Question

好的，所以出于某种原因，我只得到$ this-＆gt;信息被发送到mysql数据库

ie：应该是这个

"UPDATE `shop_products` SET shopid = 1 WHERE id = 42"

但我得到了这个

"UPDATE `shop_products` SET 1 WHERE 42"

这是我正在使用的代码。我不确定为什么这不起作用

public function addstoreproduct($id,$shopid,$categoryid,$productname,$description,$price,$rcb)
    {
        $this->id = $id;
        $this->shopid = $shopid;
        $this->categoryid = $categoryid;
        $this->productname = $productname;
        $this->descriptionid = $description;
        $this->price = $price;
        $this->rcb = $rcb;

        mysql_query("UPDATE `shop_products` SET shopid = "+$this->shopid+" WHERE id = "+$this->id+"");

        //$this->updatedb('shop_products','shopid='+$this->shopid+'','"id='+$this->id+'"');


    }

Answer 1

尝试

mysql_query("UPDATE shop_products SET shopid = " . $this->shopid . " WHERE id = " . $this->id);

Answer 2

首先构建它时是否显示正确的查询...

public function addstoreproduct($id,$shopid,$categoryid,$productname,$description,$price,$rcb)
{
    $this->id = $id;
    $this->shopid = $shopid;
    $this->categoryid = $categoryid;
    $this->productname = $productname;
    $this->descriptionid = $description;
    $this->price = $price;
    $this->rcb = $rcb;
    $query = "UPDATE `shop_products` SET shopid = "+$this->shopid+" WHERE id = "+$this->id+"";
    echo $query;
    mysql_query($query);

    //$this->updatedb('shop_products','shopid='+$this->shopid+'','"id='+$this->id+'"');


}

Stange mysql_update函数似乎只能获得$ this-＆gt;信息

2 个答案: