I trying to implement simple query using php in sql, but the value deducing is twice I don't know why.
What am trying to do is to get the current points of the user and reduce the points that are redeemed by the user and update the points. But what happening is suppose user have 500 points and am sending 100 points to reduce in the following php but the result shows user have 300 points remaining. Why double points are reducing in this table only.
Things i have already checked : -Calling php one time only, Checking the value of points by echo , make a separate table for checking it shows correct points.
Any help would be appreciated
<php
require "conn.php";
$conn = mysqli_connect($server_name, $mysql_username, $mysql_password, $db_name);
$number = mysqli_real_escape_string($conn, $_POST['number']);
$points = mysqli_real_escape_string($conn, $_POST['points']);
//echo $points ;
$sql1 = "UPDATE user_earning SET points = points - '$points' WHERE user_number = '$number'";
$result = $conn->query($sql1);
if ($conn->query($sql1) === TRUE) {echo "Current points Updated";
} else {echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
Both $points and $number are coming from android application and am not getting any error. The response i get is current points updates. So php is running fine but the value is not coming correct in mysql
You're running the query twice, once each on these two lines.
$result = $conn->query($sql1);
if ($conn->query($sql1) === TRUE) {echo "Current points Updated";
The second line, maybe, should be
if ($result === TRUE) {echo "Current points Updated";
I think your error is here
$sql1 = "UPDATE user_earning SET points = points - '$points' WHERE user_number = '$number'";
You need the following steps
1. First retrieve the points from your table and store in a variable
$q=mysqli_query($con,"select * from user_earning");
$current_points = 0;
if($q){
while($row = mysqli_fetch_assoc($q))
$current_points = $row['points'];
}
2. Subtract the points as you want`enter code here`
$new_points = $current_points - $points.
3. Then do your update
$sql1 = "UPDATE user_earning SET points = '$new_points' WHERE user_number = '$number'";
Related
I am trying to create a Secret Santa system using a PHP page and a MySQL database to store the details so if someone forgets their match they can re-request it.
Step 1: I created a random number generator based on the number of people in the list in the database.
Count Function:
$maxSQL = "SELECT COUNT(id) as total FROM secretsanta";
$maxRS = mysqli_query($conn, $maxSQL);
$maxQuery = mysqli_fetch_array($maxRS);
$maxpersons = $maxQuery['total'];
Then the Random Number Generator:
$assigned = rand(1,$maxpersons);
Step 2: Test if the random number matches the persons own id and regenerate a new number if true.
do {
$assigned = rand(1,$maxpersons);
} while ($assigned==$id);
Step 3: Write the paired id to the persons database record.
$assignSQL = "UPDATE secretsanta SET assigned = '".$assigned."' WHERE secretsanta.id = ".$id;
if (mysqli_query($conn, $assignSQL)) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . mysqli_error($conn);
}
The Problem: Now I need to check that other people aren't assigned to that person or otherwise some could miss out and others would get more than others.
I tried to implement a function that contained a query to test each record to see if that number already existed and was hoping to add it as a condition to perhaps a while or do while statement?
if (!function_exists('checkRandom')){
function checkRandom($funcid){
$Check_SQL = "SELECT assigned FROM secretsanta ORDER BY id ASC";
$Check_RES = mysqli_query($conn, $Check_SQL);
if (Check_RES) {
while ($CheckArray = mysqli_fetch_array($Check_RES, MYSQLI_ASSOC)) {
$CheckAsgn = $CheckArray['assigned'];
if ($funcid==$CheckAsgn) {return true;}else{return false;}
}
}
}
}
Then implement it into the do while statement like this:
do {
$assigned = rand(1,$maxpersons);
} while ($assigned==$id||checkRandom($assigned));
No luck so far...HELP!.. please :)
P.S. I know there are websites that already do this, I just don't trust them to give out mine and family email address' if I can make my own private version myself.
Using your method, the first few assignments will be done with no problem, but imagine the last unassigned entry and how many times it will try a random number only to find the person with that id is already assigned..
