Re: [PHP] function sql question
At 6:34 PM +0200 5/5/02, John Fishworld wrote:
John,
Your mysql_db_query should contain either "DBWEB" (with
quotes) or $DBWEB (a var). Just DBWEB is an error.
Also, from the manual re: mysql_db_query
Note: This function has been deprecated since PHP 4.0.6.
Do not use this function.
Use mysql_select_db() and mysql_query() instead.
Your SQL query is incorrect. The strings you use to identify
a column contain "_" instead of ".". Perhaps just an artifact
of your mailer, but the query as it is should be throwing
"unknown column t_city_name" errors. Try adding some "as"
clauses into your query to name these "virtual" columns.
Or replace the "_" with ".".
Hey, and it's nice to see a normalized database for a change :)
Your coding formatting style is likely to cause you problems.
You have (partially) indented the code the way you want to
behave logically, but your braces are all over the shop.
Try adopting one of the coding styles commonly in use and
that will almost certainly make your life easier. I use the
K&R "one true brace style" because I'm a C hacker from way
back, but you might like an extra line break before a "{".
As for this function, the first time through the loop,
unless the cityname is not "" you return 0. Since you test for
a blank cityname, I assume that's a possible return value from
the database. In this case you shouldn't return 0 then, but
instead loop though to the next result in the result set.
Try this:
function check_city_from_plz($str) {
// connect and select database
$city_query = "select distinct t_city.name AS t_city_name,
t_city.id_city AS t_city_id_city from
t_city,t_zipcodecity where
(t_city_id_city = t_zipcodecity_id_city) and
(t_zipcodecity.zipcode like '$str')";
$city_result = mysql_db_query(DBWEB, $city_query);
if ($city_result) {
while ($q = mysql_fetch_array($city_result)) {
$city_id = $q["t_city_id_city"];
$cityname = $q["t_city_name"];
if ($cityname != "") {
r["city_id"] = $city_id;
r["cityname"] = $cityname;
return $r;
} // close if
} // close while
// return 0 if no matches after iterating through data
return 0;
} // close function
...R.
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RE: [PHP] function sql question
It will work; you're just doing something wrong. Do you want to fix it,
or keep doing it this way?
---John Holmes...
> -Original Message-
> From: John Fishworld [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, May 05, 2002 10:55 AM
> To: [EMAIL PROTECTED]; 'PHP-General'
> Subject: Re: [PHP] function sql question
>
> duh !
> yeah that makes sense !
> but it doesn't work !
> I've no idea why but it refuses to give any information back !
>
> I've managed to get round this by now no loger declaring it a function
> but just actually requiring it where i need it
>
> require ("get_city_1.inc");
> print_r(array_values ($city));
>
> I've also chopped the return part
> and just return the values
>
> $city[0] = $city_id;
> $city_name[1] = $city_name;
>
> Strange
>
>
> > >return 1;
> > >return $city_id;
> > >return $cityname;
> >
> > You can't return 3 values...try returning an array, instead.
> >
> > $ret[0] = $city_id;
> > $ret[1] = $cityname;
> > return $ret;
> >
> > or...
> >
> > $ret["City_ID"] = $city_id;
> > $ret["CityName"] = $cityname;
> > return $ret;
> >
> > ---John Holmes...
> >
> >
>
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Re: [PHP] function sql question
duh !
yeah that makes sense !
but it doesn't work !
I've no idea why but it refuses to give any information back !
I've managed to get round this by now no loger declaring it a function
but just actually requiring it where i need it
require ("get_city_1.inc");
print_r(array_values ($city));
I've also chopped the return part
and just return the values
$city[0] = $city_id;
$city_name[1] = $city_name;
Strange
> >return 1;
> >return $city_id;
> >return $cityname;
>
> You can't return 3 values...try returning an array, instead.
>
> $ret[0] = $city_id;
> $ret[1] = $cityname;
> return $ret;
>
> or...
>
> $ret["City_ID"] = $city_id;
> $ret["CityName"] = $cityname;
> return $ret;
>
> ---John Holmes...
>
>
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RE: [PHP] function sql question
>return 1; >return $city_id; >return $cityname; You can't return 3 values...try returning an array, instead. $ret[0] = $city_id; $ret[1] = $cityname; return $ret; or... $ret["City_ID"] = $city_id; $ret["CityName"] = $cityname; return $ret; ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] function sql question
I've got a small function which checks city name with post codes
And because this needs to be checked in several scripts I want to have it in
my common functions file
and include it !
require ("common_funtions.inc");
and then use it
if (!check_city_from_plz($t_zipcode)){
echo "Your city does not exist with the
plz!
";
}
It doesn't work though, I'm sure it something simple but can someone please
point me in the right direction !
This is the function itself
function check_city_from_plz($str) {
connect and select database
$city_query = "select distinct t_city_name,t_city_id_city from
t_city,t_zipcodecity where
(t_city_id_city = t_zipcodecity_id_city) and (t_zipcodecity_zipcode like
'$str')";
$city_result = mysql_db_query(DBWEB, $city_query);
if ($city_result) {
while ($q = mysql_fetch_array($city_result)) {
$city_id = $q["t_city_id_city"];
$cityname = $q["t_city_name"];
if ($cityname !="") {
return 1;
return $city_id;
return $cityname;
} else {
return 0;
}
} // close if
} // close while
} // close function
Thanks in advance !
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