Jad*_*dar 36 php mysql database
我正在尝试显示数据库中的所有表.我试过这个:
$sql = "SHOW TABLES";
$result = $conn->query($sql);
$tables = $result->fetch_assoc();
foreach($tables as $tmp)
{
echo "$tmp <br>";
}
Run Code Online (Sandbox Code Playgroud)
但它只给我一个我知道的数据库中的一个表名2.我做错了什么?
Jas*_*Heo 78
SHOW TABLESmysql> USE test;
Database changed
mysql> SHOW TABLES;
+----------------+
| Tables_in_test |
+----------------+
| t1 |
| t2 |
| t3 |
+----------------+
3 rows in set (0.00 sec)
Run Code Online (Sandbox Code Playgroud)
SHOW TABLES IN db_namemysql> SHOW TABLES IN another_db;
+----------------------+
| Tables_in_another_db |
+----------------------+
| t3 |
| t4 |
| t5 |
+----------------------+
3 rows in set (0.00 sec)
Run Code Online (Sandbox Code Playgroud)
mysql> SELECT TABLE_NAME
FROM information_schema.TABLES
WHERE TABLE_SCHEMA = 'another_db';
+------------+
| TABLE_NAME |
+------------+
| t3 |
| t4 |
| t5 |
+------------+
3 rows in set (0.02 sec)
Run Code Online (Sandbox Code Playgroud)
你只拿了一排.修复如下:
while ( $tables = $result->fetch_array())
{
echo $tmp[0]."<br>";
}
Run Code Online (Sandbox Code Playgroud)
我认为,information_schema会比 SHOW TABLES
SELECT TABLE_NAME
FROM information_schema.TABLES
WHERE TABLE_SCHEMA = 'your database name'
while ( $tables = $result->fetch_assoc())
{
echo $tables['TABLE_NAME']."<br>";
}
Run Code Online (Sandbox Code Playgroud)