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 $DBH code) again;
It appears that prepare is quite dumb. It doesn't rely query entirely into the MySQL side, by this, I mean, if in your query, you have a table that happens to have the same name of a keyword, say "user", "order", ..., it just doesn't recognize it as a table, but rather as what the keyword commands actually do, so the query turns out to be a mess and the prepare just fail.
To fix this is simple, you have to type it in the "correct" way adding "`" in both sides of the table name. Example:
`user`, `order`, `...`
It's correct, yet, I find it silly from prepare to have this behavior.
I am trying to help other people with little experience in PHP like me.
In my case, this error occurred because I had an SQL syntax error. The console stack trace did not show the problem.
When I fixed the SQL, the error was gone.
Check the permissions of the user in database. User without "insert" permission causes "Call to a member function bind_param() on a non-object" message error too, when trying to insert.
This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 6 years ago.
I am trying to bind a variable in this prepared statement, but i keep receiving the error:
Call to a member function bind_param() on a non-object
The function is called, and variables are passed to it. When i change the function to just echo the variable, the variable prints on the page fine, but if i try to bind it here i receive the error. can anyone help?
//CALL FROM PAGE ONE
check($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.