I am trying to write a simply query that will delete an event based off its event_id. That event_id could be in 1 of 3 tables in a database however.
Leagues
Tournaments
Trainings
Each event_id is generated by a randomizer function so each id is unique.
I have tried a few different syntaxs for SQL but none of them will actually delete the event from the DB table. Any suggestions?
Here is what I have:
if(isset($_POST['delete'])){
$query = "delete from leagues where event_id= '$event_id', delete from tournaments where event_id= '$event_id', delete from trainings where event_id= '$event_id'";
$result = mysqli_query($conn, $query);
header('location: index_admin.php');
} else {
}
I have also tried using the union command, using multiple delete queries, putting a semicolon after each delete command. What else can I try?
No reason to execute 3 separate DELETEs.
DELETE leagues, tournaments, trainings
FROM (SELECT $event_id AS event_id) src -- use parameter instead of immediate insert
LEFT JOIN leagues USING (event_id)
LEFT JOIN tournaments USING (event_id)
LEFT JOIN trainings USING (event_id)
Of course you'd not insert the value into the query like above - you'd use prepared statement.
if(isset($_POST['delete'])){
$stmt = $pdo->prepare('delete from leagues where event_id = :event_id');
$stmt->execute([ ':event_id' => $event_id]);
$res = $stmt->fetch();
header('location: index_admin.php');
}else{
}
Related
I have 3 tables:
Users(id, username, sum_score, level)
user_badge(user_id, badge_id)
badges(badge-id, badge_name)
user_id and badge_id from user_badge are foreign keys for the ids in other two table.
I would like to update the user_badge table when the following condition is satisfied.
$selectscore= "SELECT sum_score from users WHERE id= '$id'";
$selectscorequery=mysqli_query($db,$selectscore);
while($set=mysqli_fetch_array($selectscorequery)){
if ($set['sum_score']>=0) {
$newbie="UPDATE user_badge a JOIN users u ON a.user_id = u.id JOIN badges b ON a.badge_id = b.badge_id SET a.badge_id = b.badge_id WHERE user_id='".$set["id"]."'" AND b.badge_id='7';
mysqli_query($db,$newbie)or die(mysqli_error($db));
}
I tried insert ignore query at first but it was not working, it was just adding a new row every time. So i used the update query. I know there is something wrong with the query. Could someone help me out with this problem. I never used update on many to many relationship before.
If I understand it right you will give a user the badge_id=7 when his sum_score is greater then 0 ?
In this case I don't want to work with joins, because in my opinion you only have to update the table user_badge to set the relationship
$selectscore= "SELECT sum_score from users";
$selectscorequery=mysqli_query($db,$selectscore);
while($set=mysqli_fetch_array($selectscorequery)){
if ($set['sum_score']>=0) {
// Query if badge already exists
$selectbadge= "SELECT user_id from user_badge WHERE badge_id='7' AND user_id='".$set["id"]."'";
$selectbadgequery=mysqli_query($db,$selectbadge);
if (mysqli_num_rows($selectbadgequery)<1) {
// INSERT THE BADGE
echo "insert Batch Newbie for User User: ".$set['id']."\n";
$newbie="INSERT user_badge (user_id,badge_id) VALUES ('".$set["id"]."','7')";
} else {
// UPDATE THE BADGE
echo "update Batch Newbie for User User: ".$set['id']."\n";
$newbie="UPDATE user_badge SET badge_id = '7' WHERE user_id='".$set["id"]."' AND badge_id='7'";
}
mysqli_query($db,$newbie)or die(mysqli_error($db));
}
}
I am creating a Log in and I have separate tables for Users A and Users B.
What I want to do is check first in first table if the Users that trying to Login is in the Table A,
if YES, it will not go to the Table B to check the Login credentials, if NOT, go to Table B and check the Login credentials.
Table A
SELECT * FROM tableA WHERE userId='$userId' AND password='$password'
Table B
SELECT * FROM tableB WHERE accountNumber='$accountNumber' AND password='$password'
Note: The 2 Tables has different Field Name userId and accountNumber.
I presume you are fetching the values of username and password from client side so I will tell you only what you asked for.
$getUserBasic1=$db->prepare('SELECT * FROM tableA WHERE userId="$userId" AND password="$password"');
$getUserBasic1->execute();
$user= $getUserBasic1->fetchAll();
if(count($user)>0)
{
//if yes do what you want here
}
else
{
$getUserBasic2=$db2->prepare('SELECT * FROM tableB WHERE accountNumber="$accountNumber" AND password="$password"');
$getUserBasic2->execute();
$user2= $getUserBasic2->fetchAll();
//write your code here
}
You could use an INNER JOIN and select both table results taking Table A's result first if it exists, else take Table B's result.
Assuming both tables have some sort of reference like the User ID you can use something like this:
SELECT tbla.*, tblb.* FROM tableA tbla
INNER JOIN tableB tblb ON tbla.userId = tblb.userId
WHERE userId='$userId' OR accountNumber='$accountNumber' AND password='$password'
ORDER BY userId ASC
LIMIT 1
The query above uses the cross-reference (userId in this case) and joins both tables together before querying the results. It orders the results by Table A before Table B but limits the result to 1 bringing either Table A or Table B out depending which is null.
Try combining the tables, some thing like:
SELECT * FROM tableA, tableB WHERE tableA.userId='$userId' AND tableA.password='$password' OR tableB.accountNumber='$accountNumber' AND tableB.password='$password'
I have not checked, so may not work, but see if this gets what you are looking for!
Something like this:
$sql = "SQL QUERY FOR TABLEA";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// checking if result in TABLE A
}
else{
//search in TABLE B by updating your sql value.
}
I hope that you want to check for the registered user, the best way to do that is to keep one table and just search there itself keeping the userID as the primary key.
How would I go about deleting a row from the table 'subjects' that has a primary id 'subject_id' based on the number of rows in another table named 'replies' that uses a 'subject_id' column as a reference.
Example in pseudo code:
If ('subject' has less than 1 reply){
delete 'subject'}
I don't know much about SQL triggers so I have no clue if I would be able to incorporate this directly in the database or if I'd have to write some PHP code to handle this...
To delete any subjects that have had no replies, this query should do the trick:
DELETE s.* FROM subjects AS s
WHERE NOT EXISTS
(
SELECT r.subject_id
FROM replies AS r
WHERE r.subject_id = s.subject_id
);
Demo: DB Fiddle Example
One of the MySQL gurus will need to weigh in on whether or not you can do this directly, but in PHP you could...
$query = "SELECT subject_id FROM subjects WHERE subject='test'";
$return = mysqli_query($mysqli, $query);
$id = mysqli_fetch_assoc($return);
$query = "SELECT reply_id FROM replies WHERE subject_id='".$id[0]."'";
$return = mysqli_query($mysqli, $query);
if(mysqli_num_rows($return) < 1){
$query = "DELETE FROM subjects WHERE subject_id='1'";
$return = mysqli_query($mysqli, $query);
}
This example assumes the "subject" is unique. In other words, SELECTing WHERE subject='test' will only ever return one subject_id. If you were doing this as a periodic cleaning, you would grab all the subject_id values (no WHERE clause) and loop through them to remove them if no replies.
You can achieve this in one query by selecting all (unique) subject-ids from the replies table, and delete all subjects that doesn't have a reply in there. Using SELECT DISTINCT, you don't get the IDs more than once (if a subject has more than one reply), so you don't get unnecessary data.
DELETE FROM subjects
WHERE subject_id NOT IN (SELECT DISTINCT subject_id FROM replies)
Any subject that doesn't have a reply should be deleted!
So you want to delete all subjects with no replies:
DELETE FROM subjects WHERE subject_id NOT IN
(SELECT subject_id FROM replies);
I think this is what you want...
Scenario:
I am working on a PHP/MySQLi aplplication where I have got 2 tables attendance and students .
students has fields: student_id, fullname, phone,email,gender, department and level.
attendance table has fields: attendance_id, student_id, department_id, level_id.
I was able to fetch all students whose records are in the attendance table according to their department and level.
Question:
Let's assume that I was able to fetch all students whose records are in the attendance table and are in 200L (with level_id, 2) computer science (with department_id, 4) department, if the list of the students present are much and it was paginated and I want to search for a particular student's fullname that's in attendance table in reference to student's table.
How will the SQL query be like? I tried the following query which didn't work.
$search_query = mysqli_query($db_connect, "SELECT *FROM attendance WHERE student_id=\"SELECT student_id FROM students WHERE fullname LIKE '%$student_fullname%'\";
Please help.
Try this query:
SELECT attendance.*
FROM `students`
INNER JOIN `attendance`
ON `attendance`.`student_id` = `students`.`student_id`
WHERE `students`.`fullname` LIKE `%$student_fullname%`
I know that may look back-to-front at first, but I prefer to structure the SQL to show the strong selector (the LIKE filter) in the WHERE clause. If you do not like that, you can get the same result like this:
SELECT attendance.*
FROM `attendance`
INNER JOIN `students`
ON `students`.`student_id` = `attendance`.`student_id`
AND `students`.`fullname` LIKE `%$student_fullname%`
Note that this second version does NOT have a WHERE clause - always put filters on the RHS of a join in the join's ON clause, because otherwise outer joins will not behave correctly.
The SQL query you are looking for is:
select * from attendance join students on attendance.student_id = students.student_id where students.fullname like '%NAME%'
so in PHP you would need something like:
$query_string = "select * from attendance join students on attendance.student_id = students.student_id where students.fullname like '%$student_fullname%'";
$search_query = mysqli_query($db_connect, $query_string);
I would recommend you to have a look at prepared statements though, to prevent SQL injection: http://php.net/manual/en/mysqli.prepare.php
/* create a prepared statement */
$query_string = "select * from attendance join students on attendance.student_id = students.student_id where students.fullname like ?";
if ($stmt = $mysqli->prepare($query_string)) {
$stmt->bind_param("s", $student_fullname);
$stmt->execute();
$result = $stmt->get_result();
while ($myrow = $result->fetch_assoc()) {
// use your $myrow array as you would with any other fetch
printf("%s found in attendance record ID: %s\n", $student_fullname, $myrow['attendance_id']);
}
$stmt->close();
}
I am trying to add data into 3 table using PHP, atm I can only view the results of the tables that are joined .
RESULTS QUERY
$sql = mysql_query("SELECT PART_ID, PART_DESC, SERIAL_NUM, PART.RACK_NUM, PART.PART_TYPE_ID, PART_TYPE_DESC, LOCATION
FROM PART
INNER JOIN PART_TYPE ON PART.PART_TYPE_ID = PART_TYPE.PART_TYPE_ID
INNER JOIN RACK ON RACK.RACK_NUM = PART.RACK_NUM
This will get all the rows from the PART table, and for each of the rows we find, match that row to a row in the PART_TYPE table (the condition being that they have the same PART_TYPE_ID). If no match between the PART and PART_TYPE tables can be found for a given row in the PART table, that row will not be included in the result.
My Insert Query This is where im having trouble
How do I add the data to the PART_ID, PART_TYPE and RACK tables?
<?php
// Parse the form data and add inventory item to the system
if (isset($_POST['PART_ID'])) {
$id = mysql_real_escape_string($_POST['PART_ID']);
$PART_DESC = mysql_real_escape_string($_POST['PART_DESC']);
$SERIAL_NUM = mysql_real_escape_string($_POST['SERIAL_NUM']);
$RACK_NUM = mysql_real_escape_string($_POST['RACK_NUM']);
$PART_TYPE_ID = mysql_real_escape_string($_POST['PART_TYPE_ID']);
$LOCATION = mysql_real_escape_string($_POST['LOCATION']);
$PART_TYPE_DESC = mysql_real_escape_string($_POST['PART_TYPE_DESC']);
// See if that product name is an identical match to another product in the system
$sql = mysql_query("SELECT PART_ID FROM PART WHERE PART_ID='$id' LIMIT 1");
$productMatch = mysql_num_rows($sql); // count the output amount
if ($productMatch > 0) {
echo 'Sorry you tried to place a duplicate "Product Name" into the system, click here';
exit();
}
// Add this product into the database now
**$sql = mysql_query("INSERT INTO PART (PART_ID, PART_DESC, SERIAL_NUM, RACK_NUM, PART_TYPE_ID)
VALUES('$id','$PART_DESC','$SERIAL_NUM','$RACK_NUM','$PART_TYPE_ID')") or die (mysql_error());**
header("location: inventory_list.php");
exit();
}
?>
Micheal if I understood your problem you just need to do 2 other SQL INSERT to add data in the other table
$sql = mysql_query("INSERT INTO PART (PART_ID, PART_DESC, SERIAL_NUM, RACK_NUM, PART_TYPE_ID)
VALUES('$id','$PART_DESC','$SERIAL_NUM','$RACK_NUM','$PART_TYPE_ID')") or die (mysql_error());
$currentID = mysql_inserted_id();
$sql2 = mysql_query("INSERT INTO PART_TYPE [..]");
$sql3 = mysql_query("INSERT INTO RACK [..]");
You can use $currentID if you need the ID of the last record inersted into PART
But still I strongly suggest you to learn PDO http://php.net/pdo for sql
your table management is wrong, you never use arrows just to show that you are joining it with that table from this table, but rather from the key in first table to foreign key in the second table, that's what i would start from, maybe a better idea would be to join them using JOIN look up in google how joins are working, that may be the cause...
I agree with #yes123, that is the correct way to insert into tables, if you have a program called heidisql then use it, because there is a window to run your queries... that way to test if it is properly written also use mysql_error.
Debug, debug, and one more time debug your code.
Your tables are not correctly designed.
Try this table structures .
In your base table Part. -
The columns in this should be:
Part_id
part_desc
serial_num
The part_type should have following columns:
part_type_id
part_type_desc
part_id -> foreign key to the parent table
The rack table should be:
Rack_num
location
part_id -> foreign key to the parent table.
So your select query to get all the part related information would be:
$sql="select * from part join part_type pt on tp.part_id=part.part_id join Rack_num rn on rn.part_id=part.part_id";
With this structure the data remains normalized. And is flexible, so if the parts are on multiple racks you just go to the rack table and add and new rack number and the part id.