I have two different tables, one named users, and another named transactions. Transactions contains wallet1, wallet2, amount. Users contains user details such as firstname, lastname, and wallet. I am trying to display the corresponding first name and last name, depending on whether or not the SESSION_wallet is equal to wallet1 or wallet2 within transactions. I tried searching for a while, and came up with a solution for showing the correct display name for the first and last name making the transfer, however, I am trying to make it display the correct value for "Transfer to:"
Here is some of my code to get a better understanding of what I mean:
MySQLi Query:
$result2 = mysqli_query($link, "SELECT * FROM transactions INNER JOIN users ON transactions.wallet1 = users.wallet WHERE transactions.wallet1 = '" . $_SESSION["wallet"] . "' OR transactions.wallet2 = '" . $_SESSION["wallet"] . "' Order by transactions.id DESC LIMIT 5 ");
PHP Code:
<?php
if(mysqli_num_rows($result2) > 0)
{
while($row = mysqli_fetch_array($result2))
{
?>
The table that needs to display the transfer from, and transfer to:
<?php
if ($_SESSION["wallet"] == $row["wallet1"]) {
echo "<td>Transfer to ".$row["firstname"]." ".$row["lastname"]."</td>";
}
else if ($_SESSION["wallet"] == $row["wallet2"]) {
echo "<td>Transfer from ".$row["firstname"]." ".$row["lastname"]."</td>";
}
?>
Right now my tables are only showing the first and last name of the user that made the Transfer, however, I need it to display the first and last name of the user that the transaction is made to as well. The else if code is working correct, but the first part is not showing the corresponding value.
You will need to JOIN your transactions table to your users table twice, once to get each users name. Then to avoid duplicate column names overwriting the results in the output array, you will need to use column aliases. Something like this should work:
$result2 = mysqli_query($link, "SELECT t.*,
u1.firstname AS w1_firstname,
u1.lastname AS w1_lastname,
u2.firstname AS w2_firstname,
u2.lastname AS w2_lastname
FROM transactions t
INNER JOIN users u1 ON t.wallet1 = u1.wallet
INNER JOIN users u2 ON t.wallet2 = u2.wallet
WHERE t.wallet1 = '{$_SESSION["wallet"]}'
OR t.wallet2 = '{$_SESSION["wallet"]}'
ORDER BY t.id DESC
LIMIT 5 ");
Then you can access each user's names as $row['w1_firstname'] etc.:
if ($_SESSION["wallet"] == $row["wallet1"]) {
echo "<td>Transfer to ".$row["w2_firstname"]." ".$row["w2_lastname"]."</td>";
}
else if ($_SESSION["wallet"] == $row["wallet2"]) {
echo "<td>Transfer from ".$row["w1_firstname"]." ".$row["w1_lastname"]."</td>";
}
Note that ideally you should use a prepared query for this, for example:
$stmt = $link->prepare("SELECT t.*,
u1.firstname AS w1_firstname,
u1.lastname AS w1_lastname,
u2.firstname AS w2_firstname,
u2.lastname AS w2_lastname
FROM transactions t
INNER JOIN users u1 ON t.wallet1 = u1.wallet
INNER JOIN users u2 ON t.wallet2 = u2.wallet
WHERE t.wallet1 = ?
OR t.wallet2 = ?
ORDER BY t.id DESC
LIMIT 5");
$stmt->bind_param('ss', $_SESSION["wallet"], $_SESSION["wallet"]);
$stmt->execute();
$result2 = $stmt->get_result();
Related
I'm working on a system, and this module is supposed to echo the contents of the database.
It worked perfectly until I added some JOIN statements to it.
I've checked and tested the SQL code, and it works perfectly. What's not working is that part where I echo the content of the JOINed table.
My code looks like this:
$query = "SELECT reg_students.*, courses.*
FROM reg_students
JOIN courses ON reg_students.course_id = courses.course_id
WHERE reg_students.user_id = '".$user_id."'";
$result = mysqli_query($conn, $query);
if (mysqli_fetch_array($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
echo $row["course_name"];
echo $row["course_id"];
The course_name and course_id neither echo nor give any error messages.
UPDATE: I actually need to increase the query complexity by JOINing more tables and changing the selected columns. I need to JOIN these tables:
tutors which has columns: tutor_id, t_fname, t_othernames, email, phone number
faculty which has columns: faculty_id, faculty_name, faculty_code
courses which has columns: course_id, course_code, course_name, tutor_id, faculty_id
I want to JOIN these tables to the reg_students table in my original query so that I can filter by $user_id and I want to display: course_name, t_fname, t_othernames, email, faculty_name
I can't imagine that the user_info table is of any benefit to JOIN in, so I'm removing it as a reasonable guess. I am also assuming that your desired columns are all coming from the courses table, so I am nominating the table name with the column names in the SELECT.
For reader clarity, I like to use INNER JOIN instead of JOIN. (they are the same beast)
Casting $user_id as an integer is just a best practices that I am throwing in, just in case that variable is being fed by user-supplied/untrusted input.
You count the number of rows in the result set with mysqli_num_rows().
If you only want to access the result set data using the associative keys, generate a result set with mysqli_fetch_assoc().
When writing a query with JOINs it is often helpful to declare aliases for each table. This largely reduces code bloat and reader-strain.
