I have two different tables of the following structure:
grouprel
id | userId | pupID | groupId
pupils
id | userId | fname | lname
pupId in groulrel is equal to id in pupils.
I want to fetch pupils from a different group and then order them by fname, lname.
Now I have two queries like this:
$q = "SELECT * FROM grouprel WHERE userid = ". $userid ." AND groupId = ". $_GET['id'] ."";
$r = mysqli_query($mysqli, $q);
while ($rows = mysqli_fetch_object($r)) {
$query = "SELECT id, fname, lname FROM pupils WHERE userid = ". $userid ." AND id = ". $rows->pupId ." AND status = 0 ORDER BY fname, lname";
$result = mysqli_query($mysqli, $query);
while($row = mysqli_fetch_object($result)) {
echo stuff...
}
}
This works, but it doesn't order the names alphabetically like I want to.
How could I fix this?
This is iterating over the first query:
while ($rows = mysqli_fetch_object($r)) {
And this iterates over each instance of the second query:
while($row = mysqli_fetch_object($result)) {
So if the first query returns 1,2,3, and each iteration of the second query returns A,B, then your output would be:
1 A
1 B
2 A
2 B
3 A
3 B
The second query is ordering by the ORDER BY clause you gave it. But you are ordering the entire output by the first query.
Ultimately, why do you need these separate queries at all? Executing a database query in a loop is almost always the wrong idea. It looks like all you need is one query with a simple JOIN. Guessing on your logic, something like this:
SELECT
pupils.id, pupils.fname, pupils.lname
FROM
pupils
INNER JOIN grouprel ON pupils.id = grouprel.pupId
WHERE
pupils.userid = ?
AND grouprel.groupId = ?
AND pupils.status = 0
ORDER BY
fname, lname
It may take a little tweaking to match exactly what you're looking for, but you can achieve your goal with a single query instead of multiple separate queries. Then the results of that query will be ordered the way you told MySQL to order them, instead of the way you told PHP to order them.
Related
I have two tables and i want to echo the total call count once each user logins:
Login
Firstname | Lastname | Login ID
------------------------------
Tyler | Durden | 3
Call Count
Name | Call Count | Open | GrandTotal
------------------------------
Tyler Durden| 100 | 33 | 133
i tried:
<?php
$result = mysqli_query($mysqli, "SELECT * FROM csvdata WHERE Name=".$_SESSION['firstname']. ' ' .$_SESSION['lastname']." ");
while($res = mysqli_fetch_array( $result )) {
echo $res['Open'].' Open Calls';
echo $res['GrandTotal'].' Grand Total Calls';
}
mysqli_close($mysqli);
?>
But its not working, i think i have to join the the two tables to get it to work. What do you think?
Assuming your Call Count table is actually called csvdata, you'll want to format your SQL request string a bit by adding single quotes around the WHERE name = part.
<?php
$result = mysqli_query($mysqli, "SELECT * FROM csvdata WHERE Name='".$_SESSION['firstname']. ' ' .$_SESSION['lastname']."' ");
while($res = mysqli_fetch_array( $result )) {
echo $res['Call Count'].' Call Count';
echo $res['Open'].' Open Calls';
echo $res['GrandTotal'].' Grand Total Calls';
}
mysqli_close($mysqli);
?>
Good practice would require that you use primary keys to facilitate joins between tables and make sure two users with the same name can be differenciated.
In that case you may want to consider replacing the Name column in your Call Count table for your loginID. This way you could get your name from the Login table (as shown below). Also, as bad as it is to have duplicated data like your user's name in both tables, you do not need your GrandTotal column since you can easily get the sum of CallCount and Open to get the exact same number. In the end, your query should look more like this (assuming your tables are called Login and CallCount).
