Removing the first character of a string...How?

Anon

Hello! How would I go about removing the first character of a string, if it were a certain number? In this case, 0. Here's an example:

$num = strftime ("%m");
$array = array (1, "Jan", 2, "Feb", ..., 10, "Oct");
print $array[$num];

Unfortunately, $num will return a two digit number everytime. Say, the month is January, $num will return 01 instead of 1. How can I fix this?

Anon

Actually, there are several things you could do to that code. I would use a simple array rather than an associative array:
$array = array("Before the beginning, of time", "Jan", "Feb",
... and so forth, until you get to
"December");

Remember that arrays are zero-based, not 1-based.

And cast $num to an int before using as an array index:

print $array[(int)$num];

And if you just want to remove a leading 0 from a string, ereg_replace() would do nicely.

Anon

Well, I just used that array as an example... The real one looks like this:

$current_month = strftime ("%m");
$current_day = strftime ("%d");

$months_days = (array (1 => 31, 2 => 29, 3 => 31, 4 => 30, 5 => 31, 6 => 30, 7 => 31, 8 => 31, 9 => 30, 10 => 30, 11 => 30, 12 => 31));

$till_days = $months_days[$current_month] - $current_day;

I guess I should have just posted this one in the first place. $months_days contains the number of days per month (Jan = 1, so 31 days)... Is there something built into PHP that'll figure out how many days per month in a more oranized fashon?

"And cast $num to an int before using as an array index" I'm assuming int = integer? I'm extremely new to programming, I guess I need to get a good, easy to read PHP book. 🙂

Anon

Not sure what you're trying to do at the higher level ... wouldn't it be simple to use mktime() to construct a Unix timestamp, and then date() to extract the information you seek? The code used to create those functions is calendar-literate.

On the book issue, I know there are some good PHP books coming out, but I swear by the annotated manual at www.php.net. The commentary is the best part.

Anon

Maybe try the manual...it's got more info than you could possible need...

So you're trying to figure out how many days are left in the current month? This will work:

$till_days = date("t") - date("j");

Also, FYI, date("j") will return the day of the month without leading zeros. (date("t") returns how many days are in the month)

Love that manual...

---John Holmes...

Anon

lol... Geez, do I feel stupid now. 🙂 I sure wish I'd read over the entire date manual page...

As for the book, I just ordered PHP Essentials. It had good reviews, and was fairly new. Hopefully It'll be worth the $25. 🙂 Thanks!

Anon

$till_days = date("t") - date("j");

Sounds like "The Days of Living Dangerously". Doesn't anybody
know that doing this opens the possibility of generating incorrect
results when the two calls to date() span midnight, and the Mr.
Murphy practically guarantees that the person this happens to
will be your Most Important Customer?

Capture a single timestamp, and then use that as the optional
second argument to date().

Anon

Whoa....just call me a rebel...I like to live life on the edge...

You're right of course, and you should always plan for any possible option, no matter how small. How about this..

$string = date("t,j");
$part = explode(",",$string);
$till_days = $part[0] - $part[1];

---John Holmes...