|
PHP - Make New Fields Out Of Beginning Year And Ending Year
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.
Similar TutorialsI have PHP code that generates a long list of Birth Years in a Drop-down list, and I want to make the control "sticky". I know how to make something like this sticky - thanks to help from you guys... Code: [Select] <!-- Gender --> <label for="gender">Gender:</label> <select id="gender" name="gender"> <option value="">--</option> <option value="F"<?php echo (isset($gender) && $gender == "F") ? 'selected="selected"' : ''; ?>>Female</option> <option value="M"<?php echo (isset($gender) && $gender == "M") ? 'selected="selected"' : ''; ?>>Male</option> </select> But my Birth Year code is a little more complex and I'm racking me brain how to do it... Code: [Select] <!-- Birth Year --> <label for="birthYear">Year Born:</label> <select id="birthYear" name="birthYear"> <option value="">--</option> <?php // Display dates for Users between 18 and 100. for($i = $newestYear; $i >= $oldestYear; $i--){ echo '<option value="' . $i . '">' . $i . '</option>'; } ?> </select> How do I make this second set of code "sticky"?? Thanks, Debbie 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. 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> 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. 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.. 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 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 I'm using this code to get me the variables to a select box on a register page: $vars['user-birthdate-years'] = array(_VARS_USER_CHOOSE_ => "Please choose", _VARS_USER_NONE_ => _VARS_USER_UNDEF_); for ($i = date('Y') - 18; $i > date('Y') - 75; --$i) { $vars['user-birthdate-years'][$i] = "$i"; } This gives me this result in the page source: <select name="birthyear" class="date"><option value="-2" selected="selected">Please choose</option><option value="1997">1997</option><option value="1996">1996</option><option value="1995">1995</option><option value="1994">1994</option><option value="1993">1993</option><option value="1992">1992</option><option value="1991">1991</option><option value="1990">1990</option><option value="1989">1989</option><option value="1988">1988</option><option value="1987">1987</option><option value="1986">1986</option><option value="1985">1985</option><option value="1982">1982</option><option value="1981">1981</option><option value="1980">1980</option><option value="1979">1979</option><option value="1978">1978</option><option value="1977">1977</option><option value="1976">1976</option><option value="1975">1975</option><option value="1974">1974</option><option value="1973">1973</option><option value="1972">1972</option><option value="1971">1971</option><option value="1970">1970</option><option value="1969">1969</option><option value="1968">1968</option><option value="1967">1967</option><option value="1966">1966</option><option value="1965">1965</option><option value="1964">1964</option><option value="1963">1963</option><option value="1962">1962</option><option value="1961">1961</option><option value="1960">1960</option><option value="1959">1959</option><option value="1958">1958</option><option value="1957">1957</option><option value="1956">1956</option><option value="1955">1955</option><option value="1954">1954</option><option value="1953">1953</option><option value="1952">1952</option><option value="1951">1951</option><option value="1950">1950</option><option value="1949">1949</option><option value="1948">1948</option><option value="1947">1947</option><option value="1946">1946</option><option value="1945">1945</option><option value="1944">1944</option><option value="1943">1943</option><option value="1942">1942</option><option value="1941">1941</option></select> Problem: When running the php code at the top it miss two years in the span as you see in the page source below, the year 1983 and 1984 is completely missing in the page source. Why is this happening, could the php code be corrected or is there a workaround in php for the missing years bug I seem to have found? I really appreciate any help on this! Edited by Drezzzer, 12 January 2015 - 05:16 AM. 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 there, I stuck in a silly logic ... I will really appreciate if someone helps me... My Fiscal year starts from 01st April of current year through 31st March of next year, for ex. (01 Apr 2011 to 31st March 2012). I want to put a latefee of 6.25% on total amount, if the payment date is not done on the current fiscal year. Everything depends on the payment date. For example, A bill of $8000 was generated on 01st Apr. 2009 then its 6.25% will be $500 and the customer is paying the same on 30th August 2011, then total years passed is 3. So, the late fee will be 500 * 2 = $1000. i.e. Any bill generated and paid in the current fiscal year (stated above) is free of late fee and after that the above calculation is applicable. Please help, how to do the calculation. A function is really appreciated.
I have a given date on a webpage, that I'm scraping to insert into a DB. The date is in this format:
$date = 'Sun, Feb 9<br />3:00 PM ET':
$healthy = array("<br />", " ET", "Sun.", "Mon.", "Tue.", "Wed.", "Thu", "Fri.", "Sat.", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec");
$yummy = array(" ", "2020", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December");
$date = str_replace($healthy, $yummy, $date1);
$start = date('Y-m-d H:i:s', strtotime("$date -120 minutes"));
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.
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) 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! 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
$tableName="records"; Edited by lovephp, 02 December 2014 - 11:50 AM. Hi everybody,
So I used to be a pretty decent PHP developer, but I haven't done much with programming for the last 8 years. I switched to the healthcare industry after the big dot-com boom and I'm finishing up a doctorate in nursing this spring. I decided to try and get back into programming to pursue some online teaching projects I want to take on and wow, have I forgotten a lot!
I've got a WordPress site up and I've manged to teach myself some decent animation skills in Flash and After Effects. Now I'm looking to try and build a set of database tables to house medication calculation equations, answers, and some metadata. I've been looking around to try and find some WordPress addons that will let me edit database tables, but everything I've been able to find turns out to be a table editor in the "Microsoft Word" sense of the word, and not actually usable to manage database tables. Is there anything out there that could save me a big chunk of time?
Looking forward to some development after a long break though. I don't plan on working through a shell, so I'm going to be using a text editor on a PC. Any suggestions for making the writing and syncing easily that might make my life easier?
Thanks,
-Bart
Edited by NerdDoc, 09 September 2014 - 07:21 PM. Hello, not sure if lack of sleep or just staring at code too long. This is what I have: Code: [Select] while($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? 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 |
|||||||