bind_param number arguments not equal to variables - php

Question

I am having an error in mysql and php My code:
$stmt->prepare("INSERT `tablename` (col1,col2,col3) VALUES (?,?,?)");
$stmt->bind_param('sss',$_POST['data'],$sub,$yaxis);
$stmt->execute();
My errors:
Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement.
But I have passed 3s and 3 arguments in bind_param. Why is this error occuring?

Answer 1

use the index of parameter since you did not specify it by name.
$data = $_POST['data'];
$stmt->prepare("INSERT `tablename` (col1,col2,col3) VALUES (?,?,?)");
$stmt->bind_param(1,$data);
$stmt->bind_param(2,$sub);
$stmt->bind_param(3,$yaxis);
$stmt->execute();
but if you want it by name then you have to specify its parameter name instead of using question mark
$data = $_POST['data'];
$stmt->prepare("INSERT `tablename` (col1,col2,col3) VALUES (:sss,:par1,:par2)");
$stmt->bind_param(":sss",$data);
$stmt->bind_param(":par1",$sub);
$stmt->bind_param(":par2",$yaxis);
$stmt->execute();

Answer 2

Are you sure you have value in your $_POST['data'] , Instead of directly passing form values to bind_param you can check if it is null or not.Also i noticed your insert query is wrong
if(!empty($_POST['data'])){
$mydata=$_POST['data'];
}else{
$mydata='No values';
}
$stmt->prepare("INSERT into `tablename` (col1,col2,col3) VALUES (?,?,?)");//Query wrong
$stmt->bind_param('sss',$mydata,$sub,$yaxis);
$stmt->execute();

Answer 3

Well, either $_POST has more than one variable assigned to it, or it has none.
This will print the contents of the entire $_POST array:
print_r($_POST);

Answer 4

Okay, I know this makes me sound stupid but the error was caused by an undefined variable farther up in the script, which I have now fixed. I thought it would not affect this part of the script, but it did. All the answers posted are great, but because of my fault they do not answer the question.