使用split()并迭代拆分的what_to_look输入，如果它在title列表中，则获取其索引：

title = ['james','tom','kim']
post_ids = [5,6,9]

what_to_look = input('Enter Name: ').split()

for name in what_to_look:
    if name in title:
        post_id = post_ids[title.index(name)]
        print('Found', post_id)
    else:
        print('Not found')

输出：

Enter Name: james tom
Found 5
Found 6

Enter Name: tom jerry
Found 5
Not found

下面是一个demo on IdeOne to try：

titles = ['james','tom','kim']
locations = [5,6,9]

find_str = input('Find Names (separated by space): ') 
terms = find_str.split()   # split entered string by delimiter space into a list

print('Searching for terms:', terms)
for t in terms:  # loop through each given name and search it in titles
  if t.lower() in titles:  # compare lower-case, if found inside titles
    i = titles.index(t.lower())  # find position which is index for other list 
    loc = locations[i]  # use found position to get location number or URL
    print("Found given term:", t, "on index:", i, " has location:", loc)

我的输入（3个术语，以空格分隔）：

James t. kirk

控制台输出（包括输入行）：

Find Names (separated by space): James t. kirk
Searching for terms: ['James', 't.', 'kirk']
Found given term: James on index: 0  has location: 5

Python字符串提供了.split函数，该函数根据指定的分隔符将字符串分解为一个列表。
例如，names = what_to_look.split(" ")将在每个空格处打断what_to_look。
为便于进一步参考，这里有一个指向文档拆分的链接

我认为您希望将值列表作为输入，可以使用以下代码输入它，例如

what_to_look = input().split()

然后你可以在单行中以空格分隔的文本输入值。希望对你有帮助！

您可以通过split()实现：

title = ['james','tom','kim']
post_ids = [5,6,9]
what_to_look = input('Enter Name: ')

可以将字符串值拆分到列表中;

entered_list = what_to_look.split()

这将从输入的字符串创建一个列表。
编辑：建议您按照@hc_dev的建议使用分隔符作为输入值。这将创建更好的用户体验并获得更好的搜索结果。