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PHP - Mktime() Issue... Leap Year Problem?
Or am I just going crazy?
OK, I've been running a query that uses the snippet below to get the previous month. It's : Code: [Select] $sql = "SELECT COUNT(`id`) as `counted` FROM `click_tracking` WHERE `date` LIKE '". date('Y-m', mktime(0, 0, 0, (date('m')-1), date('d'), date('Y'))) ."-%'"; Here's a snippet for people to try: Code: [Select] date_default_timezone_set('America/Toronto'); echo date('Y-m', mktime(0, 0, 0, (date('m')-1), date('d'), date('Y'))); //prints 2012-03 (March, which is incorrect; should be February- 2012-02) That should have printed 2012-02, but instead, it prints 2012-03. If I modify the mktime() to (date('m')+1): Code: [Select] date_default_timezone_set('America/Toronto'); echo date('Y-m', mktime(0, 0, 0, (date('m')+1), date('d'), date('Y'))); //prints 2012-05 (May, which is incorrect; should be April - 2012-04) It prints 2012-05 which is incorrect for the given timezone. Should be 2012-04 I'm thinking it's a leap year bug and will work itself out tomorrow (per my timezone, anyways). So, when I use the mktime() function, it seems to be thinking we're in April already. A standard date('Y-m') prints the correct YYYY-MM (2012-03) though. Am I tripping out? Similar Tutorialsthe code was all working well but now new listings are displaying 28.958333333333 days left is this because its a leap year? Code: [Select] $days = (strtotime(date("Y-m-d")) - strtotime($info['expiredate'])) / (60 * 60 * 24); $days=(abs($days)); 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) 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! Hello, i have two fields. a beginning year and an ending year. How can i make new fields out of the years in between the beginning and ending years. i hope that makes sense. After many, many years of struggling, I would like to finally learn how to separate my website's PRESENTATION from its LOGIC. I have been working on a startup business for over a decade now. And one of the reasons it is taking me so long to build my website from scratch is because I am a horribly INEFFICIENT PROGRAMMER!! On one hand, I think I write solid business logic. But considering that each of my scripts averages between 1,000-2,000 lines of code, that is a monster to manage. (My website is over 50,000 lines of code at this point. All home-grown and hand-coded.) While it is hard to manage things while I am actively coding, what kills me even more, is when I get pulled away for months (or recently for 2 years), and then later on when I come back, it is a nightmare to load all of that code into my brain?! I mean if you read any of my scripts starting at line 1 and going through line 1,500 I think it reads pretty well. But who in the hell can keep track of all of that in one reading? It's certainly a struggle for me!! I definitely wantto strive for more of a "component archtecture" approach. (On a side note, I have heard "good" Object-Oriented programmers claim that if a class is more than 100 lines of code then it is too long?!) So anyways.... My #1 priority for v2.0 is to learn and master separating my website's PRESENTATION from its LOGIC using something like MVC. (And after that, my goal is to finally learn and master Object-Oriented Programming, but one thing at a time!!) I want to finally learn how to program like a *professional* coder. (A very successful and RICH professional programmer!!!) ? So, can all of you gurus out there, please help me get started with a solid Model-View-Controller introduction and/or resources and/or advice?? I know the Internet is littered with this topic, but like anything done well, you need to find good learning materials and at least one or two good mentors. After all, if MVC and OOP and whatever were so easy, then everyone would be doing it, and doing it well. (And of course you realize that very few people actually do things like MVC and OOP well!!) So my goal is to become a "jedi", but I will need some good teachers to get there...
