Question

我的搜索查询效果很好，但我需要它才能显示visible = true的结果。这是需要添加到查询中的内容，我只是不知道如何正确地执行它。这是查询。

$ query = $ mysqli-＆gt; query（“SELECT 身份证，姓名，公司，镇人名称LIKE'％$ mysearchString％'或公司LIKE'％$ mysearchString％'或 town LIKE'％$ mysearchString％'ORDER BY 名称DESC LIMIT 100“）;

$ mysearchString var只是一个已清理的$_GET['s']。

Answer 1

SELECT id,name,company,town 
FROM people WHERE 
(name LIKE '%$mysearchString%' 
OR company LIKE '%$mysearchString%' 
OR town LIKE '%$mysearchString%')
 AND visible = true
ORDER BY name DESC LIMIT 100

Answer 2

$query = $mysqli->query("SELECT id,name,company,town FROM people WHERE (name LIKE '%$mysearchString%' OR company LIKE '%$mysearchString%' OR town LIKE '%$mysearchString%') AND visible = true ORDER BY name DESC LIMIT 100");

搜索SQL查询，这是真的

2 个答案: