回答:SELECTa.A,a.B,a.CFROM(SELECTPARSENAME( REPLACE( a.Content , '.', '.' ), 3 ) AS A,PARSENAME( REPLACE( a.Content , '.', '.' ), 2 ) AS B,PARSENAME( REPLACE( a.Content , '.', '.' ), 1 ) AS C FROMtest AS a WHERElen( a.content ) - len(replace( a.content, '.', '' )) = 2 UNIONSELECTPARSENAME( REPLACE( a.Content , '.', '.' ), 2 ) AS A,PARSENAME( REPLACE( a.Content , '.', '.' ), 1 ) AS B,'' AS C FROMtest AS a WHERElen( a.content ) - len(replace( a.content, '.', '' )) = 1 UNIONSELECTPARSENAME( REPLACE( a.Content , '.', '.' ), 1 ) AS A,'' AS B,'' AS C FROMtest AS a WHERElen( a.content ) - len(replace( a.content, '.', '' )) = 0 )as a代码中大写字母ABC分别对应图片中的BCD列数据,A为第一层,B为第二层,C为第三层,ABC通过.间隔