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我有一個查詢,像這樣:如何從此SELECT查詢中獲得正確的輸出?
SELECT * FROM `purchases` p
JOIN `purchase_types` pt ON p.purchase_type = pt.node
當我在phpMyAdmin運行它,它返回正確的結果集,像這樣:
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
這是我的PHP代碼:
$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' => $row['amount_spent']
);
}
每$expenses
返回以下輸出:
3 | 5 | 1.5000 | 5 | Lunch
4 | 6 | 4.6600 | 6 | Dinner
第一次「午餐」會發生什麼?我怎樣才能讓PHP輸出和直接的MySQL查詢輸出一樣?