Question

if($result = mysql_query(
"SELECT `condition` FROM `table1` WHERE `id` = '1')
 UNION ALL
SELECT `condition` FROM `table2` WHERE `id` = '1')
 UNION ALL
SELECT `condition` FROM `table3` WHERE `id` = '1')"
){ 
      $num = mysql_num_rows($result);
      for($i = 0;  $i < $num; $i++){     
        $table = mysql_field_table($result, $i);
        echo  $table.' cond: '.mysql_result($result, $i, 'condition').', ';
      }
}

这里我可以获得查询结果（条件），但我也想得到表的名称，为了知道结果属于哪个表，我试过了：mysql_field_table（$ result，$ i），但它不会返回任何东西。所以请帮我拿到桌子的名字，提前谢谢你。

Answer 1

您有一组计算的列，因此您无法追溯到源表。您需要自己添加信息：

SELECT `condition`, 'table1' AS source
FROM `table1`
WHERE `id` = '1'

UNION ALL

SELECT `condition`, 'table2'
FROM `table2`
WHERE `id` = '1'

UNION ALL

SELECT `condition`, 'table3'
FROM `table3`
WHERE `id` = '1'

Answer 2

使用此

  SELECT `condition`, '1' tablnumber FROM `table1` WHERE `id` = '1'
  UNION ALL
  SELECT `condition`,'2' FROM `table2` WHERE `id` = '1'
  UNION ALL
  SELECT `condition`,'3' FROM `table3` WHERE `id` = '1'

demo here

PHP＆amp; Mysql无法使用UNION获取表的名称

2 个答案: