def test_non_polymerization_case(self):
        '''流程非聚合'''
        menu1 = '流程'
        menu2 = '审批送出设置'  # 主页打开菜单时使用
        menu3 = '流程非聚合'
        name = "流程非聚合"
        #time.sleep(0.5)
        comp = ProcessApproverPage(self.driver)
        # 判断是否要删除记录
        lp = ListViewPage(self.driver)
        lp.judge_delete(name)
        #李玲建单提交
        comp.launch_a_flowform(name)
        #time.sleep(0.5)
        #退出当前登录,切换账号并打开菜单记录
        comp.logoff_and_openrecord('weiqiang', '123456', menu1, menu2, menu3,
                                   name)
        #time.sleep(0.5)
        comp.direct_sumit()
        #退出当前登录,切换账号并打开菜单记录for表单
        comp.logoff_and_openmenu('zjl01', '123456', menu1, menu2, menu3)
        text = comp.openagain_to_getapprover(name)
        # 断言
        self.assertIn('张强', text, msg=name + "检验不通过")
        # 判断总经理此时有无提交按钮
        bb = comp.is_elementPresent()
        self.assertEqual("true", bb, msg=name + "检验不通过")
        #time.sleep(0.5)
        #提交后判断流程是否已归档
        bool = comp.is_filed_for_aftersumbit(name)
        self.assertIsNone(bool, msg=name + "检验不通过")

        def init(self):
            self.test_non_polymerization_case()
Exemple #2
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 def test_approver_iscript_case(self):
     '''流程审批人for_iscript形式'''
     menu1 = '流程'
     menu2 = '审批人设置'  # 主页打开菜单时使用
     menu3 = ''
     name = "流程审批人设置_通过iscript形式"
     #time.sleep(0.5)
     comp = ProcessApproverPage(self.driver)
     # 判断是否要删除记录
     lp = ListViewPage(self.driver)
     lp.judge_delete(name)
     #李玲新建单据
     self.lilingnew(name)
     #退出当前登录,切换账号并打开菜单记录for视图
     comp.logoff_and_openrecord('weiqiang', '123456', menu1, menu2, menu3,
                                name)
     comp.direct_sumit()  # 直接提交
     #退出切换账号,查看是否有提交按钮
     bb = comp.logoff_and_check_submitbtn('zhangqiang', '123456', menu1,
                                          menu2, menu3, name)
     self.assertEqual("true", bb, msg=name + "检验不通过")
     #time.sleep(0.5)
     #表单是否已归档
     bool = comp.is_filed_for_aftersumbit(name)
     self.assertIsNone(bool, msg=name + "检验不通过")
    def test_Hang_case(self):
        '''流程挂起'''
        menu1 = '流程'
        menu2 = '流程通知'  # 主页打开菜单时使用
        menu3 = '流程挂起'
        name = "流程挂起"
        #time.sleep(0.5)
        comp = ProcessApproverPage(self.driver)
        # 判断是否要删除记录
        lp = ListViewPage(self.driver)
        lp.judge_delete(name)
        #李玲建单提交
        comp.launch_a_flowform(name)
        #退出登录
        comp.goback()
        #切换张强账号登陆,
        comp.switch_account("zhangqiang", "123456")
        comp.close_message()
        # 打开菜单
        comp.open_m(menu1, menu2, menu3)
        #time.sleep(0.5)
        #检查有没有提交按钮
        bb = comp.is_submit_existed(name)
        self.assertEqual("true", bb, msg=name + "检验不通过")
        #time.sleep(0.5)
        comp.click_hangbtn()
        #time.sleep(0.5)
        bb2 = comp.is_elementPresent()
        self.assertEqual("false", bb2, msg=name + "检验不通过")
        #time.sleep(0.5)
        comp.click_retracementbtn()
        #time.sleep(0.5)
        bb3 = comp.is_elementPresent()
        self.assertEqual("true", bb3, msg=name + "检验不通过")
        #time.sleep(0.5)
        comp.direct_sumit()
        #time.sleep(0.5)
        #'''表单是否已归档'''
        bool = comp.is_filed_for_openagin(name)
        self.assertIsNone(bool, msg=name + "检验不通过")

        def init(self):
            self.test_Hang_case()
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 def test_polymerization_case(self):
     '''流程聚合'''
     menu1 = '流程'
     menu2 = '审批送出设置'  # 主页打开菜单时使用
     menu3 = '流程聚合'
     name = "流程聚合"
     #time.sleep(0.5)
     comp = ProcessApproverPage(self.driver)
     #判断是否要删除记录
     lp = ListViewPage(self.driver)
     lp.judge_delete(name)
     #李玲建单提交
     comp.launch_a_flowform(name)
     #time.sleep(0.5)
     #退出当前登录,切换账号并打开菜单记录for视图
     comp.logoff_and_openrecord('weiqiang', '123456', menu1, menu2, menu3,
                                name)
     #time.sleep(0.5)
     comp.direct_sumit()
     #再次打开记录获取流程状态处理人
     text = comp.openagain_to_getapprover(name)
     self.assertIn('张强', text, msg=name + "检验不通过")
     #退出切换账号,查看是否有提交按钮
     bb = comp.logoff_and_check_submitbtn('zjl01', '123456', menu1, menu2,
                                          menu3, name)
     self.assertEqual("false", bb, msg=name + "检验不通过")
     # 退出当前登录,切换账号并打开菜单记录for视图
     comp.logoff_and_openrecord('zhangqiang', '123456', menu1, menu2, menu3,
                                name)
     #time.sleep(0.5)
     comp.direct_sumit()
     #time.sleep(0.5)
     # 退出当前登录,切换账号并打开菜单记录for视图
     comp.logoff_and_openrecord('zjl01', '123456', menu1, menu2, menu3,
                                name)
     #time.sleep(0.5)
     #表单是否已归档
     bool2 = comp.is_filed_for_aftersumbit(name)
     self.assertIsNone(bool2, msg=name + "检验不通过")
Exemple #5
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    def test_retreat_case(self):
        '''流程回撤'''
        menu1 = '流程'
        menu2 = '流程通知'  # 主页打开菜单时使用
        menu3 = '流程回撤'
        name = "流程回撤"
        #time.sleep(0.5)
        comp = ProcessApproverPage(self.driver)
        # 判断是否要删除记录
        lp = ListViewPage(self.driver)
        lp.judge_delete(name)
        #李玲建单提交
        comp.launch_a_flowform(name)
        #退出切换账号,查看是否有提交按钮
        bb = comp.logoff_and_check_submitbtn("zhangqiang", "123456", menu1,
                                             menu2, menu3, name)
        self.assertEqual("true", bb, msg=name + "检验不通过")
        #time.sleep(0.5)
        #退出切换账号,查看是否有提交按钮
        bb2 = comp.logoff_and_check_submitbtn("liling", "123456", menu1, menu2,
                                              menu3, name)
        self.assertEqual("false", bb2, msg=name + "检验不通过")
        #time.sleep(0.5)
        bool = comp.is_btn_existed("act_flow_retracement")
        self.assertTrue(bool, msg=name + "检验不通过")
        #time.sleep(0.5)
        #点击回撤按钮
        comp.click_retreatbtn()
        #time.sleep(0.5)
        text = comp.get_approver()
        self.assertIn("李玲", text, msg=name + "检验不通过")
        #time.sleep(0.5)
        comp.direct_sumit()
        text2 = comp.openagain_to_getapprover(name)
        self.assertIn("张强", text2, msg=name + "检验不通过")

        def init(self):
            self.test_retreat_case()