MySQL Prepared statements with a variable size variable list

Question

How would you write a prepared MySQL statement in PHP that takes a differing number of arguments each time? An example such query is:

SELECT `age`, `name` FROM `people` WHERE id IN (12, 45, 65, 33)

The IN clause will have a different number of ids each time it is run.

I have two possible solutions in my mind but want to see if there is a better way.

Possible Solution 1 Make the statement accept 100 variables and fill the rest with dummy values guaranteed not to be in the table; make multiple calls for more than 100 values.

Possible Solution 2 Don't use a prepared statement; build and run the query checking stringently for possible injection attacks.

Answer 1

I can think of a couple solutions.

One solution might be to create a temporary table. Do an insert into the table for each parameter that you would have in the in clause. Then do a simple join against your temporary table.

Another method might be to do something like this.

$dbh=new PDO($dbConnect, $dbUser, $dbPass);
$parms=array(12, 45, 65, 33);
$parmcount=count($parms);   // = 4
$inclause=implode(',',array_fill(0,$parmcount,'?')); // = ?,?,?,?
$sql='SELECT age, name FROM people WHERE id IN (%s)';
$preparesql=sprintf($sql,$inclause);  // = example statement used in the question
$st=$dbh->prepare($preparesql);
$st->execute($parms);

I suspect, but have no proof, that the first solution might be better for larger lists, and the later would work for smaller lists.

---

To make @orrd happy here is a terse version.

$dbh=new PDO($dbConnect, $dbUser, $dbPass);
$parms=array(12, 45, 65, 33);
$st=$dbh->prepare(sprintf('SELECT age, name FROM people WHERE id IN (%s)',
                          implode(',',array_fill(0,count($parms),'?'))));
$st->execute($parms);

---

Answer 2

There is also the FIND_IN_SET function whose second parameter is a string of comma separated values:

SELECT age, name FROM people WHERE FIND_IN_SET(id, '12,45,65,33')

Answer 3

decent sql wrappers support binding to array values. i.e.

$sql = "... WHERE id IN (?)";
$values = array(1, 2, 3, 4);
$result = $dbw -> prepare ($sql, $values) -> execute ();

Answer 4

Please take #2 off the table. Prepared statements are the only way you should consider protecting yourself against SQL injection.

What you can do, however, is generate a dynamic set of binding variables. i.e. don't make 100 if you need 7 (or 103).

Answer 5

i got my answer from: http://bugs.php.net/bug.php?id=43568
this is my working solution to my problem. Now i can dynamically use as many parameters as i want. They will be the same number as i have in an array or as in this case I am passing the ids from the last query ( which found all the ids where email = 'johndoe@gmail.com') to the dynamic query to get all the info about each of these id no matter how many i end up needing.

<?php $NumofIds = 2; //this is the number of ids i got from the last query
    $parameters=implode(',',array_fill(0,$NumofIds,'?')); 
    // = ?,? the same number of ?'s as ids we are looking for<br />
    $paramtype=implode('',array_fill(0,$NumofIds,'i')); // = ii<br/>
    //make the array to build the bind_param function<br/>
    $idAr[] = $paramtype; //'ii' or how ever many ?'s we have<br/>
    while($statement->fetch()){ //this is my last query i am getting the id out of<br/>
        $idAr[] = $id;  
    }

    //now this array looks like this array:<br/>
    //$idAr = array('ii', 128, 237);

    $query = "SELECT id,studentid,book_title,date FROM contracts WHERE studentid IN ($parameters)";
    $statement = $db->prepare($query);
    //build the bind_param function
    call_user_func_array (array($statement, "bind_param"), $idAr);
    //here is what we used to do before making it dynamic
    //statement->bind_param($paramtype,$v1,$v2);
    $statement->execute();
?>

Answer 6

If you're only using integer values in your IN clause, there's nothing that argues against constructing your query dynamically without the use of SQL parameters.

function convertToInt(&$value, $key)
{
    $value = intval($value);
}

$ids = array('12', '45', '65', '33');
array_walk($ids, 'convertToInt');
$sql = 'SELECT age, name FROM people WHERE id IN (' . implode(', ', $ids) . ')';
// $sql will contain  SELECT age, name FROM people WHERE id IN (12, 45, 65, 33)

But without doubt the solution here is the more general approach to this problem.

Answer 7

I had a similiar problem today and I found this topic. Looking at the answers and searching around the google I found a pretty solution.

Although, my problem is a little bit more complicated. Because I have fixed binding values and dynamic too.

This is the solution.

$params = array()
$all_ids = $this->get_all_ids();

for($i = 0; $i <= sizeof($all_ids) - 1; $i++){
    array_push($params, $all_ids[$i]['id']);
}

$clause = implode(',', array_fill(0, count($params), '?')); // output ?, ?, ?
$total_i = implode('', array_fill(0, count($params), 'i')); // output iiii

$types = "ss" . $total_i; // will reproduce : ssiiii ..etc

// %% it's necessary because of sprintf function
$query = $db->prepare(sprintf("SELECT * 
                                FROM clients    
                                WHERE name LIKE CONCAT('%%', ?, '%%') 
                                AND IFNULL(description, '') LIKE CONCAT('%%', ?, '%%')
                                AND id IN (%s)", $clause));

$thearray = array($name, $description);
$merge    = array_merge($thearray, $params); // output: "John", "Cool guy!", 1, 2, 3, 4

// We need  to pass variables instead of values by reference
// So we need a function to that
call_user_func_array('mysqli_stmt_bind_param', array_merge (array($query, $types), $this->makeValuesReferenced($merge))); 

And the function makeValuesreferenced:

public function makeValuesReferenced($arr){
    $refs = array();
    foreach($arr as $key => $value)
        $refs[$key] = &$arr[$key];
    return $refs;
}

Links for getting this 'know-how': https://bugs.php.net/bug.php?id=49946, PHP append one array to another (not array_push or +), [PHP]: Error -> Too few arguments in sprintf();, http://no2.php.net/manual/en/mysqli-stmt.bind-param.php#89171, Pass by reference problem with PHP 5.3.1