CODEWITHSUNDEEP
phpmysqlarrayswhile-loop
TheLettuceMaster
TheLettuceMaster

Reputation: 15734

Storing Data into PHP Array

total_starsHere is the code, it is the best explanation of what I am trying to do:

$total_stars = array();
$query = "SELECT items_id FROM items";
$result = mysql_query($query);

while ($row = mysql_fetch_array($result)) {
    $item_id = $row['items_id'];
    $sql = "Select Count(item_id) FROM ratings WHERE item_id = '{$item_id}'";
    $result = mysql_query($sql);
    $total_stars[] = mysql_fetch_array($result);
}

//  To see the output by entering the URL I want to print_r  
print_r($total_stars);

// In the end, I am trying to JSON encode this and will later output in Java for Android
print(json_encode($total_stars));

(So as not to confuse, the items_id is in items table, item_id (without an 's') is in ratings table)

Now in plain English:

I am looping through each row in the items table. As I do that, in each loop I am referring to another table called "Ratings", and getting a count of how many ratings for that item.

For example I want an array to look like (65, 43, 55, 43, etc). Sample numbers of course. But each represent the total amount of ratings for each item in a table. How can I get print__r to display the results of the SQL statement inside the loop?

UPDATED CODE TRYING TO USE JOIN:

$query = "SELECT items_id FROM items";

$q = 'SELECT 
    items.items_id, 
    COUNT(ratings.item_id) AS rate 
    FROM `items`
    LEFT JOIN ratings ON (ratings.item_id = items.items_id)
    GROUP BY items_id';
$result = mysql_query($q);

while ($row = mysql_fetch_array($result)) {

    $total_stars = mysql_fetch_array($result);

}

print_r($total_stars);

print_r outputs this: Array ( [0] => 783 [items_id] => 783 [1] => 0 [rate] => 0 )

Upvotes: 0

Views: 3706

Answers (2)

Sven van Zoelen
Sven van Zoelen

Reputation: 7229

Why not use a JOIN as suggested in the comments? If you want an array with all the ratings of the items, then this is a way (didn't test the PHP db code, but i did test the query).

$total_stars = array();

$q = 'SELECT 
        items.item_id,
        COUNT(ratings.item_id) AS rate 
      FROM `items`
      LEFT JOIN ratings ON (ratings.item_id = items.item_id)
      GROUP BY item_id';
$result = mysql_query($q);

$total_stars = mysql_fetch_array($result);

You should use MYSQLi or PDO_MYSQL instead. more info MYSQLi, more info PDO

And here you can find more info about JOINS: Info about MYSQL joins

Upvotes: 1

Jonathan
Jonathan

Reputation: 3034

$total_stars = array();
while ($row = mysql_fetch_array($result))
{
    $item_id = $row['items_id'];
    $sql = "Select Count(item_id) FROM ratings WHERE item_id = '{$item_id}'";
    $result = mysql_query($sql);
    $row = mysql_fetch_row($result);
    $total_stars[] = row[0];
}

This should do it.

Don't forget that mysql_fetch_array returns an array with index for the columns. So mysql_fetch_array($result) would be an array that you want the first element in (result from MySQL function COUNT()), not the count itself.

But as I said in my comment, use a join query instead. You will reduce the amount of queries to the mysql server from the number of rows in your first query+1 (which could be alot!) to just one, which is a huge improvement.

Update: Here's how to do it with a join query (query by Sven):

$q = 'SELECT 
    items.items_id, 
    COUNT(ratings.item_id) AS rate 
    FROM `items`
    LEFT JOIN ratings ON (ratings.item_id = items.items_id)
    GROUP BY items_id';
$result = mysql_query($q);

$total_stars = array();
while ($row = mysql_fetch_array($result))
{
    $total_stars[] = $row['rate']; // Stars count is stored in $row['rate']
    echo "Item id ".$row['items_id']." has rating ".$row['rate'];

}

Upvotes: 3

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