I'm gonna give you another approach to your problem: for each user that you want to assign a santa to, make a new SELECT statement with a WHERE clause that lets you select only those users that are not assigned yet.
check out my code and see if that helps you. I just typed this and didnt test it so there could be some mistakes.
// load all unassigned users into an array
$unassignedUsers = [];
$query = "SELECT id, assigned FROM secretsanta WHERE assigned is NULL";
$res = mysqli_query($conn, $query);
while($row = mysqli_fetch_assoc($res){
$unassignedUsers[] = $row;
}
if(count($unassignedUsers) == 1){
echo 'There is only 1 unassigned user. Therefore he cannot be matched';
} else {
// for loop for each user in DB that is not assigned yet
//for ($i = 1;$i <= count($unassignedUsers); $i++){
$i = 0;
foreach($unassignedUsers as $user)
// if its the second-to-last iterations of the for-loop, check for legality of the last one
if(count($unassignedUsers) - $i == 1){
$lastUserID = $unassignedUsers[count($unassignedUsers)-1]['id'];
$query = "SELECT id FROM secretsanta WHERE assigned is NULL AND id = ".$lastUserID;
$res = mysqli_query($conn, $query);
$rowcount = mysqli_num_rows($res);
if ($rowcount){
// last user is still unassigned
$query = "UPDATE secretsanta SET assigned = '".$lastUserID."' WHERE id = ".$user['id'];
if(mysqli_query($conn, $query)){
echo "Record with id ".$user['id']." updated successfully";
} else {
echo "Error updating record: ".mysqli_error($conn);
}
}
} else {
// select all unassigned users
$unassignedIDs = [];
$query = "SELECT id FROM secretsanta WHERE assigned is NULL AND id <> ".$user['id'];
$res = mysqli_query($conn, $query);
while($row = mysqli_fetch_assoc($res){
$unassignedIDs[] = $row['id'];
}
// get a random id from $unassignedIDs
$randomIndex = rand(0, count($unassignedIDs)-1);
$randomID = $unassignedIDs[$randomIndex];
// assign $randomID to user
$query = "UPDATE secretsanta SET assigned = '".$randomID."' WHERE id = ".$user['id'];
if(mysqli_query($conn, $query)){
echo "Record with id ".$user['id']." updated successfully";
} else {
echo "Error updating record: ".mysqli_error($conn);
}
}
$i++;
}
}
last edit: refactored whole code so it is able to be run multiple times and only assigns new users who are not assigned yet.
Step 1 is dependent on have a contiguous set of ids for the people. Think what happens if '3' leaves the company and it hires 6 to replace them....1,2,4,5,6 ($maxpersons=5)
"Now I need to check" - no you are still trying to solve the problem by guessing then seeing if your guess worked. Use an algorithm which is always going to return a correct result. The method below requires the addition of a temporary field 'sequence' of type float.
mysqli_query($conn,"UPDATE secretsanta SET sequence=RAND()");
$first=false;
$prev=false;
$all=mysqli_query($conn, "SELECT * FROM secretsanta ORDER BY sequence, id");
while ($r=mysqli_fetch_assoc($all)) {
if (false===$first) {
$first=$r['id'];
} else {
save_pair($prev, $r['id']);
}
$prev=$r['id'];
}
save_pair($prev, $first);
(but with better error checking)
I am taking a users input and storing it in a database, however I want to be able to update the records if a user adds more information. So I want to search the database find the server with the same name and update the the last downtime and the number of downtimes.
