Im trying to select and echo a user in the database that his ticket1 <= $rand and ticket2 >= $rand but its not working and i don't know why. This is the code:
<?php
$sql = "SELECT t1, t2 FROM user";
$result = mysqli_query($conn, $sql);
$sqlt = "SELECT t2 FROM user ORDER BY id DESC LIMIT 1";
$resultt = mysqli_query($conn, $sqlt);
$rowt = mysqli_fetch_assoc($resultt);
$rand = rand(1, $rowt['t2']);
echo $rand.'<br>';
while($row = mysqli_fetch_assoc($result)){
echo $row['t2'].'<br>';
$t1 = $row['t1'];
$t2 = $row['t2'];
$sqls = "SELECT uid FROM user WHERE $t1 <= $rand && $t2 >= $rand";
$results = mysqli_query($conn, $sqls);
$rows = mysqli_fetch_row($result);
echo $rows[0];
}
?>
First of all we can't help you because you didn't let us see what do you have in your table.
$sql = "SELECT t1, t2 FROM user";
$result = mysqli_query($conn, $sql);
//This is for getting t1 and t2 from user table
$sqlt = "SELECT t2 FROM user ORDER BY id DESC LIMIT 1";
$resultt = mysqli_query($conn, $sqlt);
$rowt = mysqli_fetch_assoc($resultt);
//This is for getting only t2 from last row in user table
$rand = rand(1, $rowt['t2']);
//This is for picking up random number from 1 to t2
//which is in the last row from user table
//so it may be lower number than you want
//if you want to get number of rows its done differently
while($row = mysqli_fetch_assoc($result)){
$t1 = $row['t1'];
$t2 = $row['t2'];
//Here you are getting every t1 and t2 value for every row
$sqls = "SELECT uid FROM user WHERE $t1 <= $rand && $t2 >= $rand";
//And here you are saying I want to select userId from userTable where
//someNumber <= randomNumber and someOtherNumber>= randomNumber
//which wont work, you are looking for userId from userTable where
//randomNumber is >= than SomeNumber which is t1 and randomNumber<=t2
$results = mysqli_query($conn, $sqls);
$rows = mysqli_fetch_row($result);
echo $rows[0];
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.