Fatal error: Call to a member function bind_param() on boolean using temporary table in PDO

Question

so I have 2 tables: admin and employee with only 2 columns they share in common

one login form

and 2 separate dashboards for each

to make them in one table temporarily I've incorporated them with UNION as well as adding a temporary column as well to define the type of user and hence logging in to the right dashboard.

so for making this happen after a very long searching and quite many trail and error I've concluded that this is the best I can do and understand as a newbie in this field.

that's how my final piece looks like:

    // preparing a temporary table that unions the admin and employee tables
$TMP = "SELECT EMP_ID AS ID, EMP_EMAIL AS EMAIL, 2 AS TYPE FROM employee
                        UNION
                        SELECT ID, EMAIL, 1 AS TYPE FROM admin ";
$result = $con->prepare($TMP);                      
$result-> execute();
$result-> store_result();
$result-> fetch();

    // preparing select statement for logging in
$stmt = $con->prepare('SELECT `TYPE` FROM `".$result."` WHERE `ID` = ?');
    // Bind parameters (s = string, i = int, b = blob, etc), in our case the username is a string so we use "s"
    $stmt->bind_param('i', $_POST['ID']);
    $stmt->execute();
    // Store the result so we can check if the account exists in the database.
    $stmt->store_result();
    if ($stmt->num_rows > 0) {
    $stmt->bind_result($ID);
    $stmt->fetch();
    }
        if ($stmt['TYPE'] == 1) {
            if ($_POST['ID'] == $ID) {
        // Verification success! User has loggedin!
        // Create sessions so we know the user is logged in, they basically act like cookies but remember the data on the server.
            session_regenerate_id();
            $_SESSION['loggedin'] = TRUE;
            $_SESSION['email'] = $_POST['EMAIL'];
            $_SESSION['id'] = $ID;
            echo 'Welcome ' . $_SESSION['email'] . '!';
        } else {
        echo 'Incorrect password!';
    }
        } else if ($stmt['TYPE'] == 2) {
    if ($_POST['ID'] == $ID) {
        // Verification success! User has loggedin!
        // Create sessions so we know the user is logged in, they basically act like cookies but remember the data on the server.
        session_regenerate_id();
        $_SESSION['loggedin'] = TRUE;
        $_SESSION['email'] = $_POST['EMAIL'];
        $_SESSION['id'] = $ID;
        echo 'Welcome ' . $_SESSION['email'] . '!';
    } else {
        echo 'Incorrect password!';
    }
        }
    $stmt-> close();



$result-> close();
$con-> close();

the code used was from a tutorial meant for logging one user and it was working just fine with me but because my goal was different I tried to modify it for my needs but unfortunately fatal errors about binding the parameter keeps popping up.

I really appreciate giving me some of your time. thank you.

Your concatenation is wrong. You have single quotes on the full string, but try to break out of it with double quotes. Thus, your prepare statement failed. — aynber, Mar 20 '19 at 18:24
@aynber you're right the error has gone but I received another error (Catchable fatal error: Object of class mysqli_stmt could not be converted to string ) do you think this is has something to do with the temporary table's name? thanks for your help. — Sarah Fahad, Mar 21 '19 at 05:37
You call `$stmt->fetch()` but never assign it to a variable, then attempt to use the mysql_stmt `$stmt` as an array. That won't work. Do `$row = $stmt->fetch()` so that `$row` will contain the results of the query, which you can then use as an array `$row['TYPE']` — aynber, Mar 21 '19 at 12:27

Benjie Wheeler · Answer 1 · 2019-03-21T13:52:47.323

3

The problem is in the outer quotes ' vs "

When the prepared statement fails it will return a boolean (false).

Documentation

mysqli_prepare() returns a statement object or FALSE if an error occurred.

I think you meant to make this line like this:

$stmt = $con->prepare("SELECT `TYPE` FROM `".$result."` WHERE `ID` = ?");

edited Mar 21 '19 at 13:52

answered Mar 20 '19 at 18:25

Benjie Wheeler

540
3
16

you're right this issue is solved thank you. but do know whether my table's name is affecting the code in any way or not? I really appreciate your help. – Sarah Fahad Mar 21 '19 at 05:39
Your table name being in the `$result` variable is prone to [SQL Injection](https://en.m.wikipedia.org/wiki/SQL_injection), I would advise you yo [escape](https://secure.php.net/manual/en/mysqli.real-escape-string.php) the variable – Benjie Wheeler Mar 21 '19 at 13:50

Fatal error: Call to a member function bind_param() on boolean using temporary table in PDO

1 Answers1