<?php
$result = mysqli_query($mysqli, "SELECT l.FirstName, l.LastName, cc.CallCount, cc.Open, (cc.CallCount + cc.Open) AS GrandTotal FROM Login AS l JOIN CallCount AS cc ON l.LoginID = cc.LoginID WHERE l.FirstName LIKE \"".$_SESSION['firstname']."\" AND l.LastName LIKE \"".$_SESSION['lastname']."\"");
// ...
?>
So this is the structure of my MySQL table that I wanna work this out with:
ID type category_id amount
12 Expense 3 963.39
13 Expense 5 1200.50
14 Expense 3 444.12
15 Expense 5 1137.56
..............................
Desired output:
1407,41 (for category_id = 3)
2338,06 (for category_id = 5)
....... (and for other category_id)
What I get now:
1407,41 (only for category_id = 3)
My query does not add or display the sum of other category_id.
This is the query I am trying:
$query = "SELECT SUM(amount) AS TotalAmount FROM spendee WHERE type = 'Expense'
group by category_id having count(*) >1 ";
$expense_query = mysqli_query($connection, $query);
$expense_count = mysqli_fetch_array($expense_query);
echo $expense_count[0];
Been stuck with this for the last couple of days. Any help is very much appreciated. Thank you!
You're only calling mysqli_fetch_array() once. You need to call it in a loop to get all the totals. You should also include the category ID in the SELECT list.
$query = "SELECT category_id, SUM(amount) AS TotalAmount FROM spendee WHERE type = 'Expense'
group by category_id having count(*) >1 ";
$expense_query = mysqli_query($connection, $query);
while ($row = mysqli_fetch_assoc($expense_query)) {
echo "{$row['TotalAmount']} (for category_id = {$row['category_id']}<br>\n";
}
The query here works. It's just that you only select the first result from the $expense_count variable. $expense_count[1] will return the second category listed, $expense_count[2 will return the third one, ect...
Try echo implode(" <br>", $expense_count);
Have a nice day.
Thanks for helping, first I will show code:
$dotaz = "Select * from customers JOIN contracts where customers.user_id ='".$_SESSION['user_id']."' and contracts.customer_contract = ".$_SESSION['user_id']." order by COUNT(contracts.customer_contract) DESC limit $limit, $pocetZaznamu ";
I need to get the lists of users (customers table) ordered by count of contracts(contracts table)
I tried to solve this by searching over there, but I can't... if you help me please and explain how it works, thank you! :) $pocetZanamu is Number of records.
I need get users (name, surname etc...) from table customers, ordered by number of contracts in contracts table, where is contract_id, customer_contract (user id)..
This should do it where is the column name you are counting.
$id = $_SESSION['user_id'] ;
$dotaz = "Select COUNT(`customer_contract`) AS CNT, `customer_contract` FROM `contracts` WHERE `user_id`=$id GROUP BY `customer_contract` ORDER BY `CNT` DESC";
Depending on what you are doing you may want to store the results in an array, then process each element in the array separately.
while ($row = mysqli_fetch_array($results, MYSQL_NUM)){
$contracts[$row[1]] = $row[0];
}
foreach ($contracts AS $customer_contract => $count){
Process each user id code here
}
Not sure what you are counting. The above counts the customer_contract for a table with multiple records containing the same value in the customer_contract column.
If you just want the total number of records with the same user_id then you'd use:
$dotaz = "Select 1 FROM `contracts` WHERE `user_id`=$id";
$results = $mysqli->query($dotaz);
$count = mysql_num_rows($results);
I've got
a users table named "members"
a rooms table named "rooms"
a table that associates the user id to the ids of the rooms "membersRooms"
I should write a loop that prints a dropdown for each user with all the rooms, but that adds the attribute "selected" to rooms associated with the user
What's wrong with this loop?
