函数getResult调用
I'm working transferring my functions from MySQL to MySQLi standards.
This is my old function for getresult
function getResult($sql) {
$result = mysql_query($sql, $this->conn);
if ($result) {
return $result;
} else {
die("SQL Retrieve Error: " . mysql_error());
}
}
However, I have been following the W3schools on the mysqli_query function. Here's where I'm presently at.
function getResult($connection){
$result = mysqli_query($connection->conn);
if ($result) {
return $result;
} else {
die("SQL Retrieve Error: " . mysqli_error());
}
}
Now, the examples on W3schools are just a bit different from how I want to use mysqli_query. I'm trying to make it be directed at my DB, not some input or constants as per examples on W3S.
Notice: Trying to get property of non-object in C:\xampp\htdocs\cad\func.php on line 92
Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\xampp\htdocs\cad\func.php on line 92
Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\xampp\htdocs\cad\func.php on line 96
SQL Retrieve Error:
Thank you
Your version won't work for a number of reasons, not the least of which is that you haven't included the query you want to send. Assuming $this->conn is set up properly somehow, try this:
function getResult($sql) {
$result = mysqli_query($this->conn, $sql); //Note parameters reversed here.
if ($result) {
return $result;
} else {
die("SQL Retrieve Error: " . mysql_error());
}
}
From the PHP help files I found these function definitions
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
string mysqli_error ( mysqli $link )
mysqli_query() takes the link and the actual query as mandatory parameters, and mysqli_error takes the link as one as well.