$connect = mysqli_connect("localhost", "Username", "Password","Test_downtime");
if (!$connect)
{
die("Connection failed: " . mysqli_connect_error());
}else
{
echo "Connected successfully\n";
}
$servername = $_GET["server_name"];
$downtime = $_GET["downtime"];
$time_now = time();
$result = mysqli_query($connect, "SELECT COUNT(*) FROM `Test_downtime`.`Downtime` WHERE `Server_Name` = '$servername'");
$row = mysqli_fetch_array($result);
// If no downtime have been reported before
if ($row[0] == 0){
$sql = mysqli_query($connect, "INSERT INTO `Test_downtime`.`Downtime` (ID, Server_name, First_downtime, Last_downtime, Num_of_downtime,Total_downtime) VALUES (NULL, '$servername', '$time_now','$time_now',1,'$downtime'); ");
if ($sql==true) {
echo $servername . " has has its first downtime recorded\n";
}
}
//If users is already in the database
else{
$numdowntime = ($row["Num_of_downtime"] + 1);
$id = ($row["ID"]);
$sqlupdate = "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now() WHERE `Server_Name` = '$servername'";
if ($sqlupdate == TRUE) {
echo "Oh No! " . $servername . " has had ". $numdowntime ." of downtimes" ;
}
}
?>
The program works fine if the server is not already in the database, the problems arise if the server is already in the database. I get the message saying it has been updated yet nothing happens to the database. How do i make it so it search and updates the records for the searched item.
So nothing append since you do not execute the sql statement ^^
Take a look here :
$sqlupdate = "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now() WHERE `Server_Name` = '$servername'";
You need to use :
$sql = mysqli_query($connect, $sqlupdate);
Just after it in order to execute it.
Or at least change it to
$sqlupdate = mysqli_query($connect, "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now() WHERE `Server_Name` = '$servername'" );
Btw there is other problem but here is the main one [check out the other answer in order to found another one ]
you are fetching the result as indexed array
mysqli_fetch_array($result);
and here you are accessing results as associative array
$numdowntime = ($row["Num_of_downtime"] + 1);
change your query to
mysqli_fetch_assoc($result);
use
mysqli_num_rows($result);
to checking if you have any data
change
if ($row[0] == 0){}
to
if(mysqli_num_rows($result) ==0){}
A good approach for increasing a count in a column is using SQL to increase that.
$sqlupdate = mysqli_query($connect, "UPDATE `Test_downtime`.`Downtime` SET `Num_of_downtime` = (`Num_of_downtime` + 1), `Last_downtime` = now() WHERE `Server_Name` = '$servername'" );
This way you can skip your $numdowntime calculation, and the result is more accurate.
In your current setup, two users may fire the event at the same time, they both retrieve the same number from the database (ie. 9), both increasing it with one (ie. 10), and writing the same number in the database.
Making your count one short of the actual count.
SQL takes care of this for you by locking rows, and you are left with a more accurate result using less logic :)
You miss the mysqli_query() function, which actually queries the database.
$sqlupdate = mysqli_query("
UPDATE `Test_downtime`.`Downtime`
SET `Num_of_downtime` = $numdowntime, `Last_downtime` = now()
WHERE `Server_Name` = '$servername'"
);
i have a curious problem. When i make an update to a row, the info is updated, but not how i expect it to. i call this code
$count = $v_count+1;
mysqli_query($db_conn, "UPDATE videos SET v_count='$count' WHERE id='$vk'");
I have also tried this code as well...
if(isset($_POST['vk'])){
mysqli_query($db_conn, "UPDATE videos SET v_count='$v_count'+1 WHERE id='$vk'");
}else{
echo mysqli_error($db_conn);exit();
}
It does work, But the number is increased by 2, not the 1 i expected.This line HAS worked in other applications i have used... but now it is behaving oddly. Any help will be appreciated. Here is the full PHP block
$vk = $_GET['vk'];
if($vk != ''){
$sql = "SELECT * FROM videos WHERE id='$vk' LIMIT 1";
$query = mysqli_query($db_conn, $sql) or die (mysqli_error());
while ($row = mysqli_fetch_array($query)) {
$v_count = $row['v_count'];
}
mysqli_query($db_conn, "UPDATE videos SET v_count=v_count+1 WHERE id='$vk'");
mysqli_free_result($query);
}else{
header('location: index.php');
}
You probably have some error in your application logic such that $v_count in your PHP code doesn't equal the v_count field in your MySQL table.
If you use
mysqli_query($db_conn, "UPDATE videos SET v_count=v_count+1 WHERE id='$vk'");
it will always increase by 1 no matter what the value of your $v_count PHP variable.
