I'm coding in PHP/MySQL and have the following query to fetch products and product group data:
SELECT products.id,products.name,product_groups.id,product_groups.name
FROM products
INNER JOIN product_groups
ON products.id=product_groups.id
WHERE products.name LIKE '%foobar%'
ORDER by product_groups.id ASC
So this query fetches products and orders them by product group. What I would like to have is to display product_groups.name just once for each product grouping. So even if I have ten shoe products, the group name "Shoes" is only displayed once.
I'm using the following PHP to print out the results:
while ($data = mysql_fetch_array($result))
If you want it done in the MySQL query, it is honestly more trouble than it's worth. For one, the syntax is really wonky (as I recall) to have a group name listed at the top of each grouping. And the results are still treated as rows, so the group name will be treated like a row with all the other columns as Null, so you won't really save any time or effort in the PHP script as it has to do an if statement to catch when it hits a group name instead of the group data.
If you want it done by the PHP while loop, Johan is on the right track. I use the following for a similar situation:
$result = $sql->query($query);
$prev_group = "";
while($data = $result->fetch_assoc()){
$curr_group = $data['group'];
if ($curr_group !== $prev_group) {
echo "<h1>$curr_group</h1>";
$prev_group = $curr_group;
}
else {
echo $data;
.....
}
Obviously the echo data would be set up to echo the parts of the data the way you want. But the $prev_group/$curr_group is set up so that the only time they won't match is when you are on a new group and thus want to print a header of some sort.
while($data = mysql_fetch_assoc($result)){
if($data['product_groups.name'] != $groupname){
echo "Groupname: ".$data['product_groups.name']."<br />";
$groupname = $data['product_groups.name'];
}
echo "ID: ".$data['products.id']."<br />";
echo "Name: ".$data['products.name']."<br />";
}
Maybe you can do like this!?
Related
I'm using this to display information from a queried db in Wordpress. It displays the correct information but it loops it too many times. It is set to display from a SELECT query and depending on the last entry to the db seems to be whether or not it prints double or triple each entry.
foreach ($result as $row) {
echo '<h5><i>'.$row->company.'</i> can perform your window installation for <i>$'.$row->cost.'</i><br>';
echo 'This price includes using<i> '.$row->material.'</i> as your material(s)<br>';
echo '<hr></h5>';
}
Does anyone know what could be producing this error?
Thanks
The query powering that script is:
$result = $wpdb->get_results( "SELECT bp.*, b.company
FROM `windows_brands_products` bp
LEFT JOIN `windows_brands` b
ON bp.brand_id = b.id
JOIN Windows_last_submissions ls
JOIN windows_materials wm
JOIN Windows_submissions ws
WHERE ws.username = '$current_user->user_login'
AND bp.width = ROUND(ls.width)
AND bp.height = ROUND(ls.height)
AND bp.material IN (wm.name)
AND bp.type = ls.type
AND IF (ls.minimumbid != '0.00',bp.cost BETWEEN ls.minimumbid AND ls.maximumbid,bp.cost <= ls.maximumbid)
ORDER BY b.company ASC");
I can't seem to see the duplicate but I agree it must be there.
EDIT-- when I replace the WHERE clause to WHERE ws.username = 'password' , it still repeats. It it displaying a result for each time a result has username='password' , and displaying that set twice as well.
I think you want the following, if you're using MySQLi:
while ($row = $result->fetch_object()) {
echo '<h5><i>'.$row->company.'</i> can perform your window installation for <i>$'.$row->cost.'</i><br>';
echo 'This price includes using<i> '.$row->material.'</i> as your material(s)<br>';
echo '<hr></h5>';
}
Redundant JOIN clauses in my query which was pretty much pulling the same results from two tables (one of which was just a VIEW of the other).
I have a table with ratings for every store in a directory. I want to display the average rating from each store on a business directory list page. I have the directory business listing page finished without the average rating.
I believe I have figured out how to create an average rating, but I don't know how to get the rating to output in the loop for each store.
Here is how I calculate the rating:
$result = mysql_query("SELECT AVG(rating) FROM reviews GROUP BY store_id WHERE store_id = '$storeid'");
$rating_for_page = mysql_fetch_row($result);
$rating = $rating_for_page[0];
Use the query:
echo "<table>\n";
echo "<tr><th>Store ID</th><th>Name</th><th>Avg. Rating</th></tr>\n";
$result = mysql_query("SELECT s.store_id, s.store_name, AVG(r.rating) AS avg_rating
FROM stores as s
JOIN ratings as r ON s.store_id = r.store_id
GROUP BY s.store_id");
while ($row = mysql_fetch_assoc($result)) {
echo "<tr><td>$row[store_id]</td><td>$row[store_name]</td><td>$row[avg_rating]</td></tr>\n";
}
echo "</table>\n";
Then loop through the results of this query. Each row will be the average rating of a different store.
You should iterate through the results if I am understanding the issue correctly. Also, in your MySQL command you need to set an AS for the AVG(rating). And I have set mysql_fetch_assoc (instead of mysql_fetch_row) for the results to be returned in an associative array:
foreach ($storeids as $storeid) {
$result = mysql_query("SELECT AVG(rating) AS rating FROM reviews GROUP BY store_id WHERE store_id = '$storeid'");
while ($row = mysql_fetch_assoc($result)) {
echo $row['rating'];
}
}
EDIT Looking at your query and seeing the $storeid being set, unclear where that value is coming from but I wrapped the solution I have provided in a foreach loop which assumes you have a list—and hopefully an array—that contains all of the $storeids in to loop through.
I'm trying to do nested mysql query but I'm failing
it's as simple as I have a table called 'todo' which is for saving to-do list
when I do
well this code is just to show how I wanna do it logically , I know it doesn't work
and I think it probably needs JOIN or UNION but I couldn't do it
$result = mysqli_query($con,"SELECT type FROM todo WHERE user = '$username' GROUP BY type");
while($row = mysqli_fetch_array($result))
{
echo '<td>- '." ".$row['type'].'</td>';
echo "</br>";
$result2 = mysqli_query($con,"SELECT titleFROM todo WHERE type= '$row['type']'");
while($row = mysqli_fetch_array($result2))
{
echo '<td>-- '." ".$row['title'].'
}
</td>';
echo "</br>";
}
I want the result to like
-Work
--Mr Jack
--Company
-School
--Android Class
--Entrepreneurship
--php development
and so on...
Your inner loop overwrites the outerloop's $row variable. Try using a different variable name in the inner loop. Also, there's an error in your inner loop's SQL query (missing space between the tablename and FROM keyword).
I need to return data from 2 separate tables at the same time. The info I need from the 2nd table is determined by what is returned from the first table. Here's what I'm working with..
$query = "SELECT * FROM pending WHERE paymentid = '".$_GET['vnum']."'";
$result = mysql_query($query);
$num = #mysql_num_rows($result);
$linkid = $res['paymentid'];
if ($num==0) {
echo "Hello, ".$_SESSION['Fname']."<br />There was an error, I cannot find this payment in the records.";
} else {
$picquery = mysql_query("SELECT * FROM _uploads_log WHERE linkid = '".$linkid."'");
$numb = #mysql_num_rows($picquery);
if ($numb==0) {
echo "there is no picture"; }
else {
echo "<img src=\"".$res['log_filename']."\" width=\"100\">"; }
I don't understand how to return the results as an array, if $res[] returns the results for the first query, then what returns the results for the second one?
or is there a better way to do this entirely?
Thank you
You need to do a join, but in order to still get results from your first query even if there is no picture (I assume that's why you split it up) you want a left join.
select * from pending left join _uploads_log on pending.paymentID=$_GET['vnum'] and _uploads_log.linkid = pending.paymentID
(note: php markings removed for readability - you'll have to add them back in)
This should (untested since I don't have your tables) return the full row for your vnum variable and also include the picture data if there is one.
OK,
In this query I am displaying the sales for a particular person, matching the passport number from the current form I am working on.
What I want to do however, is to sum the total sales and display it, excluding records which have been marked as paid.
I am having trouble because "paid" does not existent in the current form I am working on as a variable, nor the table it relates to.
I can´t use row['paid'] as I need to do this query outside of the while loop.
What should I do in this situation?
$sqlstr = mysql_query(
"SELECT * FROM sales where passport = ".
"'{$therecord['passport']}'");
if (mysql_numrows($sqlstr) != 0) {
echo "<b>Sales for {$therecord['firstname']} ".
"{$therecord['lastname']}</b><br />";
echo "<table><tr>";
echo '<tr><th align="left">Name</th><th align="left">Quantity</th>".
"<th align="left">Cost</th></tr>';
while ($row = mysql_fetch_array($sqlstr)) {
echo "<td>{$row['product']}</td>";
echo "<td>{$row['quantity']}</td>";
echo "<td>{$row['cost']}</td>";
echo "</tr>";
}
}
echo "</table>";
$sqltotal = mysql_query(
"SELECT SUM(cost) as total FROM sales where passport = ".
"'{$therecord['passport']} AND {$therecord['paid']} <> 1'");
$row = mysql_fetch_array($sqltotal);
echo "<br /><b>Total Owing: {$row['total']}</b>";
You could either create a MySQL view, of look at SQL joins. I'm not sure on your database structure, but you should have a SQL query like this:
SELECT SUM(sales.cost) AS total
FROM sales, table2
WHERE sales.passport = '$passport_id'
AND sales.passport = table2.passport
AND table2.paid = '1'
Not sure as that was wrote off-hand. Again, it'd be better if we knew the structure of your tables.
you've misplaced simple quote in :
"'{$therecord['passport']} AND {$therecord['paid']} <> 1'"
it must be :
"'{$therecord['passport']}' AND {$therecord['paid']} <> 1"