$members = mysql_query("SELECT * FROM members ");
$rooms = mysql_query("SELECT * FROM rooms");
while($member = mysql_fetch_array($members)){
echo("<select>");
$roomsOfUser = mysql_query("SELECT roomID FROM membersRooms WHERE userID=".$member["id"]);
$cuArray = mysql_fetch_array($roomsOfUser);
while($room = mysql_fetch_array($rooms)){
if(in_array($room["id"],$cuArray,true))
echo("<option selected='selected'>".$room["roomName"]."</option>");
else
echo("<option>".$class["roomName"]."</option>");
}
echo("</select>");
}
To make this a little easier on you, you could try utilizing left and right joins on your database. This would significantly reduce your server load and still allow you to do the same functionality.
I believe, if I'm reading your database structure right, that you'ld want something along the lines of:
SELECT members.id as memberID, rooms.id as roomID, rooms.roomName, membersRooms.roomID as memberRoom
FROM members
LEFT JOIN membersRooms
ON members.id = membersRooms.userID
RIGHT JOIN rooms
ON membersRooms.roomID = rooms.id
Then in PHP you should be able to just keep track of when your memberID changes, and when it does, start a new select. If I didn't totally bungle that SQL (which I might have) then the resulting rows should look something like:
memberID | roomID | roomName | memberRoom
1 1 foo 1
1 2 bar 1
2 1 foo 1
2 2 bar 1
So on your loop iteration you would use roomID and roomName to build your select, and if RoomID matched memberRoom then you would select that row.
$rooms query while is dead
while runs once time in while
put this $rooms = mysql_query("SELECT * FROM rooms"); query line
in first while
OK, so you need information from 3 tables - members, rooms, and membersRooms. The rows from members and membersRooms line up 1:1, so we can get both of those with 1 query.
This method will minimize the number of queries needed - if you ever see yourself querying the database in a loop, ask yourself if there's a better way.
$member_query = mysql_query("SELECT * FROM members LEFT JOIN membersRooms ON (members.id = membersRooms.userID)");
$room_query = mysql_query("SELECT * FROM rooms");
$rooms = array();
while ($room = mysql_fetch_assoc($room_query))
$rooms[] = $room;
while ($member = mysql_fetch_assoc($member_query)) {
echo '<select>';
foreach($rooms as $room) {
echo "<option value='{$room['roomID']}' ";
if ($member['roomID'] == $room['id'])
echo 'selected="selected"';
echo ">{$room['roomName']}</option>";
}
echo '</select>';
}
It's worth noting that if members:rooms is a 1:many relation, you don't need to use a third table to join them - just add a roomId to members, and you're fine.
Mysql query and PHP code that I'm using to get users from the database that meet certain criteria is:
$sql = mysql_query("SELECT a2.id, a2.name FROM members a2 JOIN room f ON f.myid = a2.id
WHERE f.user = 1 AND a2.status ='7' UNION SELECT a2.id, a2.name FROM members a2
JOIN room f ON f.user = a2.id WHERE f.myid = 1 AND a2.status ='7' GROUP BY id")
or die(mysql_error());
while ($r = mysql_fetch_array($sql))
{
$temp[] = '"'.$r[0].'"';
}
$thelist = implode(",",$temp);
The query that follows get the list of members with new galleries by using array from the previous query.
$ft = mysql_query("SELECT id, pic1 FROM foto WHERE id IN ($thelist) AND
pic1!='' ORDER BY date DESC LIMIT 10");
while ($f = mysql_fetch_array($ft))
{
echo $f['id']." - ".$f['pic1']."<br/>";
}
These queries working fine but I need to get the name for every user listed in second query. This data is in the first query in the column name. How can I get it listed beside '$f['id']." - ".$f['pic1']'?
While I might just alter the first query to pull the galleries at the same time, or change the second query to join and get the name, you could keep the same structure and change a few things:
In the loop after the first query when building $temp[], also build a lookup table of user id to user name:
$usernames[$r[0]] = $r[1];
Then in your output loop, use the id (assuming they are the same!) from the second query to call up the user name value you stored:
echo $f['id'] . " - " . $f['pic1'] . " - " . $usernames[$f['id']] . "<br/>";