获取任何表的当前AUTO_INCREMENT值

Question

如何获取MySQL中表的当前AUTO_INCREMENT值？

Answer 1

您可以使用以下查询获取所有表数据:

SHOW TABLE STATUS FROM `DatabaseName` WHERE `name` LIKE 'TableName' ;

您可以使用此查询获得完全相关的信息:

SELECT `AUTO_INCREMENT`
FROM  INFORMATION_SCHEMA.TABLES
WHERE TABLE_SCHEMA = 'DatabaseName'
AND   TABLE_NAME   = 'TableName';

Answer 2

我相信你正在寻找MySQL的LAST_INSERT_ID()函数.如果在命令行中,只需运行以下命令:

LAST_INSERT_ID();

您还可以通过SELECT查询获取此值:

SELECT LAST_INSERT_ID();

Answer 3

如果您只想知道数字,而不是在查询中获取它,那么您可以使用:

SHOW CREATE TABLE tablename;

您应该在底部看到auto_increment

Answer 4

mysqli可执行示例代码：

<?php
    $db = new mysqli("localhost", "user", "password", "YourDatabaseName");
    if ($db->connect_errno) die ($db->connect_error);

    $table=$db->prepare("SHOW TABLE STATUS FROM YourDatabaseName");
    $table->execute();
    $sonuc = $table->get_result();
    while ($satir=$sonuc->fetch_assoc()){
        if ($satir["Name"]== "YourTableName"){
            $ai[$satir["Name"]]=$satir["Auto_increment"];
        }
    }
    $LastAutoIncrement=$ai["YourTableName"];
    echo $LastAutoIncrement;
?>  

Answer 5

如果您手动运行查询，即使methai的答案是正确的，但是当两个并发事务/连接在生产环境中的运行时实际执行此查询时（例如），仍会出现问题。

刚刚在MySQL工作台中手动尝试过，同时打开了2个连接：

CREATE TABLE translation (
  id BIGINT PRIMARY KEY AUTO_INCREMENT
);
# Suppose we have already 20 entries, we execute 2 new inserts:

交易1：

21 = SELECT `AUTO_INCREMENT` FROM INFORMATION_SCHEMA.TABLES
     WHERE TABLE_SCHEMA = 'DatabaseName' AND TABLE_NAME = 'translation';

insert into translation (id) values (21);

交易2：

21 = SELECT `AUTO_INCREMENT` FROM INFORMATION_SCHEMA.TABLES
     WHERE TABLE_SCHEMA = 'DatabaseName' AND TABLE_NAME = 'translation';

insert into translation (id) values (21);

# commit transaction 1;
# commit transaction 2;

事务1的插入是可以的：事务2的插入出错：错误代码：1062。键“ PRIMARY”的条目“ 21”重复。

一个很好的解决方案是jvdub的答案，因为对于每个事务/连接，2个插入将是：

交易1：

insert into translation (id) values (null);
21 = SELECT LAST_INSERT_ID();

交易2：

insert into translation (id) values (null);
22 = SELECT LAST_INSERT_ID();

# commit transaction 1;
# commit transaction 2;

但是我们必须在插入之后执行last_insert_id（）！而且，我们可以重复使用该ID并将其插入到需要外键的其他表中！

此外，我们无法执行以下两个查询：

insert into translation (id) values ((SELECT AUTO_INCREMENT FROM
    information_schema.TABLES WHERE TABLE_SCHEMA=DATABASE()
    AND TABLE_NAME='translation'));

因为我们实际上有兴趣在其他表中获取/重用该ID或返回它！

Answer 6

查询以检查给定模式的所有表的 AUTO_INCREMENT“使用”百分比（类型为bigint unsigned的列除外）：

SELECT 
  c.TABLE_NAME,
  c.COLUMN_TYPE,
  c.MAX_VALUE,
  t.AUTO_INCREMENT,
  IF (c.MAX_VALUE > 0, ROUND(100 * t.AUTO_INCREMENT / c.MAX_VALUE, 2), -1) AS "Usage (%)" 
FROM 
  (SELECT 
     TABLE_SCHEMA,
     TABLE_NAME,
     COLUMN_TYPE,
     CASE 
        WHEN COLUMN_TYPE LIKE 'tinyint(1)' THEN 127
        WHEN COLUMN_TYPE LIKE 'tinyint(1) unsigned' THEN 255
        WHEN COLUMN_TYPE LIKE 'smallint(%)' THEN 32767
        WHEN COLUMN_TYPE LIKE 'smallint(%) unsigned' THEN 65535
        WHEN COLUMN_TYPE LIKE 'mediumint(%)' THEN 8388607
        WHEN COLUMN_TYPE LIKE 'mediumint(%) unsigned' THEN 16777215
        WHEN COLUMN_TYPE LIKE 'int(%)' THEN 2147483647
        WHEN COLUMN_TYPE LIKE 'int(%) unsigned' THEN 4294967295
        WHEN COLUMN_TYPE LIKE 'bigint(%)' THEN 9223372036854775807
        WHEN COLUMN_TYPE LIKE 'bigint(%) unsigned' THEN 0
        ELSE 0
     END AS "MAX_VALUE" 
   FROM 
     INFORMATION_SCHEMA.COLUMNS
     WHERE EXTRA LIKE '%auto_increment%'
   ) c

   JOIN INFORMATION_SCHEMA.TABLES t ON (t.TABLE_SCHEMA = c.TABLE_SCHEMA AND t.TABLE_NAME = c.TABLE_NAME)

WHERE
 c.TABLE_SCHEMA = 'YOUR_SCHEMA' 
ORDER BY
 `Usage (%)` DESC;