RE: [PHP-DB] Selection problem
If I understand your problem correctly, it sounds like you need to get all of the values for the two select objects (which you can do with PHP, of course) then use JavaScript to handle the CLIENT SIDE update process between the two objects. If that sounds like what you're looking for, you can find numerous tutorials and code samples on the web. Just search for javascript on google and you should find a number of sites with script repositories. Pick a site and look to see what they have to satisfy your requirements. Hope that helps. Rich -Original Message- From: Stiphane RIVEZ [mailto:[EMAIL PROTECTED] Sent: Friday, September 26, 2003 2:12 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] Selection problem Hello, I'm novice in php and mysql utilization so... Here is my problem: I have a form with some fields (for example, 2: field1 and field2) My mysql database contains for example 1 table (named for example SQLTABLE) with 3 fields: ID (auto_increment), AA, BB. On my form, field1 is represented in HTML by a SELECT SIZE=1... and returns values of AA in selected in SQLTABLE. I want when I choose a value with the selection menu in field1 of the form, that the BB corresponding value of the selected AA automatically appears in field2 of my form without submitting my page, without loosing informations already encoded in the page. (An application could be a customer subsciption form, where you have to enter your name, adress, ZIP, town, for a determinated country... and you can choose the zip code in a selection menu and automatically the corresponding town is shown next to it before subitting the form...) I suppose that a lot of people can help me Thanks... Stephane, Belgium -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Selection problem
Try unset after the if statement
if ($nameb == ){
$query = SELECT * FROM recipes WHERE points $pointsb;
} else {
$query = SELECT * FROM recipes WHERE name = '$nameb';
$new_nameb = $nameb;
unset($nameb);
}
Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
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-Original Message-
From: Chris Payne [mailto:[EMAIL PROTECTED]
Sent: Sunday, September 07, 2003 5:43 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Selection problem
Hi there everyone,
I'm having a strange problem, I have a form with 2 fields, 1
is for points
(Weightwatcher points) and one is for recipe names. If you
select points
(IE: 1-3, 3-6,6-9,9 and above) it works perfectly, displays
the recipes and
allows you to select a new points range from the dropdown.
Next to it, on the same form, you can also select by recipe
name and ignore
the points instead selecting by the recipe name - this works
fine, you can
select 1, view it and then select another.
The problem comes when you select a recipe from the dropdown
and view it,
then decide you want to select points instead - the points
won't work once
you have selected a recipe. Can anyone see anything below
that is wrong? I
am sure it's the if statement, i'm not sure how to handle it
as you can
select either option to get the result.
Any help would really be appreciated, if I haven't confused
you all to death
:-)
Regards
Chris
?
if ($nameb == ){
$query = SELECT * FROM recipes WHERE points $pointsb;
} else {
$query = SELECT * FROM recipes WHERE name = '$nameb';
};
$sql_result = mysql_query($query,$connection)
or die(Couldn't execute query.);
while ($row = mysql_fetch_array($sql_result)) {
$id = $row[id];
$name = $row[name];
$points = $row[points];
$ingredients = $row[ingredients];
$instructions = $row[instructions];
};
?
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Re: [PHP-DB] Selection problem
Since you didn't include the form, I'm going to have to make a couple of
assumptions:
$nameb is the recipe field.
$pointsb is the points field.
You want to use SQL statement 1 if they select points, 2 if they select
recipies, but those aren't the only possibilities with the form. It sounds to
me like you're running into an issue when they select BOTH points and
recipies. So, you will need a 3 way if statement:
if ($nameb == $pointsb != ){ // if just points
$query = SELECT * FROM recipes WHERE points $pointsb;
} elseif ($pointsb == $nameb != ) { // if just recipe
$query = SELECT * FROM recipes WHERE name = '$nameb';
} else { // if both or neither, use the query you choose.
$query = ??
}
I may be wrong about what you intend here. You may want to add a if statement
that corresponds to both fields being empty if you don't validate with
Javascript or something. Sometimes I do it in the PHP anyways since there's
no guaruntee that the user will have Javascript capabilities.
-Micah
On Sun September 7 2003 3:43 pm, Chris Payne wrote:
Hi there everyone,
I'm having a strange problem, I have a form with 2 fields, 1 is for points
(Weightwatcher points) and one is for recipe names. If you select points
(IE: 1-3, 3-6,6-9,9 and above) it works perfectly, displays the recipes and
allows you to select a new points range from the dropdown.
Next to it, on the same form, you can also select by recipe name and ignore
the points instead selecting by the recipe name - this works fine, you can
select 1, view it and then select another.
The problem comes when you select a recipe from the dropdown and view it,
then decide you want to select points instead - the points won't work once
you have selected a recipe. Can anyone see anything below that is wrong?
I am sure it's the if statement, i'm not sure how to handle it as you can
select either option to get the result.
Any help would really be appreciated, if I haven't confused you all to
death
:-)
Regards
Chris
?
if ($nameb == ){
$query = SELECT * FROM recipes WHERE points $pointsb;
} else {
$query = SELECT * FROM recipes WHERE name = '$nameb';
};
$sql_result = mysql_query($query,$connection)
or die(Couldn't execute query.);
while ($row = mysql_fetch_array($sql_result)) {
$id = $row[id];
$name = $row[name];
$points = $row[points];
$ingredients = $row[ingredients];
$instructions = $row[instructions];
};
?
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