thinkphp count用于group就无法获取数量

<code>  $count = $order->where($con)
            ->join("LEFT JOIN x_goods ON x_goods.goods_id = x_order.order_goodsid")
            ->order("order_createtime desc")
            ->group("order_no")
            ->count();// 查询满足要求的总记录数</code>

无法获取数量

<code>  $count = $order->where($con)
            ->join("LEFT JOIN x_goods ON x_goods.goods_id = x_order.order_goodsid")
            ->order("order_createtime desc")
            ->group("order_no")
            ->select();// 查询满足要求的总记录数</code>

可以获取数据

回复内容：

<code>  $count = $order->where($con)
            ->join("LEFT JOIN x_goods ON x_goods.goods_id = x_order.order_goodsid")
            ->order("order_createtime desc")
            ->group("order_no")
            ->count();// 查询满足要求的总记录数</code>

无法获取数量

<code>  $count = $order->where($con)
            ->join("LEFT JOIN x_goods ON x_goods.goods_id = x_order.order_goodsid")
            ->order("order_createtime desc")
            ->group("order_no")
            ->select();// 查询满足要求的总记录数</code>

可以获取数据

那就直接写成

$count = count($order->where($con) ->join("LEFT JOIN x_goods ON x_goods.goods_id = x_order.order_goodsid") ->order("order_createtime desc") ->group("order_no") ->select());// 查询满足要求的总记录数

缺陷是大数据量下会影响效率，导致网络io和资源消耗过大。小业务的话就不必多虑。

我瞎猜的哈，你试试

<code class="php">$count = $order->where($con)
            ->join("LEFT JOIN x_goods ON x_goods.goods_id = x_order.order_goodsid")
            ->order("order_createtime desc")
            ->group("order_no")
            ->field("count(*)")
            ->select();//</code>

最好把执行的SQL打印出来看下，比较复杂的逻辑用原生SQL最好，没必要拘泥。