Errors in prepared statement - Undefined variable: sql and Fatal error: Call to a member function bind_param()

Question

I am getting 2 errors in my code:

Notice: Undefined variable: sql
AND
Fatal error: Call to a member function bind_param() on a non-object

My code is:

//if there is a plant name
if (isset($_POST['plant_name']) && $_POST['plant_name']) { 
$where .= "AND (common_name) LIKE ? OR (latin_name) LIKE ?";
}
$stmt = $conn2->prepare($sql . $where);
if (isset($_POST['plant_name']) && $_POST['plant_name']) { 
$stmt->bind_param('s', strtolower($_POST['plant_name']));
$stmt->bind_param('s', strtolower($_POST['plant_name'])."%");
}

//execute query
$stmt->execute();

// get the roses and do the query!
$sql = "SELECT * FROM rosename";

//do we have a 'where string' to add to this query
if ($where) {
$query .= $where;
}

$sql = mysql_query($query, $conn2);

I am basically trying to get someone to type in a plant in the plant_name field and then see if it is like any values from latin_name and common_name attributes in the DB.

Could somebody please help me out.

$sql = "SELECT *.... I presume this should read $query = "SELECT *.... Also put a space before the AND so that you don't end up with rosenameAND — bcoughlan
– bcoughlan, Commented Mar 16, 2012 at 20:36

Community · Accepted Answer · 2023-11-17 20:38:18Z

2

The $stmt object is probably set to false, due to the fact the query being prepared was erroneous.

Try this:

$stmt = $conn2->prepare($sql . $where);

if (false === $stmt) {
    var_dump($conn2->errorCode();
}

You can read up on error code and prepare via the PHP docs.

ALSO, you need to move the $sql and $where initializations above the $stmt code block. That's why you are getting the undefined error.

edited Nov 17, 2023 at 20:38

CommunityBot

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answered Mar 16, 2012 at 20:38

Mike Purcell

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Comments

Y0lk · Accepted Answer · 2013-07-26 15:21:33Z

0

You seem to be preparing a statement using a variable $sql that is undefined, that's what the notice error is saying. In your code we can't see any variables named $sql defined before the prepare statement where you try to use that variable.

answered Jul 26, 2013 at 15:21

Y0lk

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