Question

我有这样的表项目：

id_item       color           stock
1         red,green,yellow   10,20,30

这里有一个表格推车：

id id_trans     id_item    color    qty
1     123          1           red      5   
2     123          1           green    2

我想要实现的是更新表项如下：

id_item       color           stock
1         red,green,yellow   5,18,30

我所做的是：

$query = mysql_query("select a.color as color, a.stock as stock, b.id_item as id_item,b.color as colors,b.qty as qty from item as a, cart as b where a.id_item = b.id_item and b.id_trans = '123' ");
while($result = mysql_fetch_array($query)){

$colors = $result['colors'];
$color = explode(",",$result['color']);
$stock = explode(",",$result['stock']);
$flag = array_search($colors,$color);
$stock[$flag] = $stock[$flag] - $result['qty'];
$stock = implode(',',$stock);

mysql_query("update item set stock = '$stock' where id_item = '$result[id_item]'");

}

从我上面写的代码，我有一些问题。我的问题是，结果如下：5,20,30＆amp; 10,18,30。所以，当我更新到表项时，它只是改变如下：

id_item       color           stock
1         red,green,yellow   10,18,30

我想要实现的是破坏股票成为5,18,30。任何人都可以告诉我如何实现这一点，或者任何人都可以告诉我我的代码中有错误？ THX

Answer 1

现在它比你之前的尝试更清楚了;）试试这个，看它是否有效：

unset($color);unset($stock);
$query = mysql_query("select a.color as color, a.stock as stock, b.id_item as id_item,b.color as colors,b.qty as qty from item as a, cart as b where a.id_item = b.id_item and b.id_trans = '123' ");
while($result = mysql_fetch_array($query)){

    $colors = $result['colors'];
    $id = $result[id_item];
    if(!isset($color))
    {
        $color = explode(",",$result['color']);
    }
    if(!isset($stock))
    {
    $stock = explode(",",$result['stock']);
    }

    $flag = array_search($colors,$color);
    $stock[$flag] = $stock[$flag] - $result['qty'];
}

$stock = implode(',',$stock);
mysql_query("update item set stock = '".$stock."' where id_item = ".$id);

我会说实话，你使用数据库的方式很糟糕。这不是数据库的使用方式。

如何将此项目内爆并更新到数据库

1 个答案: