Python 使用def无法搜索列表中的元素

有人能帮我解决这个问题吗？我只是想做一个简单的程序，但我被卡住了，现在我很沮丧

from tabulate import tabulate
myList = []

def addnew():
    name = []

    name.append(input("Enter name: "))
    myList.append(name)

    print(myList)

def find():
    global myList
    src = input("Enter name: ")
    if src in myList:
        print("Yes, name exist")
    elif src not in myList:
        print("Name not exist")

def view():
    print(tabulate(myList, tablefmt='psql'))

while True:
    print("[1] to add")
    print("[2] to search")
    print("[3] to view")
    inp = input(">> ")
    if inp == '1':
        addnew()
    elif inp == '2':
        find()
    elif inp == '3':
        view()
    else:
        print("Wrong input")

我的

def find（）：

未按预期工作：(

根据我的发现，如果像这样的列表

['bob']

结果是好的，但是当像这样的列表

[['bob']]

结果名称不存在时

如果我删除

name=[]

并更改

myList.append（input（））

表格结果太乱了：）

结果:

Enter name: bob
['bob']   
[1] to add
[2] to search
[3] to view
>> 3
+---+---+---+
| b | o | b |
+---+---+---+
[1] to add
[2] to search
[3] to view
>>

此代码将

name

作为一个列表。当您输入“bob”时，

name

类似于

['bob']

。然后将

name

应用到

myList

，这样

myList

将是一个类似

[['bob']]

的列表。在

[['bob']

中找不到

'bob'

所以你不必把

name

列成一个列表。只需将这两行替换为：

name = input("Enter name: ")

---

输入新名称时，只需将字符串添加到

myList

。然后，当您使用

tablate（）

时，从此

myList

创建一列数据：

from tabulate import tabulate
myList = []

def addnew():
    myList.append(input("Enter name: "))    # <-- just append new string to myList
    print(myList)

def find():
    global myList
    src = input("Enter name: ")
    if src in myList:
        print("Yes, name exist")
    elif src not in myList:
        print("Name not exist")

def view():
    print(tabulate([[v] for v in myList], tablefmt='psql'))   # <-- create one column data from myList
    # or you can use:
    # print(tabulate({'name': myList}, tablefmt='psql'))

while True:
    print("[1] to add")
    print("[2] to search")
    print("[3] to view")
    inp = input(">> ")
    if inp == '1':
        addnew()
    elif inp == '2':
        find()
    elif inp == '3':
        view()
    else:
        print("Wrong input")

---

myList=[['bob']]

是一个列表列表，它是另一回事。您需要在myList中编写

['bob']才能找到它……我认为您不需要
名称
列表，只需要
myList.append（输入（“输入名称”）
@scarLópez我已经更改为
myList=['bob']
，但表格结果不太好看。您可以稍后格式化表格，以使其看起来更漂亮；更重要的是程序运行正确，而不是它漂亮。@ÓscarLópez hihi。好的，好的
name = input("Enter name: ")

from tabulate import tabulate
myList = []

def addnew():
    myList.append(input("Enter name: "))    # <-- just append new string to myList
    print(myList)

def find():
    global myList
    src = input("Enter name: ")
    if src in myList:
        print("Yes, name exist")
    elif src not in myList:
        print("Name not exist")

def view():
    print(tabulate([[v] for v in myList], tablefmt='psql'))   # <-- create one column data from myList
    # or you can use:
    # print(tabulate({'name': myList}, tablefmt='psql'))

while True:
    print("[1] to add")
    print("[2] to search")
    print("[3] to view")
    inp = input(">> ")
    if inp == '1':
        addnew()
    elif inp == '2':
        find()
    elif inp == '3':
        view()
    else:
        print("Wrong input")

[1] to add
[2] to search
[3] to view
>> 1
Enter name: Andrej
['Andrej']
[1] to add
[2] to search
[3] to view
>> 1
Enter name: John
['Andrej', 'John']
[1] to add
[2] to search
[3] to view
>> 1
Enter name: Mary
['Andrej', 'John', 'Mary']
[1] to add
[2] to search
[3] to view
>> 3
+--------+
| Andrej |
| John   |
| Mary   |
+--------+
[1] to add
[2] to search
[3] to view
>> 2
Enter name: John
Yes, name exist
[1] to add
[2] to search
[3] to view