SQL 语句优化

WBOY
Libérer: 2016-06-20 12:26:42
original
1107 Les gens l'ont consulté

$searchlist = array(    array("id" => 3, "id2" => 16, "flat" => 1),    array("id" => 5, "id2" => 6, "flat" => 1),    array("id" => 2, "id2" => 3, "flat" => 2),    array("id" => 9, "id2" => 1, "flat" => 2),    array("id" => 6, "id2" => 4, "flat" => 1),);$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'];    }}
Copier après la connexion


回复讨论(解决方案)

没有看到 SQL,只看到 php

Étiquettes associées:
source:php.cn
Déclaration de ce site Web
Le contenu de cet article est volontairement contribué par les internautes et les droits d'auteur appartiennent à l'auteur original. Ce site n'assume aucune responsabilité légale correspondante. Si vous trouvez un contenu suspecté de plagiat ou de contrefaçon, veuillez contacter admin@php.cn
Tutoriels populaires
Plus>
Derniers téléchargements
Plus>
effets Web
Code source du site Web
Matériel du site Web
Modèle frontal