MySQLi Un-prepare-d

Question

I am converting my old mysql_* queries over to mysqli except i'm having a frustrating issue

$st = $this->Sql->prepare("INSERT INTO tblPlayerOnline (SteamID,PlayerName,IPAddress,ConnectedDate) VALUES (?, ?, ?, ?)") or die($this->Sql->error);
/* 225 */ $st->bind_param("i", $SteamID);
$st->bind_param("ssi", $PlayerName, $IPAddress, $TimeNow);
$st->execute();

Fatal error: Call to a member function bind_param() on a non-object in /home/vanrust/public_html/rework/class/Players.class.php on line 225

But when I dump $this->Sql I get object(mysqli)#2 (19) { bla bla so not sure what this error is telling me.

Dump of $this->Sql http://pastebin.com/gdKAgT4D

Any guidance is appreciated!

PS Looked Everywhere -.-

fyi, you don't need multiple invocations of `bind_param`, it already supports multiple parameters for you to put into it. add an `or die($this->Sql->error)` in the prepare line — Kevin, Apr 22 '15 at 11:22
hello, I'd show you a git revision of me doing that originally but its a private repo lol :(( error is empty, errno is 0 — zanderwar, Apr 22 '15 at 11:28
hm, `Open tables: 400`, is this a normal number or is it a max level of sorts depicting my large lack of `$st->close`'s that I have around — zanderwar, Apr 22 '15 at 11:43
`$st` is most probably `false`, which means the statement could not be prepared. It is likely that you have a typo in your SQL query, maybe the name of the table or a column is wrong. — Gerald Schneider, Apr 22 '15 at 12:00

score 1 · Answer 1 · answered Apr 22 '15 at 12:16

When the prepare statement fails and no mysqli_prepare object is returned, there's a big chance you have an error inside your query.. or something worse.

For debugging purposes you can use the mysqli_report(MYSQLI_REPORT_ALL) function (which is an alias for mysqli_driver->report_mode). This can give you some more information why the prepare is failing and be helpfull for debugging.

Place the following in the beginning of your code

mysqli_report(MYSQLI_REPORT_ALL);

//Or if you just want an exception when Mysqli has an error? 
//Set the reporting level to 
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

Next to that I think it's very risky to always assume that the mysql_prepare object is returned. You should first check before using it:

if ($stmt = $this->Sql->prepare(....)) {
    //Yep, we can now start binding parameters
} else {
    //Noooooo.. got error?
}

zanderwar · Accepted Answer · 2019-06-01T11:24:20.373

0

Ok so this frustrated me for awhile but I found the answer was in front of me all along right there in $this->Sql->stat - Oddly, 400 open tables seemed like a rather nice even number however this was clearly a maximum.

I tracked back through my code and began adding $st->close() to everything and this fixed this problem for me.

So lesson learned; close it once you're done with it

Thanks to all for helping

edited Jun 01 '19 at 11:24

answered Apr 22 '15 at 12:19

zanderwar

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score -3 · Answer 3 · answered Apr 22 '15 at 11:23

-3

Looks like $st is null, meaning that the statement failed to prepare.

Try to see what's causing it with this:

if (!$st) {
     echo "Prepare failed: (" . $this->Sql->errno . ") " . $this->Sql->error;
}

answered Apr 22 '15 at 11:23

Frank

664
5
15

Hello, thank you for your answer, the response I get is `Prepare failed: (0) ` – zanderwar Apr 22 '15 at 11:26

MySQLi Un-prepare-d

3 Answers3