英文:
Part of string comparison always inexplicably returns False
问题
I have a simple Python script and I do not understand why I always get False as a return of that string comparison. What's wrong here?
inputStr = "upload test"
print(inputStr[:7] == "upload") # -> False
In return the next code returns True:
inputStr = "upload"
print(inputStr[:7] == "upload") # -> True
On the other hand when I use the keyword in, it works just fine (it prints out True):
inputStr = "upload test"
print("upload" in inputStr[:7]) # -> True
Can somebody explain this?
英文:
I have a simple Python script and I do not understand why I always get False as a return of that string comparison. What's wrong here?
inputStr = "upload test"
print(inputStr[:7] == "upload") # -> False
In return the next code returns True:
inputStr = "upload"
print(inputStr[:7] == "upload") # -> True
On the other hand when I use the keyword in, it works just fine (it prints out True):
inputStr = "upload test"
print("upload" in inputStr[:7]) # -> True
Can somebody explain this?
答案1
得分: 1
inputStr[:7] 返回的是 'upload ' 而不是 'upload'。
此外，我建议您使用 str.startswith() 而不是切片！
示例：
print(inputStr.startswith("upload")) # -> True
英文:
inputStr[:7] returns 'upload ' and not 'upload'.
As well as that, I would suggest you use str.startswith() instead of slicing!
example:
print(inputStr.startswith("upload")) # -> True
答案2
得分: 1
以下是要翻译的内容：
这段代码 inputStr[:7] 在第一个块中的结果是：
> 'upload '
而 'upload ' 不等于 'upload'
这是因为它从索引 0 到 6 进行计数
你之所以在这个输入中得到 true：inputStr = "upload" 是因为字符串中只有 6 个字符
所以如果你想让第一个块返回 true
你应该使用这段代码：
inputStr = "upload test"
print(inputStr[:6] == "upload")
英文:
the results of this code inputStr[:7] in fist block is:
> 'upload '
and 'upload ' is not equal to 'upload'
because it counts from index 0 to 6
and the reason you get true in this input: inputStr = "upload"
is that there are only 6 characters in the string
so if you want fist block to return true
you should use this block of code:
inputStr = "upload test"
print(inputStr[:6] == "upload")
答案3
得分: 0
Case 1:
inputStr = "upload test"
print(inputStr[:7] == "upload")
它获取从0到6的字符（包括空格），这与“upload”不相等。所以将inputStr[:7]更改为inputStr[:6]。
Case 2:
返回True，因为输入字符串中只有“upload”，所以inputStr[:7]仍然返回“upload”，与“upload”相等。
Case 3:
'in'关键字返回True，因为“upload”在“upload ”（inputStr[:7]）中。
英文:
Case 1:
inputStr = "upload test"
print(inputStr[:7] == "upload")
it takes characters from 0 to 6 ("upload " with space) which is not equal to "upload". so change inputStr[:7] to inputStr[:6].
Case 2: returns True because in the input string there there is only "upload" so inputStr[:7] still gives "upload" which is equal to "upload".
Case 3: the 'in' keyword returns true because "upload" is in "upload "(inputStr[:7]).
答案4
得分: 0
Python索引从0开始。inputstr[::7] 返回"upload "（末尾有一个空格），而不是"upload"。
而在第二种情况下，inputstr[::7] 返回"upload"，因为7超出了字符串的长度范围。
回答你的问题，语句应该是print(inputStr[:6] == "upload")。
英文:
Python indexing is 0 based. inputstr[::7] returns "upload " (with a space at the end) not "upload".
whereas inputstr[::7] in the 2nd case returns "upload" because 7 is outside the length scope of the string.
To answer your question, the statement needs to be print(inputStr[:6] == "upload")
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