Php 如何从这个SELECT查询中获得正确的输出?
我有这样一个问题:Php 如何从这个SELECT查询中获得正确的输出?,php,mysql,Php,Mysql,我有这样一个问题: SELECT * FROM `purchases` p JOIN `purchase_types` pt ON p.purchase_type = pt.node 当我在PHPmyAdmin中运行它时,它会返回正确的结果集,如下所示: node | purchase_type | amount_spent | node | name -------------------------------------------------- 2 | 5
SELECT * FROM `purchases` p
JOIN `purchase_types` pt ON p.purchase_type = pt.node
当我在PHPmyAdmin中运行它时,它会返回正确的结果集,如下所示:
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
这是我的PHP代码:
$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' => $row['amount_spent']
);
}
对于$expenses
上的搜索,返回以下输出:
3 | 5 | 1.5000 | 5 | Lunch
4 | 6 | 4.6600 | 6 | Dinner
第一顿“午餐”怎么办?如何使PHP输出与直接MySQL查询输出相同?在
之前调用MySQL\u fetch\u acoc
而在之前调用。不要
您还应该意识到,ext/mysql
将被弃用,并将代码升级为使用PDO或mysqli的正确参数化查询尝试使用mysql\u fetch\u数组
$purchases = array();
while($row = mysql_fetch_array($query)) {
$purchases[] = $row;
}
如果你想得到数据,就去做吧
foreach($purchases as $key => $value)
{
$name = $value['name'];
$amount_spent = $value['amount_spent'];
echo 'name : '.$name.' , amount spent '.$amount_spent.'<br />';
}
foreach($key=>$value)
{
$name=$value['name'];
$amount_-spended=$value['amount_-spended'];
回显“名称:”.$name.”,金额花费“$amount_花费”。
;
}