I'm hoping this will be a piece of pie for someone! String output is currently 12:00am for everything.
The following code from MySQL with format HH:MM:SS (hours_open, hours_closed)
$get_hours_sql = "SELECT * FROM client_hours ORDER BY day";
$get_hours_res = mysqli_query($dbConnect, $get_hours_sql) or die(mysqli_error($dbConnect));
// Establish the output variable
$hoursList = '<div class="right_bar">';
while ($productList = mysqli_fetch_array($get_hours_res)) {
$id_hours = $productList['id_hours'];
$day = $productList['product_name'];
$open = $productList['hours_open'];
$close = $productList['hours_close'];
$hoursList .= ''.date("g:ia", $open).' - '.date("g:ia", $close).'<br/>';
}
$hoursList .= '</div>';
echo $hoursList;
Output is currently
12:00am - 12:00am
looped.
I want to get the output to
11:00am - 11:00pm
which would represent the database entries.
Thanks!
I always find PHP <-> MySql date handling fiddly (got better with 5.3 though).
My guess is that the mysql query returns the date as a string and date() is expecting a time stamp.
Often, I just get mysql to format the date as a string for me and return it as an additional field, like so:
$get_hours_sql = "SELECT *,date_format(hours_open,'%h:%i %p') as hours_open_formatted, date_format(hours_close,'%h:%i %p') as hours_close_formatted FROM client_hours ORDER BY day";
then just use the formatted fields:
$hoursList .= ''.$productList['hours_open_formatted'].' - '.$productList['hours_close_formatted'].'<br/>';
Data accepts as it's second parameter a Unix timestamp, so what you're trying to do simply won't work. You could use either mysql's TIME_TO_SEC function, or php's mktime to convert the time string to a Unix timestamp.
Example:
$openHours = explode(':',$productList['hours_open']);
$timestamp = mktime($openHours[0],$openHours[1]);
$yourDate = date("g:ia",$timestamp);
Edit: I think you should try Ben's answer, I think it's a better solution than mine.
Related
I have a MySQL DB with StartDates in the format of yyyy-mm-dd and starttimes in the format of HH:MM using a 24 hour clock. What would be the easiest way to compare the difference of two days in PHP? Would using a datetime object could I set it to be with the information given and just give it to zeros for the seconds? I need to get the amount of time between both dates down to the minute. I was putting the startdate (Just the day since its always within the same month for my application) and time together concatenated together and then pulling out what I need like below, but I haven't been able to get it straight yet. Thanks for the look!
$tempvar1 = $times[$i][$j];
$tempvar2 = $times[$i][$j+1];
$day1 = $tempvar1[0].$tempvar1[1];
$day2 = $tempvar2[0].$tempvar2[1];
$hours1 = $tempvar1[2].$tempvar1[3];
$hours2 = $tempvar2[2].$tempvar2[3];
$minutes1 = $tempvar1[5].$tempvar1[6];
$minutes2 = $tempvar2[5].$tempvar2[6];
$numdays = ($day2-$day1) - 1;
$time1 = ($hours1*60)+$minutes1;
$time2 = ($hours2*60)+$minutes2;
MySQL has plenty of date/time functions:
SELECT TIMEDIFF(endtime, starttime), DATEDIFF(endtime, starttime)
FROM ...
doc links for timediff and datediff
That'll you get strings in the format of 'hh:mm:ss.ssss' for timediff, and a straight-up integer representing the days between the two dates, respectively.
I am fetching data from a database and outputting it as an XML file
I am using following format to fetch it
using for loop
$arrayName['fieldName'] which in my case is $row[publication_date]
In the database its in the format of mm/dd/yyyy ..but i just want yyyy to be outputted and I am trying the following code ::
$k = "SELECT EXTRACT(YEAR FROM'$row[publication_date]')";
if($k!=NULL){
$xml_output .="\t\t\t\t<year>" . $k . "</year>\n";
The result is
<year>SELECT EXTRACT(YEAR FROM'1/1/1995')</year>
But instead i want it to be just
<year>1995</year>
What i am doing wrong here?? plzz help me out..
Any questions plzz comment..:)
You can't just run SQL without executing a query in a database: you need to use PHP's functions.
$timestamp = strtotime($row['publication_date']); // convert date string to unix timestamp
$year = date('Y', $timestamp); // extract only the year
Store $year in your XML.
Reference:
http://php.net/strtotime
http://php.net/date
I would just pull the date normally without trying to format it on the database server and format it using PHP.
$date = // code here to gather the date
$date = date('Y', strtotime($date));
echo $date; // should return the 4 digit year
http://us.php.net/manual/en/function.date.php
Well, there's a few potential issues, but the first is that you need to actually use your database connection to run the EXTRACT command.
Alternately, with less overhead, you could use a string slice or regex to pull out the date.
How about:
$xml_output .="\t\t\t\t<year>" . preg_replace('#.*/#', '', $row[publication_date]) . "</year>\n";
I have a problem where I need to handle dates where the month and day parts are optional. For example, the year will always be known but sometimes the day or month and day will be unknown.
In MySQL I can create a table with a date field and while I can't find any reference in the MySQL Manual it will accept the following as valid:
(YYYY-MM-DD format):
2011-02-10 // Current date
2011-02-00 // Day unknown so replaced with 00
2011-00-00 // Day and month unkown so replaced with 00-00
Test calculations from within the database work fine so I can still sort results easily. In the manual it says that month needs to be between 01 and 12, and day between 01 and 31 - but it does accept 00.
First question: Am I going to run into trouble using 00 in the month or day parts or is this perfectly acceptable?
Next question: Is there a PHP function (or MySQL format command) that will automatically format the following dates into the required format string?
2011 becomes 2011-00-00
2011-02 becomes 2011-02-00
Or do I need write a special function to handle this?
The following doesn't work:
<?php
$date = date_create_from_format('Y-m-d', '2011-00-00');
echo date_format($date, 'Y-m-d');
// Returns 2010-11-30
$date = date_create_from_format('Y-m-d', '2011-02-00');
echo date_format($date, 'Y-m-d');
// Returns 2011-01-31
?>
Third question: Is there a PHP function (or MySQL command) to format the dates for use in PHP?
Finally, is this the best approach? Or is there a 'best practise' method?
EDIT:
Here is what I'm currently doing:
A date field can accept a date in the format YYYY, YYYY-MM, or YYYY-MM-DD and before sending to the database it is processed in this function:
/**
* Takes a date string in the form:
* YYYY or
* YYYY-MM or
* YYYY-MM-DD
* and validates it
*
* Use date_format($date, $format); to reverse.
*
* #param string $phpDate Date format [YYYY | YYYY-MM | YYYY-MM-DD]
*
* #return array 'date' as YYYY-MM-DD, 'format' as ['Y' | 'Y-m' | 'Y-m-d'] or returns false if invalid
*/
function date_php2mysql($phpDate) {
$dateArr = false;
// Pattern match
if (preg_match('%^(?P<year>\d{4})[- _/.]?(?P<month>\d{0,2})[- _/.]?(?P<day>\d{0,2})%im', trim($phpDate), $parts)) {
if (empty($parts['month'])) {
// Only year valid
$date = $parts['year']."-01-01";
$format = "Y";
} elseif (empty($parts['day'])) {
// Year and month valid
$date = $parts['year']."-".$parts['month']."-01";
$format = "Y-m";
} else {
// Year month and day valid
$date = $parts['year']."-".$parts['month']."-".$parts['day'];
$format = "Y-m-d";
}
// Double check that it is a valid date
if (strtotime($date)) {
// Valid date and format
$dateArr = array('date' => $date, 'format' => $format);
}
} else {
// Didn't match
// Maybe it is still a valid date
if (($timestamp = strtotime($phpDate)) !== false) {
$dateArr = array('date' => date('Y-m-d', $timestamp), 'format' => "Y-m-d");
}
}
// Return result
return $dateArr;
}
So it pattern matches the input $phpDate where it must begin with 4 digits, then optionally pairs of digits for the month and the day. These are stored in an array called $parts.
It then checks if months or days exist, specifying the format string and creating the date.
Finally, if everything checks out, it returns a valid date as well as a format string. Otherwise it returns FALSE.
I end up with a valid date format for my database and I have a way of using it again when it comes back out.
Anyone think of a better way to do this?
I have a problem where I need to handle dates where the month and day parts are optional.
For example, the year will always be known but sometimes the day or month and day will be
unknown.
In many occasions, we do need such 'more or less precise' dates, and I use such dates as 2011-04-01 (precise), as well as 2011-04 (= April 2011) and 2011 (year-only date) in archives metadata. As you mention it, MySQL date field tolerates '2011-00-00' though no FAQs tell about it, and it's fine.
But then, I had to interface the MySQL database via ODBC and the date fields
are correctly translated, except the 'tolerated' dates (Ex: '2011-04-00' results empty in the resulting MySQL-ODBC-connected ACCESS database.
For that reason, I came to the conclusion that the MySQL date field could be converted in a plain VARCHAR(10) field : As long as we don't need specific MySQL date functions, it works fine, and of course, we can still use php date functions and your fine date_php2mysql() function.
I would say that the only case when a MySQL date field is needed
is when one needs complex SQL queries, using MySQL date functions in the query itself.
(But such queries would not work anymore on 'more or less precise' dates!...)
Conclusion : For 'more or less precise' dates,
I presently discard MySQL date field and use plain VARCHAR(10) field
with aaaa-mm-jj formated data. Simple is beautiful.
Since the data parts are all optional, would it be tedious to store the month, day, and year portions in separate integer fields? Or in a VARCHAR field? 2011-02-00 is not a valid date, and I wouldnt't think mysql or PHP would be excited about it. Test it out with str_to_time and see what kind of results you get, also, did you verify that the sorting worked right in MySQL? If the docs say that 1 through 31 is required, and it is taking 00, you might be relying on what is, in essence, a bug.
Since 2011-02-00 is not a valid date, none of PHP's formatting functions will give you this result. If it handled it at all, I wouldn't be surprised if you got 2001-01-31 if you tried. All the more reason to either store it as a string in the database, or put the month, day, and year in separate integer fields. If you went with the latter route, you could still do sorting on those columns.
I have also encountered this problem. I ended up using the PEAR Date package. Most date classes won't work with optional months or optional days, but the PEAR Date package does. This also means you don't need custom formatting functions and can use the fancy formatting methods provided by the Date package.
I have found this link in a textbook. This states that month and day values can be zero to allow for the possiblity of storing incomplete or unknown data
http://books.google.co.uk/books?id=s_87mv-Eo4AC&pg=PA145&lpg=PA145&dq=mysql+date+of+death+when+month+unknown&source=bl&ots=tcRGz3UDtg&sig=YkwpkAlDtBP1KKTDtqSyZCl63hs&hl=en&ei=Btf5TbL1NIexhAfkveyTAw&sa=X&oi=book_result&ct=result&resnum=8&ved=0CFMQ6AEwBw#v=onepage&q&f=false
If you pull your date in pieces from the database you can get it as if it's 3 fields.
YEAR(dateField) as Year, MONTH(dateField) as Month, DAY(dateField) as DAY
Then pushing those into the corresponding fields in the next bit of PHP will give you the result you're looking for.
$day = 0;
$month = 0;
$year = 2013;
echo $datestring;
$format = "Y";
if($month)
{
$format .= "-m";
if($day)
$format .="-d";
else
$day = 1;
}
else
{
$month = 1;
$day = 1;
}
$datestring = strval($year)."-".strval($month)."-".strval($day);
$date = date($format, strtotime($datestring));
echo $date; // "2013", if $month = 1, "2013-01", if $day and $month = 1, "2013-01-01"
I have a date returned from an sql query (a datetime type field) and want to compare it to today's date in PHP. I have consulted php manual and there are many ways to do it. I finally came up with a solution comparing strings, but I would like to know if there are either any 'better' (best practice), cleaner or faster ways to do it. This is my solution:
// $sql_returned_date='2008-10-17 11:20:04'
$today = new DateTime("now");
$f_today=$today->format('Y-m-d'); //formated today = '2011-03-09'
$sql_date=substr($sql_returned_date,0,9); //I get substring '2008-10-17'
if($f_today==$sql_date)
{
echo "yes,it's today";
}else{
echo "no, it's not";
}
thanks
Seriously guys?
//$mysql_date_string= '2013-09-20' OR '2013-09-20 12:30:23', for example
$my_date = new DateTime($mysql_date_string);
if($my_date->format('Y-m-d') == date('Y-m-d')) {
//it's today, let's make ginger snaps
}
You could factor this into the data returned from your database query:
SELECT `DateOnDB`,
DATE(`DateOnDB`) = DATE(CURDATE()) AS isToday
FROM `dbTable`
and simply use PHP to test the value of the isToday column
Excuse me for being a question-digger, but I was trying to achieve the same thing, and I found a simple solution - if you want to select only rows with today's date you can do :
WHERE DATE(datetime_column)=CURDATE()
in your mySQL query syntax.
You'd have three solutions :
Working with strings, like you are doing ; which seems like a solution that works ; even if it doesn't feel clean.
Working with timestamps, using strtotime() and time() ; which is a bad idea : UNIX Timestamps only work for dates that are greater than 1970 and lower than 2038
Working with DateTime everywhere ; which would both work and feel clean.
If I need to make any calculation on the PHP-side, I would probably go with the third solution -- but the first one would be OK in most cases, I suppose.
As a sidenote : instead of formating your date to Y-m-d, you could check if it's :
Greater of equal than today
Less than tomorrow.
If SQL returned date is in this format 2011-03-09 (date format without timing),
$sqlret = "2011-03-05";
$curdate = date('Y-m-d');
echo $diff = strtotime($curdate) - strtotime($sqlret);
echo $no_diff = $diff/(60*60*24);
If the date with time like:
$sqlret = "2011-03-05 12:05:05",
Just make your current date format also like that:
$curdate = date('Y-m-d H:i:s');
If it doesn't satisfies your need, ask your question with some example.
You can use new DateTime php Object that way.
$date1 = new DateTime('2012-01-21');
$date2 = new DateTime ( 'now');
$interval = $date1->diff($date2);
if( $interval->format('%R%a ') == 0){
echo 'it s today';
}
I'd do that:
# SQL
SELECT DATE_FORMAT(date_col, "%Y-%m-%d") AS created_at FROM table
# PHP
if ( date('Y-m-d') == $sql_date ) { // assuming $sql_date is SQL's created_at
echo 'today';
}
$time = //your timestamp
$start = mktime(0,0,0,date("j"),date("n"),date("Y"));
$end = mktime(23,59,0,date("j"),date("n"),date("Y"));
if($time > $start && $time < $end){
//is today
}
I have in a MySQL table a DATE column that represents the date in this format: YYYY-MM-DD.
I wanto to retrieve the date from the database using PHP but display it like this: DD Month, YYYY.
From '2009-04-13' to '13 April, 2009' for example.
Witch is the best way to do it?? ( I know how to get the date from the DB. I only need to know how to convert it)
I also need to display the month names in Spanish. There is a way to do it without translating each month using strplc or something like that??
I'm new to programming, please be detailed.
Thanks!!!
Refer to DATE_FORMAT() function in MySQL. I guess that's the best way for you to do it.
Also, you can make this:
Fetch your date from DB
Use strtotime in PHP, to convert to unix time
Then format the time using date.
By using date() you'll be able to get months names in Spanish when you set your locale in PHP with setlocale.
You could also skip the strtotime() part by using UNIX_TIMESTAMP(date) in your MySql select. But remember that this is a MySQL specific function and may not be be portable in the future.
Execute following MySQL queries:
SET lc_time_names = 'es_ES';
SELECT DATE_FORMAT(t.date,'%e de %M, %Y') FROM your_table t ...
With MySQLi it'll be:
$mysqli->query("SET lc_time_names = 'es_ES'");
$stmt = $mysqli->prepare("SELECT DATE_FORMAT(t.date,'%e de %M, %Y') FROM your_table t ...where id = ?");
...
Another option not yet mentioned:
SQL:
SELECT UNIX_TIMESTAMP(date) FROM table
PHP:
print date('your format', $timestamp_from_the_db);
Personally, I like to use integer data types in MySQL for date storage in the UNIX timestamp format. I leave all the processing of that integer up to PHP. Keeping tables and queries as simple as possible has always served me well. Predominantly, in the code I write, dates have some sort of calculation done to them. This is all done on the PHP side and always in the UNIX timestamp format. Storing or retrieving the dates in anything other than the UNIX timestamp format just means another step for errors to creep in and makes the query less modular. How a date is formatted is best left up until the last minute before it's displayed. It's just my opinion, but unless there are extreme circumstances where you can't process the DB value after extraction, a date shouldn't be formatted SQL-side.
A simplified example:
<?php
$date = now();
$dueDate = $date + 60*60*24*7; // One week from now
$sqlInsert = "INSERT INTO reports SET `dueDate` = $date";
$resInsert = mysql_query( $sqlInsert );
$sqlSelect = "SELECT `dueDate` FROM reports";
$resSelect = mysql_query( $sqlSelect );
$rowSelect = mysql_fetch_array( $resSelect );
$DB_dueDate = $rowSelect['dueDate'];
$daysUntilDue = ( $DB_dueDate - now() ) / 60*60*24;
$formattedDueDate = date( "j F, Y", $DB_dueDate );
?>
The report is due on <?=$formattedDueDate?>. That is <?=$daysUntilDue?> from now.
Simplest way is to use the strtotime() function to normalize the input to UNIX timestamp.
Then use the date() function to output the date in any format you wish. Note that you need to pass the UNIX timestamp as the second argument to date().
This will help you to convert as you want:
$dob ='2009-04-13';
echo date('d M Y', strtotime($dob));
$origDate = "2018-04-20";
$newDate = date("d-m-Y", strtotime($origDate));
echo $newDate;