不匹配时返回NULL

Question

以下代码获取输入并将输入拆分为值“ HttpOnly”，然后在满足“ if”条件的情况下返回该值。

如果条件在split（）本身失败，如何使值返回为NULL或“ 123”？

from soaptest.api import *
from com.parasoft.api import *

def getHeader(input, context):

    headerNew = ""
    strHeader = str(input).split("HttpOnly")
    for i in strHeader:
        if "com.abc.mb.SSO_GUID" in i:
            Application.showMessage(i)
            headerNew = i

    return headerNew

编辑

输入-“ abcdefgHttpOnly”

输出-“ abcdefg”

输入-“ abcdefg”

输出-“ 123”

Answer 1

您可以仅测试'HttpOnly'是否首先输入in并返回'123'。

def getHeader(input):
    if 'HttpOnly' not in str(input):
        return '123'

    headerNew = ""
    strHeader = str(input).split("HttpOnly")

    # Not using i as variable since it is usually used as an index
    for header in strHeader:
        if "com.abc.mb.SSO_GUID" in header:
            # Application.showMessage(header)
            headerNew = header

    return headerNew
print(getHeader('com.abc.mb.SSO_GUIDabcdefgHttpOnly')) # com.abc.mb.SSO_GUIDabcdefg
print(getHeader('com.abc.mb.SSO_GUIDabcdefg')) # 123