是否有一种快速的方法从MySQL中所有表中获得所有列名,而不必列出所有表?


当前回答

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()。

其他回答

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;
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()。

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>";

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'
<?php
        $table = 'orders';
        $query = "SHOW COLUMNS FROM $table";
        if($output = mysql_query($query)):
            $columns = array();
            while($result = mysql_fetch_assoc($output)):
                $columns[] = $result['Field'];
            endwhile;
        endif;
        echo '<pre>';
        print_r($columns);
        echo '</pre>';
?>