Question

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.

Answer 1

Transactions only work if the table engine supports them, e.g. InnoDB.