This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 6 years ago.
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.
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.
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.
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 $DBH code) again;
It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.
To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:
`user`, `order`, `...`
It's correct, yet, I find it silly from prepare to have this behavior.
I am trying to help other people with little experience in PHP like me.
In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.
When I fixed the SQL, the error was gone.
Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.
Related
This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 6 years ago.
I am trying to bind a variable in this prepared statement, but i keep receiving the error:
Call to a member function bind_param() on a non-object
The function is called, and variables are passed to it. When i change the function to just echo the variable, the variable prints on the page fine, but if i try to bind it here i receive the error. can anyone help?
//CALL FROM PAGE ONE
check($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.
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.
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.
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 $DBH code) again;
It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.
To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:
`user`, `order`, `...`
It's correct, yet, I find it silly from prepare to have this behavior.
I am trying to help other people with little experience in PHP like me.
In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.
When I fixed the SQL, the error was gone.
Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.
This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 6 years ago.
I am trying to bind a variable in this prepared statement, but i keep receiving the error:
Call to a member function bind_param() on a non-object
The function is called, and variables are passed to it. When i change the function to just echo the variable, the variable prints on the page fine, but if i try to bind it here i receive the error. can anyone help?
//CALL FROM PAGE ONE
check($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.
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.
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.
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 $DBH code) again;
It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.
To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:
`user`, `order`, `...`
It's correct, yet, I find it silly from prepare to have this behavior.
I am trying to help other people with little experience in PHP like me.
In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.
When I fixed the SQL, the error was gone.
Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.
This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 6 years ago.
I am trying to bind a variable in this prepared statement, but i keep receiving the error:
Call to a member function bind_param() on a non-object
The function is called, and variables are passed to it. When i change the function to just echo the variable, the variable prints on the page fine, but if i try to bind it here i receive the error. can anyone help?
//CALL FROM PAGE ONE
check($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.
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.
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.
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 $DBH code) again;
It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.
To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:
`user`, `order`, `...`
It's correct, yet, I find it silly from prepare to have this behavior.
I am trying to help other people with little experience in PHP like me.
In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.
When I fixed the SQL, the error was gone.
Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.
This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(2 answers)
Closed 6 years ago.
I am trying to bind a variable in this prepared statement, but i keep receiving the error:
Call to a member function bind_param() on a non-object
The function is called, and variables are passed to it. When i change the function to just echo the variable, the variable prints on the page fine, but if i try to bind it here i receive the error. can anyone help?
//CALL FROM PAGE ONE
check($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.
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.
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.
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 $DBH code) again;
It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.
To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:
`user`, `order`, `...`
It's correct, yet, I find it silly from prepare to have this behavior.
I am trying to help other people with little experience in PHP like me.
In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.
When I fixed the SQL, the error was gone.
Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.
This is a terrible question because I don't have a simple way to reproduce it. However, I'm using the Zend Framework to connect to my MySQL database on OS X. Sometimes a call to the prepare function on a mysqli object returns null. The stated return values for the prepare function are false or a statement object.
I can't figure out where else to look for info on why the prepare statement is failing. Is there any way to get visibility into the prepare process to see why it is failing? All of my problems are coming up while a transaction is open.
Sorry for the lack of specifics, but I really can't nail down why this is happening.
Just to correct ToughPal, you should be using:
mysqli_query($db, "INSERT INTO table (variable1, variable2) VALUES (hello, mynameis);
Remember that you need to have the db connection defined and stated in the query first, before your actual SQL.
Remember to enclose the table name, column names and value data in backtick escapes.
Example prepared statement
$result = $db->query( 'INSERT INTO server (key, value) VALUES (:key, :value)',
array('key' => $foo, 'value' => $bar)
Can you let us know your DB query?
Try and execute your DB query with test data and see if the query works fine to start with. If the query is ok then we can look why the code fails.
Well I managed to find the issue over the weekend but was really only able to fix the symptoms and not the cause.
I didn't include any SQL in the original issue because the problem was happening randomly, the same code would sometimes work and sometimes not. The issue looks like it was a memory pointer problem. Whenever I had a problem Zend Debugger told me that I had a mysqli object. I believe this because otherwise I would've gotten an error when trying to run the prepare function on it. I have a singleton object that acts as a container for my mysqli connection but whenever the prepare function failed, === showed that the mysqli being used was not the same as the mysqli connection in my singleton object.
In the end, Zend Framework's only issue is that it doesn't fail if the the prepare function returns null. If you are seeing this problem use === to verify that the connection is actually the same as the one that you've previously initiated.
if you're doing something like this
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
$query = "...";
$mysqli->prepare($query);
then you can inspect mysqli::$error next to see useful errors about why prepare() failed
print_r($mysqli->error);