PHP mySQL Prepared Statements Error [duplicate]

Question

Possible Duplicate:
Call to a member function bind_param() on a non-object

I'm getting the following error:

Call to a member function bind_param() on a non-object

Here's the prepared statement:

include('/path/to/connection/variable.php');

This file is verified to be working it's just creates an instance of the mysqli class, example:

$mysqli = new mysqli("localhost", "user", "password", "db");

So I know that's not the issue...

$stmt = $mysqli->prepare("INSERT INTO `users` VALUES (?,?,?,?,?,?,?)"); $stmt->bind_param("sssssss", $firstname, $lastname, $email, $subscribed, $signup_date, $unsubscribe_date, $signup_source); $stmt->execute(); $stmt->close(); $mysqli->close();

Variable types are as follows:

$firstname = string $lastname = string $email = string $subscribed = char (Y or N) $signup_date = DATE - date('Y-m-d') $unsubscribe_date = DATE - 0000-00-00 Entered Initially $signup_source = string

I've tried to find all the usual suspects, checked the connection, basically I wrote a separate SELECT statement and it works. "USERS" is a valid table. Permissions for the connection are root, so that's not the issue. I've switched the types for dates between "s" and "d", and even tried everything with dummy variables - no difference.

I'm hoping it's something simple - because I've been racking my brain for the past hour now, and I can't see anything wrong with the statement above.

Yea - I'm receiving FALSE on var_dump($stmt), which is what's confusing me, because the statement has the correct syntax, it's a correct table, and the field count is correct, etc. — William Orazi
– William Orazi, Commented Sep 4, 2012 at 16:20
Edit: I found the issue - I was missing a field on the INSERT - damn I hate when it's something so simple.... On a related since I'm not the best with mySQL is there a way I can target INSERT for only a couple fields - i.e. exclude trailing fields in the statement? — William Orazi
– William Orazi, Commented Sep 4, 2012 at 16:23

Marc B · Accepted Answer · 2012-09-04 16:16:58Z

4

That usually means your ->prepare() call has failed, and returned a boolean FALSE instead of statement object. Check for that:

$stmt = $mysqli->prepare("INSERT INTO `users` VALUES (?,?,?,?,?,?,?)") or die($mysqli::error);

answered Sep 4, 2012 at 16:16

Marc B

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3 Comments

William Orazi Over a year ago

It is - I did a var_dump($stmt) and it is returning FALSE, so I'm at a loss because the statement appears correct.

Marc B Over a year ago

Check $mysqli::error for the reason why.

William Orazi Over a year ago

I found the issue - I was missing a field on the INSERT - damn I hate when it's something so simple....

DaOgre · Accepted Answer · 2012-09-04 16:23:19Z

1

In response to your question you can do the following:

 INSERT INTO tableName (column1, column2, column3,...) VALUES (value1, value2, value3,...)

answered Sep 4, 2012 at 16:23

DaOgre

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2 Comments

William Orazi Over a year ago

There looks to be 7 to me, the first "variable" holds the field types.

William Orazi Over a year ago

Ah..thank you sir - I knew there was a way!

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