Edited February 24, 2019 by SaranacLake Type-o's and added some more stuff... Hi Im having a bit of trouble with my leap_year() function. If I enter in any year it prints "Leap Year" even if I enter in a year that is not a leap year. Below is my form and php...Please assist Code: [Select] <form Method = "POST" ACTION = "leap.php"> <input type = "text" name = "year" /> <br><br> <input type ="submit" name= "check_year" value="Check" /> </form> <?php error_reporting(0); // turn error messages off //Declare variables $Check = $_POST['check_year']; $year = $_POST['year']; if (isset($_POST['check_year'])) { leap_year(); } function leap_year() { if((($year % 4) == 0) && ((($year % 100) != 0) || (($year % 400) == 0))){ echo "Leap Year"; } else{ echo "Not a leap year"; } } ?> My friend a I have to different numbers when we echo mktime(0,0,0,5,12,1990). How is this possible? I thought that only, we you leave some parameters blank, the local this is used, else, it's standardized. thanks in advance how do I use strtotime() instead of mktime()? $matchTimea = mktime($mHour, $mMinute, 0, $mMonth, $mDay, $mYear); if($matchTimea-time() < $cutoffTimea*60) { New user of PHP here! I'm using it to display entries from my MYSQL database. Image below. This is the code snippet that displays information from the 'name', 'author' and 'date' columns of my database. I learned this technique from w3schools. $result = mysql_query("SELECT * FROM levels ORDER BY date DESC"); while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['author'] . "</td>"; echo "<td align='right'>" . $row['date'] . "</td>"; echo "</tr>"; } I would like to use the date() function to format the date differently. But the date() function requires that I convert my time into a timestamp. I know you have to use the mktime() function for this, but I don't know how. Can someone show me how to pass the contents from $row[date] into the mktime() function? I'm an extreme newbie and have this current error on my site. The error states: Warning: mktime() expects parameter 4 to be long, string given in featured_product.php on line 75 <?php for ($i = 0; $i < $num_rows; $i++) { $id = mysql_result($result,$i,"id"); $title = mysql_result($result,$i,"title"); $featured = mysql_result($result,$i,"featured"); $feature_date = mysql_result($result,$i,"feature_date"); $feature_date_arr = explode("-",$feature_date); $feat_date = mktime(0,0,0,$feature_date_arr[0],$feature_date_arr[1],2000+$feature_date_arr[2]); if ( ($feat_date+($featured*24*60*60))<time() ) { $db2->query("UPDATE product_catalog SET featured = 0 WHERE id='$id'"); $featured = 0; } else { $featured = 1; $db2->query("UPDATE product_catalog SET featured = 1 WHERE id='$id'"); } ?> Any ideas on how to correct this? Thanks! I'm practising how to do date and time in PHP. I wanted to see if I can get all of the date/time variables from user submission and format them correctly. I went with ISO 8601 date... // From the Manual: // date('c', mktime(1, 2, 3, 4, 5, 2006)); // Prints something like: 2006-04-05T01:02:03+00:00 ...but my attempts always produce a slightly different result for the year, month and timezone... Code: [Select] $inputTime = date('c', mktime((int)$timeHour, (int)$timeMins, (int)$timezone, (int)$dateMonth, (int)$dateDay, (int)$dateYear)); echo $inputTime; // Should return: 2013-11-02T02:30:01+00:00 // Instead returns: 2012-12-02T02:30:01-08:00 The timezone value coming from HTML... <option value="1.0">(GMT +1:00)</option> I noticed; 1. It always picks a year less than whatever year I select. 2. It always selects 12 as the month. 3. It always adds -08:00 after the picking the correct timezone. How come? Thanks in advance. Hi, I'm getting this error: Warning: mktime() expects parameter 5 to be long, string given in /home/user/public_html/include/functions.php on line 58 The code is: if($Date_Format == 1) #For dd-mm-yyyy { $Date_Output = date("d-m-Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 2) #For dd/mm/yyyy { $Date_Output = date("d/m/Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 3) #For mm-dd-yyyy { $Date_Output = date("m-d-Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 4) #For mm/dd/yyyy { $Date_Output = date("m/d/Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 5) #For Mon Day YYYY { $Date_Output = date("M D Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 6) #For Month Day YYYY { $Date_Output = date("F D Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 7) #For Date"nd" Month YYYY { $Date_Output = date("d\\t\h F Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 