How to get the right output from this SELECT query? - php

Question

I have a query like so:
SELECT * FROM `purchases` p
JOIN `purchase_types` pt ON p.purchase_type = pt.node
When I run it in PHPmyAdmin it returns the proper result set like so:
node | purchase_type | amount_spent | node | name
--------------------------------------------------
2 | 5 | 8.5000 | 5 | Lunch
3 | 5 | 1.5000 | 5 | Lunch
4 | 6 | 4.6600 | 6 | Dinner
This is my PHP code:
$sql = "SELECT * FROM `purchases` p
JOIN `purchase_types` pt ON p.purchase_type = pt.node";
$query = mysql_query($sql);
$result = mysql_fetch_assoc($query);
$purchases = array();
while($row = mysql_fetch_assoc($query)) {
$purchases[] = array(
'name' => $row['name'],
'amount_spent' => $row['amount_spent']
);
}
for eaching over $expenses returns the following output:
3 | 5 | 1.5000 | 5 | Lunch
4 | 6 | 4.6600 | 6 | Dinner
What happens to the first "Lunch"? How can I have the PHP output be the same as the direct MySQL query output?

Answer 1

You call mysql_fetch_accoc before while. Don't.
You should also realize that ext/mysql will be deprecated and upgrade your code to use properly parameterized queries with PDO or mysqli

Answer 2

try with mysql_fetch_array,
$purchases = array();
while($row = mysql_fetch_array($query)) {
$purchases[] = $row;
}
if you want to get data just do
foreach($purchases as $key => $value)
{
$name = $value['name'];
$amount_spent = $value['amount_spent'];
echo 'name : '.$name.' , amount spent '.$amount_spent.'<br />';
}