Mark Hensler — Fri, 01 Nov 2002 09:01:11 +0000

I prefer to through as much work at the mysql DB as (resonably) possible.

Question 2/3:
datestamp? That's not a mySQL data type. I'm guessing timestamp(. Anyway...

SELECT if(date_field=0, 'unknown', DATE_FORMAT(date_field, '%m/%d/%Y')) as date_field;

/* OR */

SELECT if(date_field=0, 'unknown', DATE_FORMAT(date_field, '%M %e, %Y')) as date_field;

Question 4:
The solution above is specific to mySQL. It's using regex to search the records. The section isn't documented well in the mySQL docs (at least, not that I've found).

fifeclub — Fri, 01 Nov 2002 05:08:14 +0000

I got the answer to #4

Instead of using "LIKE", the query should be

<?php
$query = \"SELECT * FROM alumni WHERE lastname RLIKE '^[$lastname]'\";
$result = mysql_query ($query);
?>

And with switching from LIKE to RLIKE, the variable $lastname changes from A% (for just beginning with A) to either ABCD or A-D! Note: the variable is actually [A-D] with brackets but I chose to keep the brackets out of the url part of it.

I love it when things actually work out!

Re: 3 simple questions, plus 1 more

necrotic — Thu, 31 Oct 2002 20:04:08 +0000

Quote: Originally posted by fifeclub
QUESTION #1:

Out of curiosity, what is the significance of \n before the "; at the end of a a print statement? It prints either way, so what does the \n do?

\n denotes a new line. It's like but it only changes how it's displayed in the source. I recommend using it so your source isn't cluttered when people want to view it.

Quote:
QUESTION #2:

I use a datestamp( field which produces dates like this "20021031". Is there an easy way to make this print/display in any human-fiendly format? Either "10/31/2002" (US format) or "October 31, 2002"?

You can use SUBSTR($date, 0, 4) for the year (2002), SUBSTR($date, 4,2) for the month, and SUBSTR($date, 6, for day. In PHP it would look like this:

<?php
$date = 20021031;
$year = substr($date, 0, 4);
$month = substr($date, 4, 2);
$day = substr($date, 6, 8);

// 10/31/02
$date = $month.\"/\".$day.\"/\".$year;
echo $date;

// October 31, 2002
$months = array(\"January\", \"Feburary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\", \"October\", \"November\", \"December\");
$month = $month - 1;
$dmon = $months[$month];

echo (\"$dmon $day, $year\");
?>

More on Substr

Quote:
QUESTION #3:

Related to #2, is there an easy way to make a datestamp( value of 00000000 print as the text "unknown" or to not print anything at all? I suppose this could be for any field value result. Is it possible to do something like 'if fieldname=00000000 then make other-value' (where it would print that other-value instead)?

<?php
if ($date == '00000000') {
  echo (\"Unknown\");
} else {
  // ... Date code from above
}
?>

Quote:
QUESTION #4 (not as simple as the others):

From my link above, my "Locations" script uses the following method for the query to include multiple non-intiger values (the key being "IN"):

<?php $query = \"SELECT * FROM alumni WHERE state IN (\"$location\")\"; ?>

My "Last Name Starting With The Letter x" script uses the following method for the query (the keys being "LIKE") and the string variable is formatted as "x%":

<?php $query = \"SELECT * FROM alumni WHERE lastname LIKE '$lastname'\"; ?>

What I want to do is kind of combine these concepts to display last names beginning with the letters A%,B%,C%,D%,E% (beginning with any of these multiple values). My first attempt failed and I just wanted to know if this can possibly work if I just fiddle around enough with the syntax and formatting? Am I on the right track or is this just all wrong and there's a different way to do this? [/B]

Look into arrays and the for function some, I asked a similar question. If you want to see the code I got just ask and I'll give you a link to it.

Hope I helped you some, good luck!