Reputation: 465
What are parameters in php?
I have looked online, and there are many different definitions, which is confusing for a newbie like myself.
The reason i ask is because i have the following error:
Warning: mysqli_query() expects at least 2 parameters, 1 given in /Applications/MAMP/htdocs/PhpProject2/Area_Rest_page.php on line 22
Based on this code:
$sql=mysqli_query("SELECT Rest_Details.Resturant_ID,
Rest_Details.Resturant_name, Rest_Details.Res_Address_Line_1, Rest_Details.City_name,
Rest_Details.Avg_Del,Delivery_Pcode.Pcode
FROM Rest_Details
INNER JOIN Delivery_Pcode
ON Delivery_Pcode.Restaurant_ID=Rest_Details.Restaurant_ID
WHERE Delivery_Pcode.Pcode LIKE '%$searchq'") or die ("could not search!");
Line 22 is:
WHERE Delivery_Pcode.Pcode LIKE '%$searchq'") or die ("could not search!");
Upvotes: 1
Views: 63
Reputation: 59511
It means that mysqli_query
needs 2 variables passed to it. The one is the sql query, which you have provided, but it also needs to know to which mysql connection to query against.
Have a read on the official docs for mysqli_query, as well as function arguments.
Information may be passed to functions via the argument list, which is a comma-delimited list of expressions. The arguments are evaluated from left to right.
Here's an example using the function you're trying to use:
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* Create table doesn't return a resultset */
if (mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City") === TRUE) {
printf("Table myCity successfully created.\n");
}
In the code above, we're first connecting to the database with mysqli_connect
, and we save the return value of that function to the $link
variable.
Then, to run our query with mysqli_query
we pass in the connection variable ($link
) as well as the query.
Upvotes: 1