I am trying to pull information from my database, using an option list and choosing the row to the according selection.
table: sales
columns: year, q1, q2, q3, q4
values: 2011, 127.24, 106.54, 88.04, 120.89
When trying to return a $result query from my database. It returns two warnings:
mysql_query() expects parameter 1 to be string
mysql_error() expects parameter 1 to be resource
This is my code:
<?php
$mysqli = mysqli_connect("localhost","admin","admin","testDB");
$year= $_POST['year'];
$taryear= filter_input(INPUT_POST, 'year');
$yearsql= "SELECT year FROM sales WHERE year = '".$taryear."'";
$result= mysql_query($mysqli, $yearsql) or die(mysql_error($mysqli));
?>
<html>
<body>
<form action = "salespie.php" method = "POST">
<select name = "year">
<option value = "2011">2011</option>
<option value = "2012">2012</option>
<option value = "2013">2013</option>
</select>
<input type = "submit" value = "Submit">
</form>
</body>
</html>
I have tried to echo the $year itself, and it returns 2011.
I also echo'd $yearsql with the result of: SELECT year FROM sales WHERE year = '2011'
I don't quite understand what i'm doing wrong to get those two warnings.
Any guidance / help would be greatly appreciated!
Thanks,
As Fred mentioned in the comment, you're mixing mysql_* and mysqli_* functions.
They are not the same thing.
Given that you're passing your connection link, you probably mean to be using mysqli_query instead.
mysqli_connect uses the new object oriented mysql classes and returns a new object.
The query should work if you do
$result = $mysqli->query($yearsql);
if(!$mysqli->errno) //Check if an error occured * fixed small typo with an extra o {
echo 'Error: '.$mysqli->error;
}
The problem is you are using mysqli objects with old(deprecated) mysql methods.
There are mysqli procedural functions for the mysqli to work. Where you use mysql_query or mysql_error replace those with mysqli_ equivalents.