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

Question

I am trying to bind a variable in this prepared statement, but i keep receiving the error:

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

The function is called, and variables are passed to it. When i change the function to just echo the variable, the variable prints on the page fine, but if i try to bind it here i receive the error. can anyone help?

//CALL FROM PAGE ONE check($username); //FUNCTION ON PAGE 2 function check($username){ $DBH = getDBH(); $qSelect = $DBH->prepare("SELECT * FROM users WHERE username = ?"); $qSelect->bind_param("s", $username); }

i know the function is not completely written here, but that shouldn't be a problem. I don't understand why i am receiving this error.

var_dump($qSelect) to check if it is what you think it is. — Mchl
– Mchl, Commented Dec 20, 2010 at 8:43
Some $stmt before this need close(). Like $stmt.close(); or $qSelect.close(); PS. var_dump($qSelect); write nothing interesting, only true mysqli connection — Peter Mlich
– Peter Mlich, Commented Sep 10 at 7:59

Ivar · Accepted Answer · 2018-07-31 07:30:16Z

73

Well, one reason prepare() can fail is if the sql statement sent to it is not valid in the current DB.

prepare() will then return false.

Eg - if the table name is not correct or one or more field in the query does not exist.

edited Jul 31, 2018 at 7:30

Ivar

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answered Jun 14, 2011 at 8:58

Stefan

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1 Comment

user273824 Jan 7 at 20:17

Thank you. This helped me after a long bout of trying to figure out my error. As usual, just a little nothing error causes the most trouble.

Glenn Dayton · Accepted Answer · 2012-04-02 04:06:48Z

as the error-message says, $qSelect seems to be not an object. try to debug this by using var_dump($qSelect); right after your prepare-call. also check if getDBH() returns what you need.

sounds like the prepare-call fails (don't know why) and so it returns false - false is not an object, so you can't call bind_param() on that.

EDIT: you havn't given the info, but it looks like you're using PHP's PDO. In that case, take a look at the documentation.

If the database server successfully prepares the statement, PDO::prepare() returns a PDOStatement object. If the database server cannot successfully prepare the statement, PDO::prepare() returns FALSE or emits PDOException (depending on error handling).

You should configure your server to return those PDO-Exceptions, which would tell you why the prepare call fails.

yeah its qSelect is returning bool(false) but this doesnt make any sense because i have a function that is practically the same and it works fine.
Hm. I was also in the situation of having a function that is practically the same. Turns out, I had copied-and-pasted it in order to add a field. But I forgot the comma before the new field! Stupid mistake, but it just goes to show that it might just be a typing error on your part.

Tisho · Accepted Answer · 2012-07-18 12:46:53Z

i'm using the mysqli approach as well and got the same error when I created another instance of mysqli before closing the first instance. So its important to use close() before starting the same piece of code. For example:

$DBH = getDBH(); $qSelect = $DBH->prepare("SELECT * FROM users WHERE username = ?"); $qSelect->bind_param("s", $username); $qSelect->close(); // <--- use close before calling the same function( wich contains $DBH code) again;

and if he uses such function many times, i think, he should use reset() instead of close(), and move prepare() outside of funcion, for example, into constructor, an put statement into property like this: $this->qSelect .

PHP Hater · Accepted Answer · 2014-04-13 02:37:56Z

It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.

To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:

`user`, `order`, `...`

It's correct, yet, I find it silly from prepare to have this behavior.

David Kariuki · Accepted Answer · 2019-11-15 10:49:08Z

I am trying to help other people with little experience in PHP like me.

In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.

When I fixed the SQL, the error was gone.

I found I was missing the column in the DB that I was referencing in the query.

itsites · Accepted Answer · 2016-05-30 19:51:20Z

Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.

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