I have a form which puts together an estimate based on the inputs you fill in and select. One of these inputs is a group of checkboxes that determines what finishes the project receives. They go into the database as an array (name="finishes_id[]"). They are put into a table called finishes_used which looks like the following:
used_id | estimate_id | finish_id
This table links together the finishes table with the estimates table. So for example, if 3 different finishes were chosen for one estimate, it would look like this:
used_id | line_id | estimate_id | finish_id
1 1 1000 2
2 1 1000 6
3 1 1000 7
I am now making an edit page for the estimate and I am having trouble figuring out how to pre-select the finishes checkboxes that were used. It is showing ONLY the checkboxes that were selected. I need the option to check the others as well.
My code for the checkboxes part of my form looks like the following. How can I get the desired results above?
<?php
$getid = $_GET['estimate_id']; // The current estimate number
$getLineid = $_GET['line_id']; // The current line item on the estimate
?>
<label for="finish_id">Finishing</label>
<div class="checkbox_group">
<?php
if ($select = $db -> prepare("SELECT f.finish_id, f.finish_name, u.estimate_id, u.line_id FROM finishes AS f INNER JOIN finishes_used AS u ON u.finish_id = f.finish_id WHERE u.line_id = ? ORDER BY f.finish_name ASC"))
{
$select -> bind_param('s', $getLineID);
$select -> execute();
$select -> bind_result($finish_id, $finish_name, $used_estimate_id, $used_line_id);
while ($select -> fetch())
{
echo '<div class="checkbox">';
echo '<input type="checkbox" name="finish_id[]" id="edit_'.$finish_name.'" value="'.$finish_id.'" ';
echo '/><label for="edit_'.$finish_name.'">'.$finish_name.'</label>';
echo '</div>';
}
$select -> close();
}
?>
</div>
It seems to me you have to use left outer join instead of inner join to achieve this.
Here is a great explanation about joins: What is the difference between "INNER JOIN" and "OUTER JOIN"?
Related
I have two database tables. My comments table looks like this:
comid | comuserid
10 7
20 7
30 9
My replies table looks like this:
repid | repcomid | repfromuser | reptouser
1 10 22 7
2 10 22 7
This is what my index.php page currently looks like:
comid10 (commented by comuserid1)
[repid1] repfromuser22 to comuserid1
comid10 (commented by comuserid1)
[repid2] repfromuser22 to comuserid1
My problem is that my code is ONLY echoing out comments rows that contain repliers, but completely skips over the comments rows that doesn't contain repliers. But I want ALL my comments rows to be echoed out even if they contain no repliers. I also want my repliers grouped together based on the commenter they've commented under. This is the real result I want achieved:
comid10 (commented by comuserid1)
[repid1] repfromuser22 to comuserid1
[repid2] repfromuser22 to comuserid1
comid20 (commented by comuserid1)
comid30 (commented by comuserid2)
How would I achieve that with my current code? Please help:
<?php
$rqid = (int)$_GET['rqid'];
$query = $conn->query("
SELECT comments.*, replies.*, users.username
FROM comments
INNER JOIN replies
ON comments.comid = replies.repcomid
INNER JOIN users
ON comments.comid = users.userid
WHERE comments.comrqid = {$rqid}
");
$comreps = [];
while($row = $query->fetch_object()) {
$comreps[] = $row;
}
?>
<?php foreach($comreps as $comrep): ?>
<div class="rqdivc">
<?php echo $comrep->username; ?>
</div>
<div class="fxrp">
<div>
<?php echo $comrep->repfromuser; ?> -->
<?php echo $comrep->reptouser; ?>
</div>
</div>
<?php endforeach; ?>
Suggested by Strawberry - Use a LEFT JOIN - resolved my issue. Thank you person!
First loop table
user_id | fname | lname
1 | first | emp
2 | second| emp
3 | third | emp
Second loop table
shift_id | employee_id
1 | 1
2 | 2
3 | 2
if($employees)
{
foreach ($employees as $employee)
{
if($employee['user_id'] == $shift['employee_id'])
{
echo ucwords($employee['fname']. ' ' .$employee['lname']);
}
}
}
I am getting the right result but I think there is some better way of writing this.
You can use joins in table. Left join means that the user line has to exists (because: LEFT) and the shifts enty is optional.
SELECT user.user_id, user.fname, user.lname, shifts.shift_id
FROM yourUserTable AS user
LEFT JOIN yourShiftsTable AS shifts ON(user.user_id = shifts.employee_id)
Now you get it in your initial array, as if you'd select it as one row from a table and no longer need to do tricks in PHP to combine information. If you can, always try to get the database to manage data, it does that way faster than PHP can.
Please note, the query could be a little off, I just wrote this out of the top of my head.
