0

我正在尝试匹配具有包含所有给定元素的 json 数组的行。

示例搜索项：

['gym', 'sofa']

预期的匹配行：

['gym', 'sofa']
['gym', 'sofa', 'bed']
['pool', 'gym', 'sofa']
['pool', 'gym', 'sofa', 'bed']

不应该匹配：

['pool', 'gym', 'bed']
['pool', 'sofa', 'bed']

我将项目存储在作为 json 属性索引的 json 文本中。

表格示例：

ID  ITEMS
1   {'items': ['gym', 'sofa']}
2   {'items': ['gym', 'sofa', 'bed']}
3   {'items': ['pool', 'gym', 'bed']}
4   {'items': ['pool', 'sofa', 'bed']}

我的 sphinx.conf 是这样的：

source srcItems
{
    type            = mysql

    sql_host        = localhost
    sql_user        = root
    sql_pass        =
    sql_db          = items
    sql_port        = 3306    # optional, default is 3306

    sql_query        = \
        SELECT id, items \
        FROM items

    sql_attr_json        = items
}

index items
{
    source          = srcItems
    path            = /opt/local/var/sphinx/data/items
}

indexer
{
    mem_limit        = 128M
}

searchd
{
    listen          = 9312
    listen          = 9306:mysql41
    log             = /opt/local/var/sphinx/log/searchd.log
    query_log       = /opt/local/var/sphinx/log/query.log
    read_timeout    = 5
    max_children    = 30
    pid_file        = /opt/local/var/sphinx/log/searchd.pid
    max_matches     = 1000
    seamless_rotate = 1
    preopen_indexes = 1
    unlink_old      = 1
    workers         = threads # for RT to work
    binlog_path     = /opt/local/var/sphinx/data
}

我尝试使用以下没有结果：

SELECT id, 
    ALL(var='gym' AND var='sofa' FOR var IN items.items) as i 
FROM items 
WHERE i=1;

SELECT id, 
    ANY(var='gym' AND var='sofa' FOR var IN items.items) as i 
FROM items 
WHERE i=1;

还尝试了以下方法，它返回错误的结果：

SELECT id, 
    ALL(var='gym' OR var='sofa' FOR var IN items.items) as i 
FROM items 
WHERE i=1;

SELECT id, 
    ANY(var='gym' OR var='sofa' FOR var IN items.items) as i 
FROM items 
WHERE i=1;

当我这样做时，我得到了预期的结果：

SELECT id, 
    IN(items.items, 'gym') AS gym, 
    IN(items.items, 'sofa') AS sofa 
FROM offers 
WHERE gym = 1 AND sofa = 1

但它会大大减慢查询速度，并使查询的构建更加复杂。

我究竟做错了什么？

在 Sphinx 中执行此查询的正确方法是什么？

4

2 回答 2

1

我认为

SELECT id FROM offers  
  WHERE items.items IN('gym')
  AND items.items IN('sofa')

可能会奏效。如果没有，请尝试

SELECT id, 
    IN(items.items, 'gym')+IN(items.items, 'sofa') AS i
FROM offers 
WHERE i = 2

否则

SELECT id, 
   ANY(var='gym' FOR var IN items.items)+ANY(var='sofa' FOR var IN items.items) as i 
FROM items 
WHERE i=2;
于 2014-11-19T13:18:38.330 回答
0

最终使用：

SELECT id, 
    IN(items.items, 'gym') AS gym, 
    IN(items.items, 'sofa') AS sofa 
FROM offers 
WHERE gym = 1 AND sofa = 1

它工作正常，额外的时间不是什么大问题，可以通过更好的硬件来缓解。

谢谢

于 2015-01-06T15:30:30.260 回答