Parse the date into integer

[momogi]

I' making an database application. In one form (Input data Form), I need to fill the text field with the reference number automatically. This number must be start with date of today. For the example, 2007092401. And the text field must be filled with the largest number in database (the largest of that day).
Firstly, the form will check if there's a number start with today's date. If it's not NULL, the form takes the biggest one. But if it returns NULL, I wanna make a number that start with today's date. But I don't know how to join the date (I take the date with time's function) and the '01'...
(PS : the type of my reference number's field in MySQL is integer)

Here's a piece of my code.

$now=time();
$ref=date("Ymd",$now);
 
$refku=implode($ref,"01");  // I need some help here
 
$peri="select max(noreff) as A from tblref where noreff like '$ref%' ";
$hasil=mysql_query($peri);
$row=mysql_fetch_array($hasil);
 
if($row!=NULL)
{
echo (" <input name=\"noreff\" type=\"text\" value=\"$row[A]\" maxlength=10></td> ");
}
 
else
{
echo (" <input name=\"noreff\" type=\"text\" value=\"$refku\" maxlength=10 /></td> ");
}    
 

---Thank You---

[Atli]

Hi momogi.

You could simply do this:

$refku= $ref ."01";
 

$refku would technically be a string, but PHP is smart enough to convert it into whatever type it needs to be in various situations.

For example:

$refku = date("Ymd", time()) . "01";
# Now $refku is a string "2007092401"
 
echo "refku plus one: ". ($refku + 1);
# Will output "refku plus one: 2007092402"
 

[momogi]

Hi momogi.

You could simply do this:

$refku= $ref ."01";
 

$refku would technically be a string, but PHP is smart enough to convert it into whatever type it needs to be in various situations.

For example:

$refku = date("Ymd", time()) . "01";
# Now $refku is a string "2007092401"
 
echo "refku plus one: ". ($refku + 1);
# Will output "refku plus one: 2007092402"
 

Hi Atli...

thx 4 your reply... :)

But I have to say that it doesn't work.
My text field still empty,,, (except I enter the today's reference manually in database).

Is there another solution? thx b4.

[code green]

But I have to say that it doesn't work.Is there another solution?

Atli's solution will work.
You have obviously asked the wrong question which was

Parse the date into integer

My text field still empty,,,

Is this you problem? If so, what text field and where?

[momogi]

Atli's solution will work.
You have obviously asked the wrong question which was Is this you problem? If so, what text field and where?

Hi...

The text field is a field where I wanna put my reference number automatically. its placed in the top of my form.
if the user haven't fill the form at the day, the form will generate the reference number automatically which started with the date.

hope that was clear 4 you. I'm sorry if I was ask the wrong question...

gbu.

[Atli]

Try something like this:

# Execute query and get row
$ref = date("Ymd", time());
$query = "SELECT MAX(noreff) AS 'A' FROM tblref WHERE noreff LIKE '$ref%' ";
$result=mysql_query($query) or die("Query failed");
$row=mysql_fetch_assoc($result);
 
# Create reference
if(@isset($row['A'])) {
  $refku = $row['A'];
}
else {
  $refku = $ref . "01";
}
 
# Print reference
echo "<input name=\"noreff\" type=\"text\" value=\"$refku\" maxlength=10 /></td> "
 

[momogi]

Try something like this:

# Execute query and get row
$ref = date("Ymd", time());
$query = "SELECT MAX(noreff) AS 'A' FROM tblref WHERE noreff LIKE '$ref%' ";
$result=mysql_query($query) or die("Query failed");
$row=mysql_fetch_assoc($result);
 
# Create reference
if(@isset($row['A'])) {
  $refku = $row['A'];
}
else {
  $refku = $ref . "01";
}
 
# Print reference
echo ("<input name=\"noreff\" type=\"text\" value=\"$refku\" maxlength=10 /></td> "
}    
 

whoaaa... it works!!!

thanks...

thanks... :)

and thanks... :D