如何检查MySQL中的表字段是否存在索引?

时间:2008-09-24 13:35:14

标签: mysql indexing

我需要Google几次,所以我正在分享我的问答。

11 个答案:

答案 0 :(得分:125)

像这样使用SHOW INDEX

SHOW INDEX FROM [tablename]

文档:https://dev.mysql.com/doc/refman/5.0/en/show-index.html

答案 1 :(得分:33)

尝试:

SELECT * FROM information_schema.statistics 
  WHERE table_schema = [DATABASE NAME] 
    AND table_name = [TABLE NAME] AND column_name = [COLUMN NAME]

它会告诉您某个列上是否存在任何类型的索引,而无需知道为索引指定的名称。它也可以在存储过程中工作(而不是显示索引)

答案 2 :(得分:9)

SHOW KEYS FROM  tablename WHERE Key_name='unique key name'

您可以找到表格中是否存在唯一键

答案 3 :(得分:6)

show index from table_name where Column_name='column_name';

答案 4 :(得分:2)

使用以下语句:SHOW INDEX FROM your_table

然后检查字段的结果:row [" Table"],row [" Key_name"]

确保你写了#34; Key_name"正确

答案 5 :(得分:1)

只是从cli看一下表格布局。你会做的

  

desc mytable

  

show table mytable

答案 6 :(得分:0)

如果您需要功能(如果存在作为数据库功能的列索引(在这里排在第一位)),则可以使用/采用此代码。 如果要检查索引是否存在而不管多列索引中的位置如何,那么只需删除“ AND SEQ_IN_INDEX = 1”部分即可。

DELIMITER $$
CREATE FUNCTION `fct_check_if_index_for_column_exists_at_first_place`(
    `IN_SCHEMA` VARCHAR(255),
    `IN_TABLE` VARCHAR(255),
    `IN_COLUMN` VARCHAR(255)
)
RETURNS tinyint(4)
LANGUAGE SQL
DETERMINISTIC
CONTAINS SQL
SQL SECURITY DEFINER
COMMENT 'Check if index exists at first place in sequence for a given column in a given table in a given schema. Returns -1 if schema does not exist. Returns -2 if table does not exist. Returns -3 if column does not exist. If index exists in first place it returns 1, otherwise 0.'
BEGIN

-- Check if index exists at first place in sequence for a given column in a given table in a given schema. 
-- Returns -1 if schema does not exist. 
-- Returns -2 if table does not exist. 
-- Returns -3 if column does not exist. 
-- If the index exists in first place it returns 1, otherwise 0.
-- Example call: SELECT fct_check_if_index_for_column_exists_at_first_place('schema_name', 'table_name', 'index_name');

-- check if schema exists
SELECT 
    COUNT(*) INTO @COUNT_EXISTS
FROM 
    INFORMATION_SCHEMA.SCHEMATA
WHERE 
    SCHEMA_NAME = IN_SCHEMA
;

IF @COUNT_EXISTS = 0 THEN
    RETURN -1;
END IF;


-- check if table exists
SELECT 
    COUNT(*) INTO @COUNT_EXISTS
FROM 
    INFORMATION_SCHEMA.TABLES
WHERE 
    TABLE_SCHEMA = IN_SCHEMA
AND TABLE_NAME = IN_TABLE
;

IF @COUNT_EXISTS = 0 THEN
    RETURN -2;
END IF;


-- check if column exists
SELECT 
    COUNT(*) INTO @COUNT_EXISTS
FROM 
    INFORMATION_SCHEMA.COLUMNS
WHERE 
    TABLE_SCHEMA = IN_SCHEMA
AND TABLE_NAME = IN_TABLE
AND COLUMN_NAME = IN_COLUMN
;

IF @COUNT_EXISTS = 0 THEN
    RETURN -3;
END IF;

-- check if index exists at first place in sequence
SELECT 
    COUNT(*) INTO @COUNT_EXISTS
FROM 
    information_schema.statistics 
WHERE 
    TABLE_SCHEMA = IN_SCHEMA
AND TABLE_NAME = IN_TABLE AND COLUMN_NAME = IN_COLUMN
AND SEQ_IN_INDEX = 1;


IF @COUNT_EXISTS > 0 THEN
    RETURN 1;
ELSE
    RETURN 0;
END IF;


END$$
DELIMITER ;

答案 7 :(得分:0)

尝试使用此功能:

SELECT TRUE
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE 
WHERE TABLE_SCHEMA = "{DB_NAME}" 
AND TABLE_NAME = "{DB_TABLE}"
AND COLUMN_NAME = "{DB_INDEXED_FIELD}";

答案 8 :(得分:0)

添加到 GK10 建议的内容:

<块引用>

使用以下语句:SHOW INDEX FROM your_table

然后检查字段的结果:row["Table"], 行["Key_name"]

确保正确写入“Key_name”

人们可以将其应用到 PHP(或其他语言)中,并包裹在 sql 语句中以查找索引列。基本上你可以将 SHOW INDEX FROM 'mytable' 的结果拉入 PHP,然后使用列 'Column_name' 来获取索引列。

制作您的数据库连接字符串并执行如下操作:

$mysqli = mysqli_connect("localhost", "my_user", "my_password", "world");

$sql =  "SHOW INDEX FROM 'mydatabase.mytable' WHERE Key_name = 'PRIMARY';" ;
$result = mysqli_query($mysqli, $sql);

while ($row = $result->fetch_assoc()) {
    echo $rowVerbatimsSet["Column_name"];
}

答案 9 :(得分:-1)

您可以使用以下SQL语句来检查表上的给定列是否已编入索引

select  a.table_schema, a.table_name, a.column_name, index_name
from    information_schema.columns a
join    information_schema.tables  b on a.table_schema  = b.table_schema and
                                    a.table_name = b.table_name and 
                                    b.table_type = 'BASE TABLE'
left join (
 select     concat(x.name, '/', y.name) full_path_schema, y.name index_name
 FROM   information_schema.INNODB_SYS_TABLES  as x
 JOIN   information_schema.INNODB_SYS_INDEXES as y on x.TABLE_ID = y.TABLE_ID
 WHERE  x.name = 'your_schema'
 and    y.name = 'your_column') d on concat(a.table_schema, '/', a.table_name, '/', a.column_name) = d.full_path_schema
where   a.table_schema = 'your_schema'
and     a.column_name  = 'your_column'
order by a.table_schema, a.table_name;

由于连接是针对INNODB_SYS_ *的,因此匹配索引仅来自INNODB表

答案 10 :(得分:-2)

您无法运行特定的show index查询,因为如果索引不存在,它将引发错误。因此,如果要避免任何SQL错误,必须将所有索引都捕获到数组中并循环遍历它们。

继承人我是怎么做到的。我从表中获取所有索引(在本例中为leads),然后在foreach循环中检查列名称(在本例中为province)是否存在。

$this->name = 'province';

$stm = $this->db->prepare('show index from `leads`');
$stm->execute();
$res = $stm->fetchAll();
$index_exists = false;

foreach ($res as $r) {
    if ($r['Column_name'] == $this->name) {
        $index_exists = true;
    }
}

这样你就可以真正缩小索引属性。执行print_r $res以了解您可以使用的内容。