = $workingdays - $absent = ($counter / $workingdays) * 100; the second line i missing the number of days for (Fridays and Saturdays) that should be excluded from calculations. The third line i did calculate the actual working days ($counter) for the employee so then i can get percentage of absent days. How to exclude the weekend days from calculations? Edited April 1, 2020 by ramiwahdanmistake in dates How can I check in php if a timestamp was x days ago For example the timestamp 1287677304 , how can I check if it was more than 2 days ago? Thanks lots Jake |
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