how to converting mysql into mysqli - php

Question

i have the following problem with a login script. at the moment i refresh my site and would like to change mysql into mysqli. i have a working code, that works with mysql like it should to. now i get in trouble with changing that into mysqli, which doesn't work.
here is the original mysql code:
while($row = mysql_fetch_array($sql)){
$id = $row["id"];
session_register('id');
$_SESSION['id'] = $id;
mysql_query("UPDATE tableA SET time=now(), x4=x4 + 1 WHERE id='$id'");
mysql_query("DELETE FROM tableB WHERE (NOW() - INTERVAL 1 DAY) > Date AND ID='$id'");
$result = mysql_query("SELECT COUNT(*) AS val FROM tableB WHERE ID='$id'");
$count = mysql_fetch_assoc($result);
var_dump($count);
if ($count [val] <xy){
mysql_query("INSERT INTO tableB (Date, ID) VALUES (now(),'$id') ");
mysql_query("UPDATE tableA SET x7=x7 + 1 WHERE id='$id'");
and here is the mysqli version, that wont work and i dont know why:
$time = gmdate("M d Y H:i:s", time());
$id = '".$row["id"]."';
$get_id = "SELECT id FROM tableA WHERE x1='".$x1."' AND x2='".$x2."'";
$result = mysqli_query($db, $get_id);
if ($result === false) {
printf("Errormessage 1");
exit();
}
$row = $result->fetch_array(MYSQLI_ASSOC);
$update = "UPDATE tableA SET time=now(), x4=x4 + 1 WHERE id='".$row["id"]."'";
$result2 = mysqli_query($db, $update);
if ($result2 === false) {
printf("Errormessage 2");
exit();
}
$reset = "DELETE FROM tableB WHERE (NOW() - INTERVAL 1 DAY) > Date AND ID='".$row["id"]."'";
$result3 = mysqli_query($db, $reset);
if ($result3 === false) {
printf("Errormessage 4");
exit();
}
$count = "SELECT COUNT(*) AS val FROM tableB WHERE ID='".$row["id"]."'";
$result4 = mysqli_query($db, $count);
if ($result4 === false) {
printf("Errormessage 5");
exit();
}
$sum = $result4->fetch_assoc($count);
var_dump($sum);
if ($count [val] <xy){
$insert = "INSERT INTO TableB (Date, ID) VALUES(?,?) ";
if($query = $db->prepare($insert)){
$query->bind_param('ss', $time, $id);
$query->execute();
$update_x = "UPDATE tableA SET x7=x7 + 1 WHERE id='".$row["id"]."'";
$result5 = mysqli_query($db, $update_x);
if ($result5 === false) {
printf("Errormessage 5");
exit();

Answer 1

You haven't mentioned the actual error but it seems that problem occurs from here:
$row = $result->fetch_array(MYSQLI_ASSOC);
you didnot use a loop here like your mysql version of code.

Answer 2

You are attempting to insert $time into what appears to be a DATETIME column, based on your old mysql version, but it is improperly formatted.
$time = gmdate("M d Y H:i:s", time());
Based on your use of NOW() in the old code, we assume TableB.Date to be a DATETIME type:
mysql_query("INSERT INTO tableB (Date, ID) VALUES (now(),'$id') ");
So in your new code, since you don't use NOW() for the TableB insert, you should be creating $time as YYYY-MM-DD:
// Should be YYYY-MM-DDD H:i:s for MySQL
$time = gmdate("Y-m-d H:i:s", time());
// It gets inserted into TableB here
$insert = "INSERT INTO TableB (Date, ID) VALUES(?,?) ";
if($query = $db->prepare($insert)){
$query->bind_param('ss', $time, $id);
$query->execute();
Or, just use MySQL's NOW() in the new code unless you have a reason to specify the time in PHP code:
$insert = "INSERT INTO TableB (Date, ID) VALUES(NOW() ,?) ";
if($query = $db->prepare($insert)){
$query->bind_param('s', $id);
$query->execute();