以下是一个示例代码，用于找到两个链表的共同节点，并按照它们出现的顺序进行排序：

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def getIntersectionNode(headA, headB):
    # 计算链表A的长度
    lenA = 0
    currA = headA
    while currA:
        lenA += 1
        currA = currA.next

    # 计算链表B的长度
    lenB = 0
    currB = headB
    while currB:
        lenB += 1
        currB = currB.next

    # 将两个链表的指针移动到相同的起点
    currA, currB = headA, headB
    if lenA > lenB:
        for _ in range(lenA - lenB):
            currA = currA.next
    else:
        for _ in range(lenB - lenA):
            currB = currB.next

    # 遍历链表，找到共同节点
    while currA and currB:
        if currA == currB:
            return currA
        currA = currA.next
        currB = currB.next
    
    # 如果没有共同节点，则返回None
    return None

使用示例：

# 创建链表A
a1 = ListNode(1)
a2 = ListNode(2)
a3 = ListNode(3)
a4 = ListNode(6)
a5 = ListNode(7)
a1.next = a2
a2.next = a3
a3.next = a4
a4.next = a5

# 创建链表B
b1 = ListNode(4)
b2 = ListNode(5)
b1.next = b2
b2.next = a4  # 共同节点

# 找到共同节点并按出现顺序排序
intersection = getIntersectionNode(a1, b1)
if intersection:
    result = []
    while intersection:
        result.append(intersection.val)
        intersection = intersection.next
    print(result)  # 输出 [6, 7]
else:
    print("No intersection")

以上代码中，首先计算两个链表的长度，然后将指针移动到相同的起点，最后遍历链表，找到共同节点。输出结果为按出现顺序排序的共同节点。如果没有共同节点，则输出"No intersection"。