Joining two different tables containing csv and display them on a webpage - php

Question

I have been stuck with a problem and I am newbie in mysql and php, Here is the code first , so that I can explain in detail:
$metros = array(1,263);
foreach($metros as $metro_id) {
$sql = "SELECT cuisine_id, cuisine_name_en FROM poi_restaurant_cuisines";
$result = mysql_query($sql);
$cuisine_id = array();
$cusine_name = array();
while($row = mysql_fetch_assoc($result)) {
$cuisine_id[] = $row['cuisine_id'];
$cuisine_name[] = $row['cuisine_name_en'];
}
foreach ($cuisine_id as $cuisine) {
$sql = "
SELECT COUNT(*)
FROM poi AS p
LEFT JOIN poi_restaurant AS pr USING (poi_id)
WHERE p.poi_address_prefecture_id = '$metro_id'
AND pr.poi_restaurant_cuisine_id_array
AND find_in_set('$cuisine', poi_restaurant_cuisine_id_array)
AND p.poi_status = 1";
$result = mysql_query($sql);
$count_cuisine = array();
while($row = mysql_fetch_array($result)) {
$count_cuisine[$metro_id][$cuisine] = $row['COUNT(*)'];
}
echo "<table border = 1 cellpadding= 5 cellspacing= 5 width= 100>";
echo "<tr><th>CuisineID</th><th>Count</th></tr>";
echo "<tr><td>";
echo $cuisine;
echo "</td><td>";
echo $count_cuisine[$metro_id][$cuisine];
echo "</td><td>";
echo "</tr>";
echo "</table>";
}
}
The poi_restaurant_cuisine_id_array contains csv values. I am able to produce the count and the cuisine ID on the web page. I want to replace the cuisine ID with the name of the cuisine. I am not very good at sql or either PHP as I am a beginner. I hope I am being clear enough. Any help is highly appreciated ...Thank you.

Answer 1

Try this:
echo '<table border="1" cellpadding="5" cellspacing="5" width="100">';
echo "<tr><th>Cuisine</th><th>Count</th></tr>";
$metros = array(1,263);
foreach($metros as $metro_id) {
$sql = "SELECT cuisine_id, cuisine_name_en FROM poi_restaurant_cuisines";
$result = mysql_query($sql);
$cuisines = array();
while($row = mysql_fetch_assoc($result)) {
$cuisines[] = array(
'id' => $row['cuisine_id'],
'name' => $row['cuisine_name_en'],
);
}
foreach ($cuisines as $cuisine) {
$sql = "
SELECT COUNT(*)
FROM poi AS p
LEFT JOIN poi_restaurant AS pr USING (poi_id)
WHERE p.poi_address_prefecture_id = '$metro_id'
AND pr.poi_restaurant_cuisine_id_array
AND find_in_set('{$cuisine['id']}', poi_restaurant_cuisine_id_array)
AND p.poi_status = 1";
$result = mysql_query($sql);
$count_cuisine = array();
while($row = mysql_fetch_array($result)) {
$count_cuisine[$metro_id][$cuisine['id']] = $row['COUNT(*)'];
}
echo "<tr>
<td>{$cuisine['name']}</td>
<td>{$count_cuisine[$metro_id][$cuisine['id']]}</td>";
</tr>";
}
}
echo "</table>";

Answer 2

Assuming cuisine_id is a unique identifier, then just use it as the array index....
while($row = mysql_fetch_assoc($result)) {
$cuisines[$row['cuisine_id']] = $row['cuisine_name_en'];
}
....
foreach ($cuisines as $cuisine_id=>$cuisine_name_en) {
However storing multiple values in a single column is a very bad idea.
Generating and running queries in a loop is another very bad idea.
It is possible to reduce this to a single query, declared and invloked outside the inner loop but because your data is not mormalized, this is rather complex.