SELECT * FROM information_schema.columns
WHERE table_schema = DATABASE()
ORDER BY table_name, ordinal_position

因为我没有足够的代表来评论，这里有一个小的改进(在我看来)尼克rulez的优秀答案:替换WHERE table_schema = 'your_db'与WHERE table_schema = DATABASE()。

您可以使用information_schema简单地获取一个表的所有列。列只需添加group_concat以获得逗号分隔的列列表。

select group_concat( column_name ) as ColumnNames from information_schema.columns where table_schema = 'ur_db_name' and table_name = 'ur_tbl_name'

您可以使用此查询查询特定数据库中每个表的所有列

select
    TABLE_SCHEMA,
    TABLE_NAME,
    COLUMN_NAME
from
    information_schema.columns
where
    table_schema = 'database_name'
order by 
    table_name,
    ordinal_position;

问题是:

是否有一个快速的方法，从MySQL所有表中获得所有的列名，而不必列出所有的表?

SQL来获取每列的所有信息

select * from information_schema.columns
where table_schema = 'your_db'
order by table_name,ordinal_position

获取所有的列名

select COLUMN_NAME from information_schema.columns
where table_schema = 'your_db'
order by table_name,ordinal_position

类似于@suganya发布的答案，这并没有直接回答问题，但对于单个表来说是一个更快的选择:

DESCRIBE column_name;

在Nicola的回答上加上一些可读的php

$a = mysqli_query($conn,"select * from information_schema.columns
where table_schema = 'your_db'
order by table_name,ordinal_position");
$b = mysqli_fetch_all($a,MYSQLI_ASSOC);
$d = array();
foreach($b as $c){
    if(!is_array($d[$c['TABLE_NAME']])){
        $d[$c['TABLE_NAME']] = array();
    }
    $d[$c['TABLE_NAME']][] = $c['COLUMN_NAME'];
}
echo "<pre>",print_r($d),"</pre>";