I have two tables. One table is the matches table (e2wedstrijden) and another table is my scoring table with the points earned etc. (e2teams).
Now I have that I can delete a match from the e2wedstrijden table. And this is working fine.
But I want that if I delete a match from that table. It also add or decrease points to the table ("e2teams"). I tried to compare the tables but this is not working.
So I want for example:
If($row['thuisscore'] == $row['uitscore']) what are to row names in my e2wedstrijden table. So if these two are the same (like 0-0 or 1-1 or something) Than it needs to decrease 1 point from the table e2teams. But only by the teams that are the same as the rows "Thuisteam" and "Uitteam" in my e2wedstrijden table. So the Row Thuisteam (in "e2wedstrijden") Needs to find the same result in ("e2teams") row Team. And this needs to be done the same with the Row Uitteam (in "e2wedstrijden") Needs to find the same result in ("e2teams")
Thuisteam and Uitteam = Dutch for hometeam and awayteam. I think my fault is that the system can't link the 'Thuisteam' from e2wedstrijden to the Team in e2teams but don't know how to solve it
This is my deletematches.php, It deletes the match but doesn't decrease or adds points:
<?php
if(!isset($_COOKIE['E2ingelogd'])) {
header("location:../../index.php");
}
include "../../connect.php";
$dbhandle = mysql_connect($hostname, $username, $password) or die("Could not connect to database");
$selected = mysql_select_db("login", $dbhandle);
$result = mysql_query("SELECT * FROM e2wedstrijden WHERE ID = ".$_GET['del']."");
while($row = mysql_fetch_assoc($result)){
if( $row['thuisscore'] == $row['uitscore']){
echo $row['thuisscore'];
mysql_query("UPDATE e2teams SET Punten = Punten-1 WHERE Team ='".$row['Thuisteam']."'");
mysql_query("UPDATE e2teams SET Gespeeld = Gespeeld-1 WHERE Team = ('".$row['Thuisteam']."'");
mysql_query("UPDATE e2teams SET Verloren = Gelijk-1 WHERE Team ='".$row['Uitteam']."'");
echo "Team is deleted";
}else{
echo 'Update Error!';
}
}
$table_1_delete = mysql_query("DELETE FROM e2wedstrijden WHERE ID = ".$_GET['del']."");
?>
This is my e2teams table:
And this is my E2wedstrijden table:
So i need something like:
UPDATE e2teams SET Punten = Punten-1 WHERE Team = Look in table ("e2wedstrijden) deleted Thuisteam and deleted Uitteam
Hope you can help
You've placed an extra parentheses in the 2nd query for "gespeeld" right after the equal sign:
mysql_query("UPDATE e2teams SET Gespeeld = Gespeeld-1
WHERE Team = ('".$row['Thuisteam']."'");
Is this what isn't updating?
Without being 100% sure on how your data model works, it might make sense at refactoring what you have. Something that might be useful would be to create a view of the summary table and just update the data from the child/master table.... aggregating in the view layer. Views in mysql can be seen here.
If you are stuck with the data model you have (legacy application, etc.) you can possibly look at triggers if you have to modify data in two tables you might want to consider stored procedures or triggers, discussed here and here.
The third thing that comes to mind, is around correlated sub-queries and how you could reference the another table in a sort of update-from. However, you're ID's aren't surrogate keys in this situation.
Also, have a look at sql injection; I haven't looked at PHP in a while but those sql statements kind of look like they are created with sting composition
Good luck,
Related
I have an over time sheet that gets printed when there is over time from an employee. The overtime format goes like "B.Eng." And then the name of the employee. Now I need it to check the name of the employee (or id) to print either "B.Eng." Or "MR.", This because there is an employee (just one) that does not have a degree. I would think the answer would be an IF condition.
Here is my code:
$db = mysql_select_db ("over_time");
$strqry = "SELECT emp_name FROM contr_acces where id_emp='".$vp_idemp."';";
$qry2 = mysql_query ($strqry);
$row2 = mysql_fetch_object ($qry2);
$vl_emp_name= "B.Eng. ".$row2->emp_name;
print $vl_emp_name;
you can do something like this
$db = mysql_select_db ("over_time");
$strqry = "SELECT emp_name FROM contr_acces where id_emp='".$vp_idemp."';";
$qry2 = mysql_query ($strqry);
$row2 = mysql_fetch_object ($qry2);
$vl_emp_name= $row2->emp_name;
if($vl_emp_name == 'name_without_the_degree){
$vl_emp_name= "Mr. ".$vl_emp_name;
}else{
$vl_emp_name= "B.Eng ".$vl_emp_name;
}
print $vl_emp_name;
It is not the best solution since you are hardcoding the condition but without knowing more about the db structure is not possible to give you a better solution. The most efficient one would be to add a field to the db with the degree type for the users and retrieve it together with the name.
See my comment under your question for the api used and the sql injection risk that this soultion doesn't address
I have a table with articles(articles) that have their contents to be updated by a team accessing them through the Internet(PHP) and assigning them automatically the articles to worked upon, based on the article status and id.
We have the id of the article(artid), its content(artcont) and its update status(artstat): Y(yes)/N(no)/B(blocked).
When one member of the team wants to work on updating articles, I want to automatically select for him the article that has its artstat "no" and the lowest articleid, setting at the same time artstat as "blocked" (to prevent it being edited at the same time by other team members).I want at the same time to return the artid that it's being edited.
For the moment, I have in PHP a select query followed immediately by an update one:
....(credentials for connecting to database)...
$conn = mysqli_connect($servername, $username, $password, $dbname);
$sql=SELECT * from articles where artstat='N' order by artid asc LIMIT 1;
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) == 1) {
// output data
$row = mysqli_fetch_assoc($result);
echo "id: " . $row["artid"]. "<br>";
$upd = "UPDATE articles SET artstat='B' where artid=" .$row["artid"];
mysqli_query($conn, $upd);
}
else {
echo "0 results";
}
Now, my question is: is it possible that a second team member (on another opened connection) who wants to work on an article makes his request exactly between the execution of the select and update statements, he will get the same article to work with (artstat is not 'B' yet), isn't it?
How to avoid this?
Thank you.
In Mysql , there are two type of locking to avoid such situation in innodb engine
table level locking
row level locking
You will need row level locking for this
You can read more about this from the below stack over flow post.
https://dba.stackexchange.com/questions/15854/innodb-row-locking-how-to-implement
I need to synchronize specific information between two databases (one mysql, the other a remote hosted SQL Server database) for thousands of rows. When I execute this php file it gets stuck/timeouts after several minutes I guess, so I wonder how I can fix this issue and maybe also optimize the way of "synchronizing" it.
What the code needs to do:
Basically I want to get for every row (= one account) in my database which gets updated - two specific pieces of information (= 2 SELECT queries) from another SQL Server database. Therefore I use a foreach loop which creates 2 SQL queries for each row and afterwards I update those information into 2 columns of this row. We talk about ~10k Rows which needs to run thru this foreach loop.
My idea which may help?
I have heard about things like PDO Transactions which should collect all those queries and sending them afterwards in a package of all SELECT queries, but I have no idea whether I use them correctly or whether they even help in such cases.
This is my current code, which is timing out after few minutes:
// DBH => MSSQL DB | DB => MySQL DB
$dbh->beginTransaction();
// Get all referral IDs which needs to be updated:
$listAccounts = "SELECT * FROM Gifting WHERE refsCompleted <= 100 ORDER BY idGifting ASC";
$ps_listAccounts = $db->prepare($listAccounts);
$ps_listAccounts->execute();
foreach($ps_listAccounts as $row) {
$refid=$row['refId'];
// Refsinserted
$refsInserted = "SELECT count(username) as done FROM accounts WHERE referral='$refid'";
$ps_refsInserted = $dbh->prepare($refsInserted);
$ps_refsInserted->execute();
$row = $ps_refsInserted->fetch();
$refsInserted = $row['done'];
// Refscompleted
$refsCompleted = "SELECT count(username) as done FROM accounts WHERE referral='$refid' AND finished=1";
$ps_refsCompleted = $dbh->prepare($refsCompleted);
$ps_refsCompleted->execute();
$row2 = $ps_refsCompleted->fetch();
$refsCompleted = $row2['done'];
// Update fields for local order db
$updateGifting = "UPDATE Gifting SET refsInserted = :refsInserted, refsCompleted = :refsCompleted WHERE refId = :refId";
$ps_updateGifting = $db->prepare($updateGifting);
$ps_updateGifting->bindParam(':refsInserted', $refsInserted);
$ps_updateGifting->bindParam(':refsCompleted', $refsCompleted);
$ps_updateGifting->bindParam(':refId', $refid);
$ps_updateGifting->execute();
echo "$refid: $refsInserted Refs inserted / $refsCompleted Refs completed<br>";
}
$dbh->commit();
You can do all of that in one query with a correlated sub-query:
UPDATE Gifting
SET
refsInserted=(SELECT COUNT(USERNAME)
FROM accounts
WHERE referral=Gifting.refId),
refsCompleted=(SELECT COUNT(USERNAME)
FROM accounts
WHERE referral=Gifting.refId
AND finished=1)
A correlated sub-query is essentially using a sub-query (query within a query) that references the parent query. So notice that in each of the sub-queries I am referencing the Gifting.refId column in the where clause of each sub-query. While this isn't the best for performance because each of those sub-queries still has to run independent of the other queries, it would perform much better (and likely as good as you are going to get) than what you have there.
Edit:
And just for reference. I don't know if a transaction will help here at all. Typically they are used when you have several queries that depend on each other and to give you a way to rollback if one fails. For example, banking transactions. You don't want the balance to deduct some amount until a purchase has been inserted. And if the purchase fails inserting for some reason, you want to rollback the change to the balance. So when inserting a purchase, you start a transaction, run the update balance query and the insert purchase query and only if both go in correctly and have been validated do you commit to save.
Edit2:
If I were doing this, without doing an export/import this is what I would do. This makes a few assumptions though. First is that you are using a mssql 2008 or newer and second is that the referral id is always a number. I'm also using a temp table that I insert numbers into because you can insert multiple rows easily with a single query and then run a single update query to update the gifting table. This temp table follows the structure CREATE TABLE tempTable (refId int, done int, total int).
//get list of referral accounts
//if you are using one column, only query for one column
$listAccounts = "SELECT DISTINCT refId FROM Gifting WHERE refsCompleted <= 100 ORDER BY idGifting ASC";
$ps_listAccounts = $db->prepare($listAccounts);
$ps_listAccounts->execute();
//loop over and get list of refIds from above.
$refIds = array();
foreach($ps_listAccounts as $row){
$refIds[] = $row['refId'];
}
if(count($refIds) > 0){
//implode into string for use in query below
$refIds = implode(',',$refIds);
//select out total count
$totalCount = "SELECT referral, COUNT(username) AS cnt FROM accounts WHERE referral IN ($refIds) GROUP BY referral";
$ps_totalCounts = $dbh->prepare($totalCount);
$ps_totalCounts->execute();
//add to array of counts
$counts = array();
//loop over total counts
foreach($ps_totalCounts as $row){
//if referral id not found, add it
if(!isset($counts[$row['referral']])){
$counts[$row['referral']] = array('total'=>0,'done'=>0);
}
//add to count
$counts[$row['referral']]['total'] += $row['cnt'];
}
$doneCount = "SELECT referral, COUNT(username) AS cnt FROM accounts WHERE finished=1 AND referral IN ($refIds) GROUP BY referral";
$ps_doneCounts = $dbh->prepare($doneCount);
$ps_doneCounts->execute();
//loop over total counts
foreach($ps_totalCounts as $row){
//if referral id not found, add it
if(!isset($counts[$row['referral']])){
$counts[$row['referral']] = array('total'=>0,'done'=>0);
}
//add to count
$counts[$row['referral']]['done'] += $row['cnt'];
}
//now loop over counts and generate insert queries to a temp table.
//I suggest using a temp table because you can insert multiple rows
//in one query and then the update is one query.
$sqlInsertList = array();
foreach($count as $refId=>$count){
$sqlInsertList[] = "({$refId}, {$count['done']}, {$count['total']})";
}
//clear out the temp table first so we are only inserting new rows
$truncSql = "TRUNCATE TABLE tempTable";
$ps_trunc = $db->prepare($truncSql);
$ps_trunc->execute();
//make insert sql with multiple insert rows
$insertSql = "INSERT INTO tempTable (refId, done, total) VALUES ".implode(',',$sqlInsertList);
//prepare sql for insert into mssql
$ps_insert = $db->prepare($insertSql);
$ps_insert->execute();
//sql to update existing rows
$updateSql = "UPDATE Gifting
SET refsInserted=(SELECT total FROM tempTable WHERE refId=Gifting.refId),
refsCompleted=(SELECT done FROM tempTable WHERE refId=Gifting.refId)
WHERE refId IN (SELECT refId FROM tempTable)
AND refsCompleted <= 100";
$ps_update = $db->prepare($updateSql);
$ps_update->execute();
} else {
echo "There were no reference ids found from \$dbh";
}
I have a database listed as $db under mysqli. This database is contains into two tables, I listed them below as table and table2 (just for this example). Table2's rows requires an id from table. This is fine, but there might be a problem adding the columns into table2 thus requiring a rollback routine. However, it doesn't seem to be working.
I started with turning off the auto-commit. I then tried to put in the rollback command even though I am using the die command to signal a failure. As far as I am concerned the transaction could be blasted into oblivion in mid operation and the database should still be stable. So I am not sure what is going on here unless the database is completely ignoring the fact that I am trying to turn off auto-commit.
The basic structure of my code is listed below:
function problem($str)
{
global $db;
mysqli_rollback($db);
die($str);
}
mysqli_autocommit($db,false);
//Basic check if exists
$sqlstr = "SELECT * FROM table WHERE name = '$name';";
$r = mysqli_query($db,$sqlstr);
if (mysqli_num_rows($r)>0){problem("A row already exists under that id");}
//Insert the row
$sqlstr = "INSERT INTO table (name,v1,v2,v3) VALUES ('$name','$v1','$v2','$v3');";
$r = mysqli_query($db,$sqlstr);
if (!$r){problem("Could not insert into the table. $sqlstr");}
//Get the generated id part 1
$sqlstr = "SELECT id FROM table WHERE name = '$name';";
$r = mysqli_query($db,$sqlstr);
if (!$r){problem("Could not add into the table. $sqlstr");}
//Get the generated id part 2
$row = mysqli_fetch_assoc($r);
$eid = $row['id'];
//A simple loop
$count = count($questions);
for ($i=1;i<=$count;$i++)
{
//This is where it typically could die.
$r = mysqli_query($db,"INSERT INTO table2 VALUES (...);");
if (!$r){problem("Could not add to the table2. $sqlstr");}
}
mysqli_commit($db);
Is there something I am missing? I tried to follow the examples I found for the auto-commit as closely as I could.
Transactions only work if the table engine supports them, e.g. InnoDB.
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.