mysqli bind_param don't work php

Question

I'm learning mysqli_. If I run the query without the bind_param it works, but if I add the bind_param my query stops working.

This is my code:

// Make a connection to database.
$user = 'test';
$sql = "SELECT * FROM `user` WHERE `user` = ?";
$querySelect = $mysqli->prepare($sql);
$querySelect->bind_param('s', $user);
$querySelect->execute();
echo 'N: '.$querySelect->num_rows.'<br>'; // Got 0, but the correct result is 1.

ERROR:

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

Errors, please. "Not working" is not a useful bit of information. You may want to [enable exceptions](http://stackoverflow.com/questions/14578243/turning-query-errors-to-exceptions-in-mysqli) to have errors bubbled up. — tadman, Jul 13 '16 at 17:55
That call isn't shown in this code, so you'll have to track it down. — tadman, Jul 13 '16 at 17:56
You should assign the execute method to a variable and work with this — JRsz, Jul 13 '16 at 17:56
Add `var_dump($querySelect);` before the call to `bind_param`. My guess is that the `prepare` statement is failing. Per the [docs](http://php.net/manual/en/mysqli.prepare.php): _"`mysqli_prepare()` returns a statement object or `FALSE` if an error occurred."_ — War10ck, Jul 13 '16 at 17:59
When you are using PDO you should use `$querySelect->bindParam('s', $user);`. — Hamid, Jul 13 '16 at 18:03

score 2 · Answer 1 · answered Jul 13 '16 at 18:09

I guess this line returns false:

$querySelect = $mysqli->prepare($sql);

try to do:

var_dump($querySelect);

in order to be sure. If return false, that means something wrong with getting data user from database (wrong table, connection, table column, ...)

but this sql is also strange:

$sql = "SELECT * FROM `user` WHERE `user` = ?";

maybe you wanted to write:

$sql = "SELECT * FROM `user` WHERE `user_id` = ?";

so, user_id instead of user or maybe only id, depends on the name of you primary key

mysqli bind_param don't work php

1 Answers1