作者:jinglongyy70 | 来源:互联网 | 2023-08-28 16:54
IhaveoverriddentheSimpleHandlertopassausernameandpasswordusingsoap4r.theproblemistha
I have overridden the SimpleHandler to pass a username and password using soap4r. the problem is that I am forced to give a QName, and this is causing the result to fail because it's not in the right format.
我已经重写了SimpleHandler以使用soap4r传递用户名和密码。问题是,我被迫给出一个QName,这导致结果失败,因为它不是正确的格式。
What soap4r is adding is something like this (the "ns1" things are dummy values):
soap4r添加的是这样的东西(“ns1”的东西是虚拟值):
someuser
topsecret
What it needs to be is this:
它需要的是:
someuser
topsecret
How can I NOT pass in a containing name?
我怎么能不传递一个包含的名字呢?
2 个解决方案