CODEWITHSUNDEEP
phpmysqlvariablespdoprepared-statement
sldk
sldk

Reputation: 374

PDO - Prepared statement doesn't handle variable right

I have a variable ($q=$_GET["q"];) that I want to run my prepared statement with. It contains a column name of my database table.

The value of the variable comes from a dropdown list. When I run the following code the output is the exact value that was chosen in the list. So it is just a column name.

$q=$_GET["q"];

$dsn = "mysql:host=$host;port=$port;dbname=$database";

$db = new PDO($dsn, $username, $password);

$sql = "SELECT DISTINCT ? FROM repertoire";
$stmt = $db->prepare($sql);
$stmt->execute(array($q));

    echo "<select>";
    while ( $row = $stmt->fetchObject() ) {
        echo "<option>";
        echo "{$row->{$q}}";
        echo "</option>";
        }
    echo "</select>";

However, When I change this line $sql = "SELECT DISTINCT ? FROM repertoire";

to $sql = "SELECT DISTINCT ".$q." FROM repertoire"; the I get the desired rows from the database...

I'm not so good with PHP, so I guess that my syntax is wrong somewhere.

Thank you in advance for your help.

Upvotes: 0

Views: 1312

Answers (1)

Stegrex
Stegrex

Reputation: 4024

In PDO, prepared statements prepare the values, not the tables.

You'll need to handle the user input and quote directly.

$q=$_GET["q"];

// Make sure you sanitize your user inputs using filter_inputs() or similar.

$dsn = "mysql:host=$host;port=$port;dbname=$database";

$colNames = new PDO($dsn, $username, $password); // Create object for getting column names.
$sql = "DESC repertoire Field"; // SQL for getting column names.
$stmt = $colNames->prepare($sql);
$stmt->execute();

$colList = $stmt->fetchAll(PDO::FETCH_COLUMN, 0); // Fetch the results into $colList array.

if (in_array($q, $colList)) { // If the value of $q is inside array $colList, then run.

    $db = new PDO($dsn, $username, $password);

    $sql = "SELECT DISTINCT $q FROM repertoire";
    $stmt = $db->prepare($sql);
    $stmt->execute(array($q));

        echo "<select>";
        while ( $row = $stmt->fetchObject() ) {
            echo "<option>";
            echo "{$row->{$q}}";
            echo "</option>";
            }
        echo "</select>";
}

Also read: Can PHP PDO Statements accept the table or column name as parameter?

Edit: I've added a way to check to make sure $q is a valid column by basically doing a SQL desc in order to get all the column names out of table repertoire.

Upvotes: 2

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