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

Question

So I'm trying to create a prepared insert statement to a database for a web app. For the registration system, I'm feeding in values using post, then redirecting to a webpage to do the processing. However, I keep getting the error 'Call to a member function bindValue() on a non-object in /path/to/file.php on line 18, the first reference to bindValue(). The code is:

<?php

 session_start();

 require "database.php";
 $db = new Database("bills.db");

 $admin_password = $_POST['admin_password'];
 $admin_email = $_POST['admin'];
 $salt = sha1(time());
 $group_name = $_POST['group_name'];
 $group_password = $_POST['password'];

 $admin_hash = sha1($salt."--".$admin_password);
 $group_hash = sha1($salt."--".$group_password);

$stmt = $db->prepare("INSERT INTO billgroup VALUES (NULL, :adminemail, :adminpassword_hash, :groupname, :password_hash, :salt)");
$stmt->bindValue(':adminemail', $admin_email, SQLITE3_TEXT);
$stmt->bindValue(':adminpassword_hash', $admin_hash, SQLITE3_TEXT);
$stmt->bindValue(':groupname', $group_name, SQLITE3_TEXT);
$stmt->bindValue(':password_hash', $group_hash, SQLITE3_TEXT);
$stmt->bindValue(':salt', $salt, SQLITE3_TEXT);
$results = $stmt->execute();




 $stmt = $db->prepare("SELECT * FROM billgroup WHERE adminemail = :adminemail");
 $stmt->bindValue(':adminemail', $admin_email, SQLITE3_TEXT);
 $users = $stmt->execute();
 $user = $users->fetchArray();
...

Yes, it takes in the value of the textbox on the previous page from post — sjwarner, Mar 11 '15 at 12:15
Yes, just checked it, correct names and correct action on the form — sjwarner, Mar 11 '15 at 12:16
Do your column counts match up? Six columns in your table? Not specifying column names is considered bad practice. — Jay Blanchard, Mar 11 '15 at 12:20
Another thing to note, is that when I comment out line 18, and put a NULL value in the INSERT statement, it then errors on line 19. I can only assume all of the bindValue() statements are broken? — sjwarner, Mar 11 '15 at 12:41
You have ***three*** columns in the tables but you're trying to insert into ***six*** columns `(NULL, :adminemail, :adminpassword_hash, :groupname, :password_hash, :salt)`. That's causing the problem. — Jay Blanchard, Mar 11 '15 at 12:52

Adrian Cid Almaguer · Accepted Answer · 2015-03-13T19:21:53.227

0

You want connect to a sqlite database but you have:

$db = new Database("bills.db");

You must know that you must have something like this:

$db = new PDO('sqlite:bills.db');

And change in all the bindValue calls the SQLITE3_TEXT to PDO::PARAM_STR like this:

$stmt->bindValue(':adminemail', $admin_email, SQLITE3_TEXT);

to:

$stmt->bindValue(':adminemail', $admin_email, PDO::PARAM_STR);

edited Mar 13 '15 at 19:21

answered Mar 13 '15 at 19:12

Adrian Cid Almaguer

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Maybe `Database` extends `PDO`... Or wraps around it. Or around an SQLite class. Who knows what's inside `Database`...!? – Rudie Mar 13 '15 at 19:15
@Rudie Can be, but the problem is because $db is not a PDO object. – Adrian Cid Almaguer Mar 13 '15 at 19:21
How do you know? If `Database extends PDO` it's a PDO object. You're probably right though. – Rudie Mar 13 '15 at 19:56

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

1 Answers1