Emiliano Boragina wrote:
Hi Chris,
Sorry... you're right!
With this I want to print the category in a html table after a search...
The search find the data, and print category and description. This category
is a number (ID), but I want to print the name of the category. My MYSQL
CATEGORY TABLE have ID and NAME, so using the SWITCH, I want to put in CASE
the ID and ECHO the NAME.
In other webs I do
switch($category) //where $category is an array fom the db
{
case 1:
echo "category1";
break;
case 2:
echo "category2";
break;
...
}
I want the 1, 2, 3... like the ID from my table, and "category1,
category2..." like the NAME from my table
Do not know if it was understood...
There are a couple of ways you can do it and no need for a switch.
1) get the category name as part of the query itself
This works nicely if an article only fits into one category. If it fits
into multiple categories, it gets a little messy (though it's still doable).
<guessing tablenames etc>
select * from articles where id='x';
becomes
select a.*, c.category_name from articles a inner join categories c on
(a.category_id=c.category_id) where id='x';
2) load all categories at once and then print the relevant one
This works if you have a small number of categories available. If you
have a lot (50+) then I'd look at the first method.
$category_list = array();
$category_query = "select * from categories order by category_name";
$category_result = mysql_query($category_query);
while ($row = mysql_fetch_assoc($category_result)) {
$cat_id = $row['category_id'];
$cat_name = $row['category_name'];
$category_list[$cat_id] = $cat_name;
}
then you can just print out the name, eg:
// article_category comes from your query to load the article.
$article_category = 'x';
// make sure it's a valid category name
if (isset($category_list[$article_category])) {
echo "Your article is in category " .
htmlspecialchars($category_list[$article_category]);
}
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