Trouble with inserting integer into MySQL Database using PHP

Question

Im having trouble inserting an integer in my DB the string variable work just fine, but the integer is getting set to 0. Here is my code.

if ($_SERVER['REQUEST_METHOD'] == 'POST' && !empty($_POST['submit'])) {
    $empId = $conn-> escape_string($_POST['empID']);
    $empIdInt = (int)$empId;
    $name_esc = $conn-> escape_string($_POST['name']);
    $username_esc = $conn-> escape_string($_POST['username']);
    $password_esc = $conn-> escape_string($_POST['password']);
    $confirm_password_esc = $conn-> escape_string($_POST['confirm_password']);
    $empAddress_esc = $conn-> escape_string($_POST['empAddress']);

    $query = "
    INSERT INTO users (empId,username,password,securityLevel,name,empAddress) 
    VALUES 
    ('$empIdInt','$username_esc','$password_esc','user','$name_esc','$empAddress_esc')";

    if ($password_esc == $confirm_password_esc)
    {
        $result = $conn->query($query) or die(mysqli_error($conn)); 
    } else {
        echo("Passwords do not match.");
    }
}

<label>Employee ID: </label><input type = "text" name="empId"  /><br><br>

Which column? What is the structure of that column? Can you give an example of data? — El_Vanja, Apr 09 '20 at 17:15
You're also wide open to [SQL injection](https://stackoverflow.com/questions/332365/how-does-the-sql-injection-from-the-bobby-tables-xkcd-comic-work). You should use [prepared statements](https://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php) instead. — El_Vanja, Apr 09 '20 at 17:16
thanks for the prepared statement suggestion i will look into that. My DB Structure is 6 columns like int,varchar,varchar,varchar,varchar,varchar — CallMeAaron, Apr 09 '20 at 17:29
Yeah so basically the issues is the number i enter get changed into a 0 so the initial entry validates all fields but the id becomes a 0. when i go to enter another employee i use empId = 2 i get an error duplicate entry 0 for primary key. — CallMeAaron, Apr 09 '20 at 17:52
I found my mistake in the POST empID vs empId. thanks for your time. — CallMeAaron, Apr 09 '20 at 17:57

score -1 · Accepted Answer · answered Apr 09 '20 at 17:48

-1

If i see correctly, you are trying to insert a "string" into an "integer" column. see here: '$empIdInt' => the ' .. ' makes it a string. drop them and give it a go.

answered Apr 09 '20 at 17:48

Kosem

373
2
11

thanks for the suggestion. I thought that might be an issue as well, but it did not make a difference. – CallMeAaron Apr 09 '20 at 17:52
yet you should not enter a string to a int column. even if it didn't solve your problem, i advise to fix that :) – Kosem Apr 09 '20 at 17:54
in addition, please provide output of $_POST. do: print_r($_POST); and show the array. You might not be getting the value of the integer at all (and maybe that's why it inserts as zero) – Kosem Apr 09 '20 at 17:55
1

Yes, Thanks. I did make that change and found the silly mistake i made. In my post i was have empID instead of empId. Thanks for your time. – CallMeAaron Apr 09 '20 at 17:56

Trouble with inserting integer into MySQL Database using PHP

1 Answers1