The below script inputs data to a database this takes some information from a form then stores them in to the database. And I'm also using uplodify to upload a image file and store the file name in the database but my issue is this data processing script keeps updating the row ID one never jumps to the second line I tried every thing can some one help me with this or show me what I'm doing wrong.
Also this checks the ID and if it's not equal to 1 then does an insertion if it's equal then update it but this not happening.
The ID is auto incrementing.
My script
<?php
/**
* #author SiNUX
* #copyright 2013
*/
include ('connect.php');
$getId = mysql_query("SELECT ID FROM poiinfo ORDER BY ID DESC LIMIT 1");
$row = mysql_fetch_array($getId);
$poiName = $_REQUEST['Name'];
$poiDes = $_REQUEST['Descrip'];
$poiCon = $_REQUEST['ConInfo'];
//$poiId = $_REQUEST['pID'];
if($row['ID'] != "1"){
$dbData = "INSERT INTO poiinfo(`Name`, `Des.`, `Contact`) VALUES ('$poiName','$poiDes','$poiCon')";
$putData = mysql_query($dbData);
if ($putData){
echo "Data inserted";
}else {
echo "Not Done";
}
}else {
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon'";
$updDone = mysql_query($updLn);
if ($updDone){
echo "Data inserted";
}else {
echo "Not Done";
}
}
?>
I tried u r suggestions but it's still the same now my code for the update is looks like this.
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon' WHERE `ID`='".$row['ID']."'";
But still it keeps up dating the ID 1 not moving on to the next one.
Your update query is missing a WHERE clause. Try this:
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon' WHERE ID = '".$row['ID']."'";
Also be beware of MySQL Injections: http://en.wikipedia.org/wiki/SQL_injection
To check why your update failed, you should call mysql_error in your last else clause :
} else {
echo mysql_error();
}
As for the first problem : if you never insert a new record (I don't see how that could happen, provided your code), you will never have a record whose ID is 2.
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon'";
You need a where clause in this sql to specify a record to update. Currently it is updating all records.
$updLn = "UPDATE `poiinfo` SET `Name`='$poiName',`Des.`='$poiDes',`Contact`='$poiCon' WHERE `ID` = ".$row['id']";";
You will need to set an $id variable for this to work.
Before you assume I didn't establish a database connection, I did. the only portion of the code that does not update is the if empty statements.
All the values can be echoed out correctly, it's just that query doesn't work.
This is in directory config and named stuff.php
$user = $mysqli->real_escape_string($_SESSION['username']);
$user_query = "SELECT * FROM users WHERE username = '$user'";
$result = $mysqli->query($user_query);
$row = $result->fetch_assoc();
$referrer = $row['ref'];
$refearn = $row['refearn'];
verify.php
include('config/stuff.php');
if (isset($_SERVER['HTTP_X_FORWARDED_FOR'])) { // Get Real IP
$IP = $_SERVER['HTTP_X_FORWARDED_FOR'];
} else {
$IP = $_SERVER['REMOTE_ADDR'];
}
if ($IP=="external server ip here") {
if (!empty($referrer)){
$mysqli->query("UPDATE users SET points=points+10, refearn = refearn+10 WHERE username='".$referrer."'") or die(mysqli_error($mysqli));
}
$mysqli->query("UPDATE users SET points=points+".$earnings.", completed = completed+1 WHERE username='".$subid."'") or die(mysqli_error($mysqli));
}
My guess is you could try to retrieve the value of points through a query then add to it so you're just updating to a simple value. However, if mysql_error() is returning an error, it should be easier to figure out.
Example:
$getPoints = mysql_query("SELECT points FROM table WHERE condition");
$points = mysql_result($getPoints, 0, "points");
$update = mysql_query("UPDATE table SET points=" . ($points+10) . " WHERE condition");
Hope that helps. Another consideration, though. Why use an endif structure unless you're breaking PHP tags to display content?
try this:
$mysqli->query("UPDATE `users` SET `points`=`points`+10, `refearn` = `refearn`+10 WHERE `username`='".$referrer."'") or die(mysqli_error($mysqli));
Hope this helps. What I think is, mysql query might be taking those as constant - not as the mysql rows. Try that