Question

需要一些查询帮助。在数据库中有大约700k的图像，每个图像都有自己的标签。我希望能够通过带有全文搜索查询的标签搜索图像。查询完全符合我的要求，但速度非常慢。有些人可以帮助我加快速度，或创建另一个。 de'image_tag'表中的id-fields和name字段都有索引。

SELECT
    image.*
FROM image
INNER JOIN (SELECT image_to_tag.image_id,
             GROUP_CONCAT(image_tag.`name`) AS tags
        FROM image_to_tag
        INNER JOIN image_tag
        ON (image_tag.id = image_to_tag.image_tag_id)
    GROUP BY image_to_tag.image_id) t ON t.image_id = image.id
WHERE MATCH (t.tags) AGAINST ('+justin* +backgrounds*' IN BOOLEAN MODE)

表：图像

id | filename 
1  | image1.jpg
2  | image2.jpg
3  | image3.jpg

表：image_to_tag

image_id | image_tag_id
1        | 1
1        | 2
2        | 3
2        | 4

表：image_tag

id | name
1  | justin bieber
2  | backgrounds
3  | justin timberlake
4  | other backgrounds

如果我搜索“justin background”，我想找到图像1和2。如果我搜索“贾斯汀·比伯背景”，我想找到图像1。

Answer 1

请尝试这个，让我知道这可能会更快一点。

select id,filename from image
where id in(
    select image_id from image_to_tag a
    inner join image_tag b
    on b.id = a.image_tag_id and b.name like '%justin%'
)

Answer 2

您可以通过将查询分为三个部分来提升性能。它将以毫秒为单位给出结果：

$tagids = '';    
$res = $mysqli->query("SELECT group_concat(id) as tagids FROM image_tag WHERE MATCH (name) AGAINST ('+justin* +backgrounds*' IN BOOLEAN MODE");
if($row = $res->fetch_object()){
  $tagids = $row->tagids;
}
$res->close();

$imageids = '';
$res = $mysqli->query("SELECT group_concat(image_id) as imageids FROM image_to_tag WHERE image_tag_id in($tagids)");
if($row = $res->fetch_object()){
   $imageids = $row->imageids;
}
$res->close();

$imgarr = array();
$res = $mysqli->query("SELECT id,filename FROM image WHERE id in($imageids)");
if($row = $res->fetch_object()){
   array_push($imgarr,$row);
}
$res->close();
echo json_encode($imgarr);

慢mysql查询全文搜索标签数据库

2 个答案: