Fatal error: Cannot pass parameter 2 by reference when using PHP and MYSQL

Question

I am using PHP PDO to insert into a MYSQL database using PHP. I am getting the error:

Fatal error: Cannot pass parameter 2 by reference in /home/sandyit/public_html/hosting/findibuzz/design2/sign-up.php on line 200

This is my code:

$ID is an auto incremented integer while the rest are varchar variables filled out as below as an example:

$db = new PDO('mysql:host=localhost;dbname=dbname;charset=utf8', 'dbusername', 'dbpass');

$FULLNAME = "David";
$PW_HASH = "sadsad";
$SALT = "adadad";
$EMAIL_ADDRESS = "david@gmail.com";


$ID=0;
$addrequest = $db->prepare("INSERT INTO FB_USERS (ID,FULL_NAME,PASSWORD,PASSWORD_SALT,EMAIL_ADDRESS) VALUES (:ID,:FULL_NAME,:PASSWORD,:PASSWORD_SALT,:EMAIL_ADDRESS)");
$addrequest->bindParam(':ID',$ID, PDO::PARAM_INT);
$addrequest->bindParam(':FULL_NAME',$FULL_NAME, PDO::PARAM_STR);
$addrequest->bindParam(':PASSWORD',$PW_HASH, PDO::PARAM_STR);
$addrequest->bindParam(':PASSWORD_SALT',$SALT, PDO::PARAM_STR);
$addrequest->bindParam(':EMAIL_ADDRESS',$EMAIL_ADDRESS, PDO::PARAM_STR);
$addrequest->execute();
$addrequest->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

I know i have something wrong, but i cannot spot the error, can i have some advise please?

Thanks

Add `$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);` right after the connection is opened. — Funk Forty Niner, Jun 19 '14 at 19:42

score 2 · Answer 1 · answered Jun 19 '14 at 19:21

2

Remove quotation marks from '$ID'

$addrequest->bindParam(':ID',$ID, PDO::PARAM_INT);

answered Jun 19 '14 at 19:21

Unlink

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Not sure but i think he could remove the ':' from ':ID' – Sebastien Jun 19 '14 at 19:23
1

@Sebastien The `:` is optional. – Funk Forty Niner Jun 19 '14 at 19:25
@Fred-ii- Thank! Didn't knew that. – Sebastien Jun 19 '14 at 19:27
@Sebastien You're welcome. TBH, I didn't know that neither till about 4-5 months ago. – Funk Forty Niner Jun 19 '14 at 19:27
That has helped, thanks for pointing out those mistakes, what happens now is that the script runs, there are no errors generated but no entries are entered into the database. – user3657758 Jun 19 '14 at 19:32
I think this is my first column is auto incremented and is a primary key, but i can't figure out what im doing wrong? – user3657758 Jun 19 '14 at 19:32
@user3657758 What `$addrequest->execute();` returns? – Unlink Jun 19 '14 at 19:36
2

If your first colomn is auto increment dont bind a parameter to it. – Sebastien Jun 19 '14 at 19:38
That is part of the problem, i do not get any errors returned. Could this be as my first column is auto incremented and has a primary key and it does not like this value when entering a new entry into the table? – user3657758 Jun 19 '14 at 19:39
@user3657758 it's all right to send 0 for this column – Your Common Sense Jun 19 '14 at 19:41
It can have, but mysql automatically replace it – Unlink Jun 19 '14 at 19:42

score 2 · Accepted Answer · edited Jun 19 '14 at 19:40

Just for reference. I know this wont help solve your problem, but you could do something like this (see code below) to achieve the same result:

$db = new PDO('mysql:host=localhost;dbname=dbname;charset=utf8', 'dbusername', 'dbpass');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

$sql = "INSERT INTO FB_USERS (FULL_NAME,PASSWORD,PASSWORD_SALT,EMAIL_ADDRESS)
        VALUES (:FULL_NAME,:PASSWORD,:PASSWORD_SALT,:EMAIL_ADDRESS)";
$stmt = $db->prepare($sql);
$params = array
(
    'FULL_NAME'=>'David',
    'PASSWORD'=>'sadsad',
    'PASSWORD_SALT'=>'adadad',
    'EMAIL_ADDRESS'=>'david@gmail.com'
);
$stmt->execute($params)

I find it easier to work with an array and than to just pass it to the statment. But I guess its just a mather of taste. Like I said this is just for reference and wont help you resolve your issue.

Fatal error: Cannot pass parameter 2 by reference when using PHP and MYSQL

2 Answers2