Just some test code I whipped up to test this from the information provided for this "Demonstration Code".
Note: I have used the mysqli class for the database (instantiating $db ) and have excluded the SQL Table setup.
What you would have had is something along the lines of this...
Case 1 - The original
$db = new mysqli('localhost', 'root', 'test', 'phptutorials_st26');
echo '<h2>Create $employees </h2>';
$query = "SELECT * FROM users";
$result = $db->query($query);
$employees = $result->fetch_all(MYSQL_ASSOC);
var_dump($employees);
echo '<h2>Create $shifts </h2>';
$query = "SELECT * FROM shifts";
$result = $db->query($query);
$shifts = $result->fetch_all(MYSQL_ASSOC);
var_dump($shifts);
echo '<h2>Using foreach on $employees and $shifts</h2>';
if ($employees) {
foreach ($employees as $employee) {
foreach ($shifts as $shift) {
if ($employee['user_id'] == $shift['employee_id']) {
echo ucwords($employee['fname'] . ' ' . $employee['lname']);
echo '<br>';
}
}
}
}
The Result from the above is
First Emp
Second Emp
Second Emp
Case 2 - Using a Join
Well using a join, as everyone has already stated, is the way to go...
$sql = "SELECT u.user_id, u.fname, u.lname, s.shift_id
FROM users AS u
JOIN shifts AS s ON(u.user_id = s.employee_id)
";
$result = $db->query($sql);
$employees = $result->fetch_all(MYSQL_ASSOC);
// To see what comes out because we always check things.
var_dump($joined_result);
(Don't ask me why I love using very abbreviated aliases for the table names! It's just "a thing".)
Then your "loop" simply becomes...
echo '<h2>Using foreach on join</h2>';
foreach ($employees as $employee) {
echo ucwords($employee['fname'] . ' ' . $employee['lname']);
echo '<br>';
}
And the result is...
First Emp
Second Emp
Second Emp
Case 2 - has reduced the code and only requires 1 Trip to the Database.
Does that help you any?
You could do it this way also. Its a little shorter.
SELECT TABLE1.FNAME, TABLE1.LNAME, TABLE2.EMPLOYEE_ID
FROM TABLE1, TABLE2
WHERE TABLE1.USER_ID = TABLE2.EMPLOYEE_ID;
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']."\"");
// ...
?>
I am working on a webpage that displays list of shops. I have 2 tables, shops and shops_sched.
+-shops-+
| id | title |
+-------------shops_sched-------------+
| id | shops_id | start_date | end_date |
Basically, the program displays the list of shops from the shops table, but if a value from shops.id is found # shops_sched.shops_id the page must output shops.title + 'coming soon'.
I understand this will be easy if I just place the date fields inside the table shops but due to programming restrictions I can't. I'm working on an existing project and I'm trying to minimize changes to existing functions. I can create new PHP functions if necessary though.
In addition, I need to get all the entries from the shops table. The Program needs to return all shops.title but for those shops whose id is found # shops_sched.shops_id, the program will have to return shops.title + "Coming Soon".
must output shops.title + 'coming soon'.
So do it like this:
$shops.title = "Donut-John";
echo $shops.title." coming soon";
To join the shops and shops_sched table
$query = SELECT `title` FROM `shops` JOIN `shops_sched` ON `shops`.`id` = `shops_sched`.`shops_id` WHERE `shops_sched`.`shop_id` = 5;
$result = mysql_query($query);
while($row = mysql_fetch_array($result) {
echo $row['title'] . 'coming soon';
}
For more about join you also can refer the following link
https://dev.mysql.com/doc/refman/5.0/en/join.html
http://www.tutorialspoint.com/mysql/mysql-using-joins.htm
Join the two tables :
SELECT shops.title
FROM shops INNER JOIN shops_sched ON shops.id = shops_sched.shops_id
The query should return only the the shops inside shops_sched
EDIT :
If I understood your question, try this :
SELECT shops.title, shops_sched.id
FROM shops LEFT JOIN shops_sched ON shops.id = shops_sched.shops_id
This will return all the titles, and the shops_sched.shops_id if shops.id = shops_sched.shops_id. In the other case, the hops_sched.shops_id will be null
Then you fetch the rows and if the second row is not null, print title + coming soon
Sample code : (Something like this)
$query = "SELECT `title`, 'shops_id' FROM `shops` LEFT JOIN `shops_sched` ON `shops`.`id` = `shops_sched`.`shops_id` WHERE `shops_sched`.`shop_id`";
$result = mysql_query($query);
while($row = mysql_fetch_array($result) {
if($row['shops_id'] != "")
{
echo $row['title'] . ' coming soon';
}
else
{
echo $row['title'];
}
}
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.