<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
<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
立即学习“PHP免费学习笔记(深入)”;
$count = count($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。
PHP怎么学习?PHP怎么入门?PHP在哪学?PHP怎么学才快?不用担心,这里为大家提供了PHP速学教程(入门到精通),有需要的小伙伴保存下载就能学习啦!
Copyright 2014-2025 https://www.php.cn/ All Rights Reserved | php.cn | 湘ICP备2023035733号