将多个MySql查询合并到一个查询中

Question

我正在努力解决第二天的问题,但还没有成功.

我的目标是:

对于某些分类术语ID(例如1),我需要获得至少一个节点的imagefield文件路径,这些路径属于该术语id(tid 1)

以下三个查询给了我我需要的东西.一切正常,但我知道它看起来很丑陋,我敢肯定,这是一个很大的性能问题.

$childterm = 10; // Taxonomy term ID
$result = db_fetch_array(db_query("SELECT node.vid FROM node JOIN 
                                   term_node ON  node.vid=term_node.vid WHERE  
       term_node.tid=$childterm AND 
                                   node.type= 'product' LIMIT 0,1"));           

$nvid = $result['vid']; // Extracting node VID by term ID that will be used futher 

$result = db_fetch_array(db_query("SELECT field_image_cache_fid FROM
                                   content_field_image_cache WHERE
       vid = '%d'", $nvid));

$fid = $result['field_image_cache_fid']; // Extracting file ID from array

$result = db_fetch_array(db_query("SELECT filepath FROM files WHERE
                                   files.fid = '%d'", $fid));           

$filepath = $result['filepath']; // Finally. Extracting file path from array

请看图片.

如何改进查询？我只能使用一个sql查询获取文件路径值吗？

提前致谢.

Answer 1

这样的事情应该有效.

$sql = "SELECT f.filepath FROM {node} AS n
INNER JOIN {term_node} AS t ON t.nid = n.nid
INNER JOIN {content_field_image_cache} as c ON c.nid = n.nid
INNER JOIN {files} AS f on f.fid = c.fid
WHERE n.type = '%s'
AND t.tid = %d;"

$ result = db_query($ sql,$ node_type,$ tid);

请注意,您应该使用内容表名称获取信息

$db_info = content_database_info(content_fields($field_name));

原因是,如果从/切换到多个值,SQL将会中断,除非您使用此方法添加额外的表和列名称.