在MySQL中，我知道我可以在数据库中列出表:
SHOW TABLES
但是，我想将这些表名插入到另一个表中，例如:
INSERT INTO metadata(table_name) SHOW TABLES /* does not work */
是否有一种方法可以使用标准的SELECT语句来获取表名，比如:
INSERT INTO metadata(table_name) SELECT name FROM table_names /* what should table_names be? */
当前回答
插入、更新和删除操作如下:
$teste = array('LOW_PRIORITY', 'DELAYED', 'HIGH_PRIORITY', 'IGNORE', 'INTO', 'INSERT', 'UPDATE', 'DELETE', 'QUICK', 'FROM');
$teste1 = array("\t", "\n", "\r", "\0", "\x0B");
$strsql = trim(str_ireplace($teste1, ' ', str_ireplace($teste, '', $strsql)));
$nomeTabela = substr($strsql, 0, strpos($strsql, ' '));
print($nomeTabela);
exit;
其他回答
还有一种更简单的方法来获取表名
SHOW TABLES FROM <database_name>
MySQL INFORMATION_SCHEMA。TABLES表包含关于两个表(不是临时表，而是永久表)和视图的数据。TABLE_TYPE列定义了这是表记录还是视图记录(对于表TABLE_TYPE='BASE table '，对于视图TABLE_TYPE=' view ')。所以如果你只想从你的schema(数据库)表中看到下面的查询:
SELECT *
FROM information_schema.tables
WHERE table_type='BASE TABLE'
AND table_schema='myschema'
SELECT table_name
FROM information_schema.tables
WHERE table_schema = 'DATABASE'
除了使用INFORMATION_SCHEMA表外，要使用SHOW TABLES插入到表中，还需要使用以下命令
<?php
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
$arrayCount = 0
while ($row = mysql_fetch_row($result)) {
$tableNames[$arrayCount] = $row[0];
$arrayCount++; //only do this to make sure it starts at index 0
}
foreach ($tableNames as &$name {
$query = "INSERT INTO metadata (table_name) VALUES ('".$name."')";
mysql_query($query);
}
?>
用于获取所有表的名称:
SELECT table_name
FROM information_schema.tables;
如果你需要为一个特定的数据库获取它:
SELECT table_name
FROM information_schema.tables
WHERE table_schema = 'your_db_name';
输出:
+--------------------+
| table_name |
+--------------------+
| myapp |
| demodb |
| cliquein |
+--------------------+
3 rows in set (0.00 sec)
