I am using a DATE field in my MYSQL table, and pulling it through on a php page. The problem is it comes out as "2011-04-23"
Is there a way I can reformat this as 23/04/2011?
Thanks :)
date("d/m/Y", strtotime("2011-04-23"));
that should do it
date()
strtotime()
DATE_FORMAT(date,format)
Look here: http://dev.mysql.com/doc/refman/5.0/es/date-and-time-functions.html
Assuming variable $date contains your MySQL data:
$date = '2011-04-23';
$timezone = 'Europe/London'; // this is optional argument
$formatted = DateTime::createFromFormat('Y-m-d', $date, new DateTimeZone($timezone));
// or without the optional timezone - where php will assume the default timezone from your OS
$formatted = DateTime::createFromFormat('Y-m-d', $date);
echo $formatted->('d/m/Y');
Related
In users table I have a column call "month".
I want to list all users that meet the condition: current month - user month <=2
Here my code
$time = new \DateTime('now');
$today = $time->format('m');
$users = Users::find()->where(['<=', 'month' - $today, 2])->all();
But this code is wrong. Please help me with this.
Hope this all make sense.
Thank you!
In Yii2 you can use different format for buil where condition
for this kind of situatio is useful use string format with param
in string format you can pass the literal string and param this way
$users = Users::find()->where('(month - :today ) <= 2' , [':today'=>$today])->all();
see this for more http://www.yiiframework.com/doc-2.0/yii-db-query.html#where()-detail
Is it possible to format the following date in an active record where clause?
I need to format channel_titles.entry_date to be in the format Y-m-d, like the $yesterday variable.
$yesterday = date('Y-m-d', strtotime('yesterday'));
$this->db->where('channel_titles.entry_date', $yesterday);
You can use DATE_FORMAT function like that:
$this->db->where("DATE_FORMAT(from_unixtime(channel_titles.entry_date), '%Y-%m-%d')", $yesterday, false);
Try this
$this->db->where("DATE_FORMAT(channel_titles.entry_date,'%Y-%m-%d') ", $yesterday);
function details_klanten($idKlant,$start,$eind){
$this->db->select(' Project.idProject,
Project.Titel,
Project.idProjecttype,
Project.Begindatum,
Project.Deadline,
Project.idKlant,
Projecttypes.idProjecttypes,
Projecttypes.Type,
Werknemer.idWerknemer,
Werknemer.Voornaam,
Statusproject.idStatusProject,
Statusproject.Soort,
Klant.Naam');
$this->db->order_by('Titel', 'asc');
$this->db->from('Project');
$this->db->join('Klant', 'Klant.idKlant = Project.idKlant');
$this->db->join('Projecttypes', 'Projecttypes.idProjecttypes = Project.idProjecttype');
$this->db->join('Werknemer', 'Werknemer.idWerknemer = Project.idWerknemer');
$this->db->join('Statusproject', 'Statusproject.idStatusProject = Project.idStatusProject');
if ($idKlant > 0){
$this->db->where('Klant.idKlant',$idKlant);
$this->db->where('Project.Begindatum >',$start);
$this->db->where('Project.Deadline <',$eind);
}
$query = $this->db->get();
if($query->num_rows()>0){
return $query->result();
}
else{
return false;
}
}
The project.Begindatum and the Project.Deadline are varchar(10). So it looks at the first two numbers not the full date. For example:
$start = '01-04-2014';
'Project.Begindatum' = (Varchar)'02-03-2014'.
Then it will be shown because
it looks only to the '01'(-04-2014) and the '02'(-03-2014)
Use mysql's STR_TO_DATE() function and also pass third parameter in where() as FALSE
$this->db->where("STR_TO_DATE(Project.Begindatum,'%d-%m-%Y') >",$start,FALSE);
$this->db->where("STR_TO_DATE(Project.Deadline,'%d-%m-%Y') <",$eind,FALSE);
Its better to change the type of your columns to store in standard format use date type to store dates in database ,Using STR_TO_DATE with format %d-%m-%Y then the value stored in table should have %d-%m-%Y format otherwise it won't work ,it should be compared to $start = '2014-04-01';,otherwise you need another function i.e DATE_FORMAT to format it like 01-04-2014
$this->db->where("DATE_FORMAT(STR_TO_DATE(Project.Begindatum,'%d-%m-%Y'),'%d-%m-%Y') >",
$start,FALSE);
$this->db->where("DATE_FORMAT(STR_TO_DATE(Project.Deadline,'%d-%m-%Y'),'%d-%m-%Y') <"
,$eind,FALSE);
Reminder: STR_TO_DATE(...) output format: Y-m-d
Example: 2020-12-24
Therefore: $start and $eind format: Y-m-d
Example: 2020-12-31
$this->db->where("STR_TO_DATE(Project.Begindatum, '%m-%d-%Y') >",$start);
$this->db->where("STR_TO_DATE(Project.Deadline, '%m-%d-%Y') <",$eind);
I hope this helps you too. Peace.
I'm having some troubles creating query conditions in Propel 1.6 using Mysql date functions, such as:
$query = PostQuery::create()->where('YEAR(Post.PublishedAt) = ?', $year)->find();
The following works:
$query = PostQuery::create()->where('YEAR(Post.PublishedAt) = 2011')->find();
But in order to handle a $year variable, I then have to write something like:
$query = PostQuery::create()->where(sprintf('YEAR(Post.PublishedAt) = %d', $year))->find();
which seems wrong.
I've also tried to specify param type by adding: PDO::PARAM_INT const:
$query = PostQuery::create()->where('YEAR(Post.PublishedAt) = ?', $year, PDO::PARAM_INT)->find();
This doesn't work.
UPD: There are no errors. Just an empty collection as a result.
Could anybody help me? Thanks in advance.
According to OP
// works
PostQuery::create()->where('YEAR(Post.PublishedAt) = 2011')->find();
// doesn't work
PostQuery::create()->where('YEAR(Post.PublishedAt) = ?', $year)->find();
This would indicate that PDO is quoting the $year variable so it's actually doing the following when using params
PostQuery::create()->where('YEAR(Post.PublishedAt) = \'2011\'')->find();
As detailed in this bug report https://bugs.php.net/bug.php?id=44639
Do either of these work
PostQuery::create()->where('YEAR(Post.PublishedAt) = CONVERT(?,UNSIGNED)', $year,PDO::PARAM_INT)->find();
or (this is slightly cleaner)
settype($year,'int');
PostQuery::create()->where('YEAR(Post.PublishedAt) = ?', $year,PDO::PARAM_INT)->find();
Seems a bit idiotic to have to do it that way but it should convert the numeric string into a MySQL UNSIGNED INT suitable for the WHERE statement. Either way the above are either having MySQL convert it into a numeric type explicitly, or alternatively trying to ensure that PHP has it explicitly as an INTEGER type before binding it.
I'm quite new to php and have been reading Larry Ullman book to develop a basic db site.
I have used the YUI Calendar pop up date picker to add a date to a text field called"date". The date format it enters is eg Thursday, 7 May 2009
I have tried many different ways to try and enter the date in to mysql db but it remains at 00 00 00 00 00 00
This is the code related to the date field I have,
// Check for a Date.
if (eregi ("^([0-9]{2})/([0-9]{2})/([0-9]{4})$", $_POST['date'],)) {
$p = escape_data($_POST['date'],);
} else {
$p = FALSE;
echo '<p><font color="red">Please enter a valid Date!</font></p>';
}
// Add the URL to the urls table.
$query = "INSERT INTO urls (url, title, description, date) VALUES ('$u', '$t', '$d', '$p')";
$result = #mysql_query ($query); // Run the query.
$uid = #mysql_insert_id(); // Get the url ID.
if ($uid > 0) { // New URL has been added.
I think I have provided all pertinent information but again apologies if this isn't helpful and I will do my best to provide yo with any other information you may require.
Thanks - Sean
If the format that your date picker is passing in is "Thursday, 7 May 2009", then the strtotime() and date() functions should work to give you a valid date to pass to MySQL:
$p = date("Y-m-d H:i:s",strtotime('Thursday, 7 May 2009'));
If your database field is a DATE, you probably only want the "Y-m-d" part. If it's a DATETIME, you'll want the "H:i:s" as well.
You probably need to format the date in the way that mysql expects it for that datatype, or else it cannot recognize it. You could use a regex or similar method to extract the component parts and format it according to the format of the MySQL DATETIME type.
EDIT: See http://dev.mysql.com/doc/refman/5.0/en/datetime.html for said format. Also, you might be better off storing the date/time as a unix timestamp, as that usually is easier to maintain.
the date format in mysql is YYYY-MM-DD, so change the way you accept data on php
if (eregi ("^([0-9]{2})/([0-9]{2})/([0-9]{4})$", $_POST['date'],)) {
$p = escape_data($_POST['date'],);
} else {
$p = FALSE;
echo 'Please enter a valid Date!';
Should be
if (eregi ("^([0-9]{4}/([0-9]{2})/([0-9]{2}))$", $_POST['date'],)) {
$p = escape_data($_POST['date'],);
} else {
$p = FALSE;
echo 'Please enter a valid Date!';