8) #For Month Day YYYY { $Date_Output = date("F d Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 9) #For Month Day YYYY { $Date_Output = date("d/m/Y H:i",mktime($DateDispsplittimehr,$DateDispsplittimemin,0,$Month,$Day,$Year)); // Line 58 } return $Date_Output; } else { return NULL; Line 58 says $Date_Output = date("d/m/Y H:i",mktime($DateDispsplittimehr,$DateDispsplittimemin,0,$Month,$Day,$Year)); I really appreciate any help on this. I want to have a couple of different formats from mktime(), which the value could be: 1292183335. I would like to have some of the formats like: Tuesday at 12:00am, Wednesday, January 12, 2011 at 12:00am Also taking different timezones in consideration. I'm not sure what to use for that. Why does this work: $dateAllowedSelect = date('Y') -18; But this doesn't: $dateAllowedSelect = date('Y') +5; I nee to get the current year plus 2010 - 2015 in a select box. Thank you. Okay I am trying to list out years in a dropdown box. It should show say how many years old. I want to start from 16 though and stop at 100. So the year has to be 1995 for 16 but then I would have to change the code every year. so I was wondering how do I modify this code to do what I explained? Code: [Select] <?php $start_year = ($start_year) ? $start_year - 1 : date('Y') - 100; $end_year = ($end_year) ? $end_year : date('Y'); for ($i = $end_year; $i > $start_year; $i -= 1) { $date=date(Y); $age = $i - $date; echo '<option value="'.$i.'">'.$age.'</option>'; } ?> Thanks in advanced I need help with a query.
I've got a table that stores references to calls that come in to our office. It has a datetime field, and for the purposes of the query, it's the only thing in this table that is important.
I've got another table that stores appointment information. The appointment start time is also a datetime field. For this query, this start time is the only thing in this table that is important.
I need to know how many times in the last year there was a day when we had at least 150 calls come in and made at least 1 appointment.
SELECT COUNT(*) FROM appointments a LEFT JOIN ( SELECT DATE(time) FROM calls WHERE <there are at least 150 calls on a day> ) c ON DATE(a.start_time) = c.time WHERE a.start_time >= CURDATE() - INTERVAL 1 YEARTo make things difficult, I guess the time would have to be converted to a day, because I'm looking for all days where there was an appointment, and the datetime field is too specific. I've been looking online for 30 mins or so, and was hoping somebody here could point me in the right direction. I'm not making much progress by myself. Edited by sKunKbad, 06 October 2014 - 01:17 PM. Gud Pm! Im trying to display a list of years (last year, current year and next year) with these simple code i made.. It displays the current and next year correctly except last year, which gives me "1970". I don't get why it gives me that since may next year works properly.. Code: [Select] $currentyear = date('Y'); $subtract_year = strtotime(date('Y', strtotime($currentyear)) . '-1 year'); $lastyear = date('Y', $subtract_year); $addyear = strtotime(date('Y', strtotime($currentyear)) . '+1 year'); $newyear = date('Y', $addyear); $year_array = array($lastyear, $currentyear, $newyear); foreach($year_array as $year) { echo "<option value='{$year}'>"; echo $year; echo "</option>"; } I'm not familiar with dates yet.. so i'm gonna read the manual while w8ting for your kind replies. I am using the following code to sort by Year. It displays the current year fine and current year also displays in selection but does give the option for the previous years selection. Code: [Select] <form action="archivednews.php" method="post"> <select name="year" id="year"> <?PHP for($i=date("Y"); $i<=date("Y"); $i++) if($year == $i) echo "<option value='$i' selected>$i</option>"; else echo "<option value='$i'>$i</option>"; ?> </select> <input type="submit" value="GO"> </form> Hello, I seem to be having a problem. I am trying to extract the year from a date Code: [Select] 2012-03-01 echo "2012"; I have tried this and it only displays 1969 $dateorig = "2012-03-01"; $new_year = date("Y", strtotime($dateorig)); echo $new_year; So I have decided to do research as my project and gain something from this. But also I will have to include programming which i prefer PHP. and will be testing these for my research. Does this sound like a good project and also If it is good sound.. Where do I start?... need a starting point and then I could go on to doing it.. |
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