Untested Code:
$query = "SELECT c.course_name, t.t_fname, t.t_othernames, t.email, f.faculty_name
FROM reg_students r
INNER JOIN courses c ON r.course_id = c.course_id
INNER JOIN faculty f ON c.faculty_id = f.faculty_id
INNER JOIN tutors t ON c.tutor_id = t.tutor_id
WHERE r.user_id = " . (int)$user_id;
if (!$result = mysqli_query($conn, $query)) {
echo "Syntax Error";
} elseif (!mysqli_num_rows($result)) {
echo "No Qualifying Rows";
} else {
while ($row = mysqli_fetch_assoc($result)) {
echo "{$row["course_name"]}<br>";
echo "{$row["t_fname"]}<br>";
echo "{$row["t_othernames"]}<br>";
echo "{$row["email"]}<br>";
echo "{$row["faculty_name"]}<br><br>";
}
}
I have 2 tables userdatafiles & users, where i merge the 2 tables together based on common ID, to find the corresponding works attached to the ID, the following PHP code i have retrieve the works from the user, it works fine as i only want to show 1 work per user, but if a user have 2-3 works for example, it will also display the different works out..
PHP
<?php
include 'dbAuthen.php';
$sql = 'SELECT * FROM userdatafiles JOIN users ON userdatafiles.UserID = users.UserID WHERE Specialisation = "Painting"';
$result = mysqli_query($con,$sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$links[] = array(
"links" => $row["Link"],
"caption" => $row["Name"],
);
}
shuffle($links);
echo json_encode($links);
} else {
echo "0 results";
}
?>
the above code works if i only uploaded 1 work per user, what should i do to improve the code so that if a user have multiple works, it will only display a random work among the many that user has? Thanks..
use rand() for random display and set limit to 1 so you will get one random record
ORDER BY RAND() LIMIT 1
so your query will be:
$sql = 'SELECT * FROM userdatafiles JOIN users ON userdatafiles.UserID = users.UserID WHERE Specialisation = "Painting" ORDER BY RAND() LIMIT 1';
Use DISTINCT to eliminate repeating userid:
$sql = 'SELECT DISTINCT(users.UserID),name,link FROM userdatafiles JOIN users ON userdatafiles.UserID = users.UserID WHERE Specialisation = "Painting" ORDER BY RAND()';
or use GROUP BY:
$sql = 'SELECT users.UserID,name,link FROM userdatafiles JOIN users ON userdatafiles.UserID = users.UserID WHERE Specialisation = "Painting" GROUP BY UserID ORDER BY RAND()';
I have two tables which both have a unique PID. I need to know how I should setup the primary/foreign keys so I can access fields from both tables in one SQL statement.
For example: With the following data structure i'd like to echo out the players name and photo but also echo out all their stats via PHP as well. I have successfully done this with just the player Stats, but I do not know how to get access to fields in another table.
Here is my database structure so far:
Players
-PID (Set as Primary Key)
-Name
-Height
-College
-Photo
Stats
-PID
-Touchdowns
-Receptions
Current PHP Code:
$query="
SELECT * FROM Stats
ORDER BY Stats.FantasyPoints DESC";
$res=mysql_query($query);
$num=mysql_numrows($res);
$i=0;
while($i< $num){
$Name = mysql_result($res, $i, "Name");
$FantasyPoints = mysql_result($res, $i, "FantasyPoints");
echo $Name . ': '. $FantasyPoints . "<br />";
$i++;
}
$sql = "
SELECT p.*, s.*
FROM Players AS p
LEFT JOIN Stats AS s ON p.PID = s.PID
ORDER BY s.FantasyPoints DESC
";
You could also use a JOIN rather than a LEFT JOIN which would limit the result to only players who have stats
Edited sql to produce result similar to your own sql.
======================================================
This is how I would go about it...
$query =
"
SELECT s.*, p.*
FROM Stats AS s
LEFT JOIN Players AS p ON p.PID = s.PID
ORDER BY s.FantasyPoints DESC
";
$res = mysql_query($query);
while ($row = mysql_fetch_assoc($res))
{
echo "{$row['name']}: {$row['FantasyPoints']}<br />";
}
$sql = "SELECT * FROM books LEFT JOIN users
ON books.readby=users.user_id WHERE users.email IS NOT NULL";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
echo $row['readby']. " - read 10 books";
} //while ends
this is the code I have so far. I am trying to retrieve the number of books read by each user
and echo the results. echo the user_id and number of books he/she read
books table is like this : id - name - pages - readby
the row readby contains the user id.any ideas/suggestions? I was thinking about using count() but Im not sure how to go about doing that.
A subquery can return the count of books read per user. That is left-joined back against the main table to retrieve the other columns about each user.
Edit The GROUP BY had been omitted...
SELECT
users.*,
usersread.numread
FROM
users
/* join all user details against count of books read */
LEFT JOIN (
/* Retrieve user_id (via readby) and count from the books table */
SELECT
readby,
COUNT(*) AS numread
FROM books
GROUP BY readby
) usersread ON users.user_id = usersread.readby
In your PHP then, you can retrieve $row['numread'] after fetching the result.
// Assuming you already executed the query above and checked errors...
while($row = mysql_fetch_array($result))
{
// don't know the contents of your users table, but assuming there's a
// users.name column I used 'name' here...
echo "{$row['name']} read {$row['numread']} books.";
}
You can use count() this way:
<?php
$count = mysql_fetch_array(mysql_query("SELECT COUNT(`user_id`) FROM books LEFT JOIN users ON books.readby=users.user_id WHERE users.email IS NOT NULL GROUP BY `user_id`"));
$count = $count[0];
?>
Hope this helps! :)
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/>";