Question

好的，我已经使用了一段时间了，我有点卡住了。

也许我一切都错了！

基本上我有搜索字段的查询。一般的想法是选择基于LIKE %%的结果，同时仍然首先放置完全匹配。

因此，例如，如果你搜索47我想要显示id，surname或company_name中47的所有内容，但是如果我输入了姓氏，那么id号为47的结果应该在顶部并且相同。

请参阅下面的代码，这可能有助于澄清我的问题。

SELECT id, 
IF(company_name IS NOT NULL AND company_name <> '', company_name, surname) AS name, 
first_name, country, phone1, isowner, isholidayrenter, isproholidayrenter, 
islongtermrenter, isprolongtermrenter, isprobuyer, isbuyer 
FROM clients 
WHERE id LIKE '$search' OR surname LIKE '$search' OR company_name LIKE '$search'
union all
SELECT id, 
IF(company_name IS NOT NULL AND company_name <> '', company_name, surname) AS name,      
first_name, country, phone1, isowner, isholidayrenter, isproholidayrenter,     
islongtermrenter, isprolongtermrenter, isprobuyer, isbuyer 
FROM clients 
WHERE id LIKE '%$search%' AND id NOT LIKE '$search' OR surname LIKE '%$search%' 
and SURNAME NOT LIKE '$search' OR company_name LIKE '%$search%' 
and company_name NOT LIKE '$search' 
LIMIT $start, $limit";

`

Answer 1

试试这个：

（选择完全匹配）联合全部（选择部分匹配省略完全匹配）

示例：

(
  SELECT
      id, 
      IF( company_name IS NOT NULL AND company_name <> '', company_name, surname ) AS name,
      first_name, country, phone1, isowner, isholidayrenter, isproholidayrenter, 
      islongtermrenter, isprolongtermrenter, isprobuyer, isbuyer 
  FROM
      clients 
  WHERE
      id LIKE '$search' OR 
      surname LIKE '$search' OR 
      company_name LIKE '$search'
)
union all
(
  SELECT 
      id,
      IF( company_name IS NOT NULL AND company_name <> '', company_name, surname ) AS name, 
      first_name, country, phone1, isowner, isholidayrenter, isproholidayrenter,
      islongtermrenter, isprolongtermrenter, isprobuyer, isbuyer 
  FROM
      clients 
  WHERE
      ( id LIKE '%$search%' AND id NOT LIKE '$search' ) OR 
      ( surname LIKE '%$search%' AND SURNAME NOT LIKE '$search' ) OR
      ( company_name LIKE '%$search%' AND company_name NOT LIKE '$search' )
)
LIMIT $start, $limit;

mysql查询顺序与多个LIKE子句完全匹配

1 个答案: