windows - 如何正确调用LsaLogonUser进行交互式登录？

我正在尝试使用LsaLogonUser创建交互式登录 session ，但它始终返回STATUS_INVALID_INFO_CLASS(0xc0000003)。从在线搜索中发现，KERB_INTERACTIVE_LOGON结构的内存布局有些棘手，但是我敢肯定我做对了。

我还尝试过使用MSV1.0代替Kerberos，并使用MSV1_0_INTERACTIVE_LOGON作为身份验证结构，并使用MSV1_0_PACKAGE_NAME作为程序包名称，但是使用STATUS_BAD_VALIDATION_CLASS(0xc00000a7)失败。

有人可以告诉我我在做什么错吗？这是代码，其中去除了大多数错误处理。显然，这不是生产质量。我只是想获取一个工作样本。

// see below for definitions of these
size_t wcsByteLen( const wchar_t* str );
void InitUnicodeString( UNICODE_STRING& str, const wchar_t* value, BYTE* buffer, size_t& offset );

int main( int argc, char * argv[] )
{
    // connect to the LSA
    HANDLE lsa;
    LsaConnectUntrusted( &lsa );

    const wchar_t* domain = L"mydomain";
    const wchar_t* user = L"someuser";
    const wchar_t* password = L"scaryplaintextpassword";

    // prepare the authentication info
    ULONG authInfoSize = sizeof(KERB_INTERACTIVE_LOGON) +
     wcsByteLen( domain ) + wcsByteLen( user ) + wcsByteLen( password );
    BYTE* authInfoBuf = new BYTE[authInfoSize];
    KERB_INTERACTIVE_LOGON* authInfo = (KERB_INTERACTIVE_LOGON*)authInfoBuf;
    authInfo->MessageType = KerbInteractiveLogon;
    size_t offset = sizeof(KERB_INTERACTIVE_LOGON);
    InitUnicodeString( authInfo->LogonDomainName, domain, authInfoBuf, offset );
    InitUnicodeString( authInfo->UserName, user, authInfoBuf, offset );
    InitUnicodeString( authInfo->Password, password, authInfoBuf, offset );

    // find the Kerberos security package
    char packageNameRaw[] = MICROSOFT_KERBEROS_NAME_A;
    LSA_STRING packageName;
    packageName.Buffer = packageNameRaw;
    packageName.Length = packageName.MaximumLength = (USHORT)strlen( packageName.Buffer );
    ULONG packageId;
    LsaLookupAuthenticationPackage( lsa, &packageName, &packageId );

    // create a dummy origin and token source
    LSA_STRING origin = {};
    origin.Buffer = _strdup( "TestAppFoo" );
    origin.Length = (USHORT)strlen( origin.Buffer );
    origin.MaximumLength = origin.Length;
    TOKEN_SOURCE source = {};
    strcpy( source.SourceName, "foobar" );
    AllocateLocallyUniqueId( &source.SourceIdentifier );

    void* profileBuffer;
    DWORD profileBufLen;
    LUID luid;
    HANDLE token;
    QUOTA_LIMITS qlimits;
    NTSTATUS subStatus;
    NTSTATUS status = LsaLogonUser( lsa, &origin, Interactive, packageId,
     &authInfo, authInfoSize, 0, &source, &profileBuffer, &profileBufLen,
     &luid, &token, &qlimits, &subStatus );
    if( status != ERROR_SUCCESS )
    {
        ULONG err = LsaNtStatusToWinError( status );
        printf( "LsaLogonUser failed: %x\n", status );
        return 1;
    }
}

size_t wcsByteLen( const wchar_t* str )
{
    return wcslen( str ) * sizeof(wchar_t);
}

void InitUnicodeString( UNICODE_STRING& str, const wchar_t* value,
 BYTE* buffer, size_t& offset )
{
    size_t size = wcsByteLen( value );
    str.Length = str.MaximumLength = (USHORT)size;
    str.Buffer = (PWSTR)(buffer + offset);
    memcpy( str.Buffer, value, size );
    offset += size;
}

最佳答案

您烦恼了LsaLogonUser()的参数之一；而不是&authInfo，您应该只传递authInfo。发生在每个人:)

关于windows - 如何正确调用LsaLogonUser进行交互式登录？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/7627750/