Home > Backend Development > PHP Tutorial > SQL 语句优化

SQL 语句优化

WBOY
Release: 2016-06-20 12:26:42
Original
1107 people have browsed it

$searchlist = array(    array("id" => 3, "id2" => 16, "flat" => 1),    array("id" => 5, "id2" => 6, "flat" => 1),    array("id" => 2, "id2" => 3, "flat" => 2),    array("id" => 9, "id2" => 1, "flat" => 2),    array("id" => 6, "id2" => 4, "flat" => 1),);$searchresult = "";foreach($searchlist as $k=>$v){    if($v['flat'] == 1){        $sql = sprintf("select insurance_entitle1 as title1 from %s_insurance where insurance_id = %d", $this->tablepre, $v['id']);        $result = $this->db->get_one($sql);        $searchresult[$k]['title1'] = isset($result['title1']) ? $result['title1'] : "";        $sql = sprintf("select insurance_encontent as content from %s_insurance_section where insurancese_section=%d order by insurancese_order", $this->tablepre, $v['id']);        $result = $this->db->get_one($sql);        $searchresult[$k]['content'] = isset($result['content']) ? $result['content'] : "";        $searchresult[$k]['flat'] = $v['flat'];        $searchresult[$k]['id'] = $v['id'];    }elseif($v['flat'] == 2){        $sql = sprintf("select insurance_entitle1 as title1 from %s_insurance where insurance_id = %d", $this->tablepre, $v['id2']);        $result = $this->db->get_one($sql);        $searchresult[$k]['title1'] = isset($result['title1']) ? $result['title1'] : "";        $sql = sprintf("select insurance_encontent as content from %s_insurance_section where insurancese_id=%d order by insurancese_order", $this->tablepre, $v['id']);        $result = $this->db->get_one($sql);        $searchresult[$k]['content'] = isset($result['content']) ? $result['content'] : "";        $searchresult[$k]['flat'] = $v['flat'];        $searchresult[$k]['id'] = $v['id'];        $searchresult[$k]['sid'] = $v['id2'];    }}
Copy after login


回复讨论(解决方案)

没有看到 SQL,只看到 php

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template