Inserting date value into MySQL - php

Question

I currently have a form which takes a date in the format m/d/y - I have then attempted to insert it into a table, but the value in the table reads 0000-00-00. I understand that the value is not being inserted due to the format of the date being inserted.
The problem is, I am unsure on how to change the format so that it is inserted in a format that MySQL will store.
Below is the function that inserts the data into the table:
public function addUser($array) {
$array['password'] = $this->hashPassword($array['password']);
$implodeArray = '"'.implode( '","', $array ).'"';
$sql = ('INSERT INTO user
(email, password, firstName, lastName, officeID, departmentID, managerID, roleID, username, contractType, startDate, endDate, totalLeaveEntitlement, remainingLeave)
VALUES
('.$implodeArray.')');
echo $sql;
die();
mysql_query($sql,$this->_db) or die(mysql_error());
mysql_close();
}
Due to the use of implodeArray, I cannot format the value of startDate and endDate to match the MySQL DATE format.

Answer 1

Why don't you use similar method to when you hashed the password? So, you just need to add another function to convert your date input into mysql date format:
public function addUser($array) {
$array['password'] = $this->hashPassword($array['password']);
$array['startDate'] = $this->mysql_date_format($array['startDate']);
$array['endDate'] = $this->mysql_date_format($array['endDate']);
$implodeArray = '"'.implode( '","', $array ).'"';
$sql = ('INSERT INTO user (email, password, firstName, lastName, officeID, departmentID, managerID, roleID, username, contractType, startDate, endDate, totalLeaveEntitlement, remainingLeave) VALUES ('.$implodeArray.')');
echo $sql;
die();
mysql_query($sql,$this->_db) or die(mysql_error());
mysql_close();
}

Answer 2

Hmmmmm
I know it looks like its easier to write queries like this (one function generates all your parameters etc etc) but I would STRONGLY advise that you prepare your statements - someone coming along to support your code will thank you for it.
That way you can use NOW(), DATE_DIFF and such other awesomes...
I know that doesn't answer your question but I do feel you should take the time to construct your queries properly - help prevent run time errors/ attacks etc etc.

Answer 3

Not sure on the specifics of your issue, but in general:
$mysql_formatted_date = date("Y-m-d", strtotime($mdy_formatted_date));

Answer 4

I think you'll want STR_TO_DATE()
STR_TO_DATE("%m/%d/%Y") is I think the right format

Answer 5

While both arrays and mysql columns have an implicit order, how do you know they are the same?
It would have been a lot more useful if you'd provided the output of 'echo $sql' rather than all the PHP code - although hte latter highlights a lot of messy programming not least:
the field order problem
quoting non-numeric values
not escaping fields properly
not trapping / handling errors
no comments
form which takes a date in the format m/d/y - I have then attempted to insert it
In the case of date fields, quoting is optional depending on the format used for the literal - but it is always ordered as per ISO 8601 - i.e. big endian

Answer 6

public function addUser($array) {
list($d,$m,$y) = explode("/",$array['startDate']);
$array['startDate'] = "$y-$m-$d";
list($d,$m,$y) = explode("/",$array['endDate']);
$array['endDate'] = "$y-$m-$d";
$array['password'] = $this->hashPassword($array['password']);
foreach($array as $key => $value){
$array[$key] = mysql_real_escape_string($value);
}
$implodeArray = implode("','", $array);
$sql = "INSERT INTO user VALUES (NULL,'$implodeArray')";
echo $sql;
die();
mysql_query($sql,$this->_db) or trigger_error(mysql_error());
}