pre-pared statement query not returning any results

Question

I need to send some database row id's to another page, then use them id's to load the rows and process. I have used a session variable which works fine.

$_SESSION['tmp'] = "50,51,52";

    if ($stmt = $mysqli->prepare("SELECT id,jpg WHERE id = ?")) {
        $stmt->bind_param("s",$_SESSION['tmp']);
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($result_id,$result_jpg);
        if ($stmt->num_rows > 0) {
            while($stmt->fetch()) {
                $image = array($result_id => $result_jpg);
                print_r($image."<br />");
            }
        } else {
            echo "Prepare failed: (" . $stmt->errno . ") " . $stmt->error;
        }
    }

The query would be

SELECT id,jpg WHERE id = 50,51,52

which should return all of them rows, but instead nothing is being displayed at all, no errors or anything. Any ideas?

######Edit#####

Updated my code:

//Generate the query
    $query = "SELECT id,jpg FROM images WHERE id IN (?";
    for ($i=1; count($_SESSION['tmp']) > $i; $i++) {
        $query = $query.",?"; 
    } $query = $query.")";

    if ($stmt = $mysqli->prepare($query)) {
        for ($i=1; count($_SESSION['tmp']) > $i; $i++) {
            $b = $_SESSION['tmp'][$i]-1;
            $stmt->bind_param("s",$b);
        }
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($result_id,$result_jpg);
        if ($stmt->num_rows > 0) {
            while($stmt->fetch()) {
                $image = array($result_id => $result_jpg);
                print_r($image);
            }
        } else {
            echo "Prepare failed: (" . $stmt->errno . ") " . $stmt->error;
        }
    }

Can't seem to do the looping of

bind_param("s",50);

as getting the error:

Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement in C:\\xampp\\htdocs\\ppa\\add_sales.php on line 39

#####Edit 2#####

Changed the way im going about this, this works fine.

$image = array();
    foreach($_SESSION['tmp'] as $x) {   
        if ($stmt = $mysqli->prepare("SELECT id,jpg FROM images WHERE id = ?")) {
            $stmt->bind_param("s",$x);
            $stmt->execute();
            $stmt->store_result();
            $stmt->bind_result($result_id,$result_jpg);
            if ($stmt->num_rows > 0) {
                while($stmt->fetch()) {
                    $image[$result_id] = $result_jpg;
                }
            } else {
                echo "Prepare failed: (" . $stmt->errno . ") " . $stmt->error;
            }
        }
    }
    print_r($image);

Answer 1

1. You have missed FROM clause
2. If you checked $mysqli->error you would have known it by yourself
3. WHERE id = 50,51,52 is not a valid mysql syntax
4. You need to use IN operator instead
5. With prepared statements it will be evaluated as WHERE id = '50,51,52' . That's it - an id column is compared with a single string.

Answer 2

SELECT id,jpg WHERE id = 50,51,52

Its wrong statement.

1) You can't write params in where so. Use where id = 50 or id = 51 or id = 52

2) you missed FROM: FROM Table

This is correct:

SELECT id,jpg FROM table WHERE id = 50 or id = 51 or id = 52

Answer 3

您可以尝试以下操作： SELECT id,jpg FROM tablename WHERE id IN (50,51,52);

Answer 4

You need to bind them all separately:

WHERE id IN (?, ?, ?)

And bind:

$stmt->bind_param("i", 50);
$stmt->bind_param("i", 51);
$stmt->bind_param("i", 52);

You will want to use explode() and a loop to do it properly. You will also need to use implode to generate the right number of question marks:

$questionMarks = '(' . implode(',', array_fill(0, count($resultOfSplit), '?')) . ')';