Prepared Statement not returning anything

Question

I know this particular query works, as I tested it with unprepared, procedural methods. Here it is:

$name = 'introduction';
$mysqli = new mysqli('localhost', 'user', 'pass', 'db') or die('There was a problem connecting to the database.');
$stmt = $mysqli->prepare("SELECT name, content FROM sections WHERE name = ?");
$stmt->bind_param('s', $name);
$stmt->execute();
$stmt->bind_result($content);
$stmt->fetch();
echo  $content;
$stmt->close();

I realized that, since I have an id column as an index in the sections table, I needed to bind that as a result as well, given the above statement at php.net, (thanks again, Bill).

Here's the new code:

$name = 'introduction';
$mysqli = new mysqli('localhost', 'user', 'pass', 'db') or die('There was a problem connecting to the database.');
$stmt = $mysqli->prepare("SELECT name, content FROM sections WHERE name = ?");
$stmt->bind_param('s', $name);
$stmt->execute();
$stmt->bind_result($id, $name, $content);
$stmt->fetch();
echo $content;
$stmt->close();

Thanks again to all who can offer suggestions. (I'm curious: I find it hard to debug when using the OOP style of prepared statements in this way. Is there, for example, an easy way to simply see the query that was actually used?)

If I do the following, just as a quick-and-dirty example:

$name = 'introduction';
@mysql_connect('host', 'user', 'pass');
@mysql_select_db('db');
$query = "SELECT name,content FROM sections WHERE name = '$name'";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_object($result)) {
    $content = $row->content;
    echo $content;
}

My data appears and all is well. If, however, I do the following:

$name = 'introduction';
$mysqli = new mysqli('localhost', 'user', 'pass', 'db') or die('There was a problem connecting to the database.');
$stmt = $mysqli->prepare("SELECT name, content FROM sections WHERE name = ?");
$stmt->bind_param('s', $name);
$stmt->execute();
$stmt->bind_result($name, $content);
$stmt->fetch();
echo  $content;
$stmt->close();

Which I believe is correct (feel free to yell if not, of course), I get nothing. What's more, with that code, when I do an html validation (just in case), I get an internal server warning (500), which I take to be a problem with the sql code. Am I just nuts?

Answer 1

I don't see anything wrong with your preparation of the statement or use of parameters, but there is something wrong in your binding results:

http://php.net/manual/en/mysqli-stmt.bind-result.php says:

Note that all columns must be bound after mysqli_stmt_execute() and prior to calling mysqli_stmt_fetch() .

(emphasis mine)

---

The above doc should be taken as all columns in your query, not all columns in your table.

Okay, I just tried this myself. If I omit the $name column, it gives this warning:

PHP Warning:  mysqli_stmt::bind_result(): Number of bind variables doesn't 
match number of fields in prepared statement in mysqli.php on line 9
PHP Stack trace:
PHP   1. {main}() /Users/bill/workspace/PHP/mysqli.php:0
PHP   2. mysqli_stmt->bind_result() /Users/bill/workspace/PHP/mysqli.php:9

But it does fetch the data.

If I bind both $name and $content to the results of the query, it works without error or warning.

So I'm forced to ask you: are you sure there's a row in the database that matches your condition? That is, where name = 'introduction' ? Keep in mind that in SQL, string comparisons are case-sensitive by default.

One mistake I see people make frequently is that they connect to a different database in their PHP script than the database they use for ad hoc queries. So you need to be absolutely sure you're verifying that the data exists in the right database.

Answer 2

Shouldn't that be

$stmt->bind_result($name, $content);

As you select 2 columns