问题描述
我有一个jQuery列表,它在php页面上返回了一个用户名列表,例如
I have a jQuery list which is returning a list of user_name on php page like
rohit,Bhalu,Ram
现在我想从数据库中筛选出不在上面列表中的用户名
Now I want to filter the user_names from the database which is not the part of above list
到目前为止,我正在尝试mysql的基本查询,例如
So far I am trying the basic query of mysql like
select * from table_name where user_name NOTIN('rohit','Bhalu','Ram');
但是上述查询的问题是,这不是包含1000个user_name的较大列表的特定解决方案,因此我想对php使用一些查询过滤器
But problem with above query is, this a not the specific solution for bigger list which contains 1000 user_name so I want to use some query filter with php
请建议我在这个阶段应该做什么?
Please suggest me what should I do in this stage ?
推荐答案
字段user_name的首次使用索引.
First use index for field user_name.
第二次使用此查询(在$ array-用户名中)
Second use this query (in $array - usernames)
$array = array('Rohit', 'Bhalu');
$comma_separated = implode("','", $array);
$comma_separated = "'".$comma_separated."'";
$query = "select * from table_name where user_name NOT IN($comma_separated)";
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