Question

我使用robotframework和appium进行UI自动化，当打开应用程序时，它将弹出不同的弹出窗口（如更新版本，重复登录等），我想要一个条件始终为true的循环，在此循环中可以在我看到弹出窗口取消按钮后立即单击它，直到所有弹出窗口消失并且我不需要关心弹出窗口的顺序我该如何在robotframework中做到这一点？

我不会降低弹出窗口的顺序，所以有时我找不到元素，当前的解决方案是我多次使用for循环，但这并不是最好的方法

:FOR  |  ${i}  |  IN RANGE  |  2

    \    ${binaryPopStatus}  |  Run Keyword And Return Status   | Page Should Contain Element  |  &{locator_binary_popup_close_btn}[${PLATFORM}]

    \    Run Keyword If |   '${binaryPopStatus}'=='True'  |  Click Element  |  &{locator_binary_popup_close_btn}[${PLATFORM}]

    \    ${sysPopStatus} |   Run Keyword And Return Status  |  Page Should Contain Element  |  &{locator_sys_popup}[${PLATFORM}]

    \    Run Keyword If  |  '${sysPopStatus}'=='True'  |  Click Element |   &{locator_sys_popup_deny}[${PLATFORM}]

    \    ${versionPopStatus}  |  Run Keyword And Return Status  |  Page Should Contain Element  |  &{locator_version_popup_cancel}[${PLATFORM}]

    \    Run Keyword If  |  '${versionPopStatus}'=='True'  |  Click Element    |&{locator_version_popup_cancel}[${PLATFORM}]

    \    ${locationPopStatus}  |  Run Keyword And Return Status  |  Page Should Contain Element  |  &{locator_enable_location_popup_notnow}[${PLATFORM}]

    \    Run Keyword If  |  '${locationPopStatus}'=='True'  |  Click Element  |  &{locator_enable_location_popup_notnow}[${PLATFORM}]

    \    ${enjoyPopStatus}  |  Run Keyword And Return Status  |  Page Should Contain Element    |id=com.sixdays.truckerpath.dev:id/rate_app_not_now

    \    Run Keyword If  |  '${enjoyPopStatus} '=='True'  |  Click Element |   id=com.sixdays.truckerpath.dev:id/rate_app_not_now

    \    ${loggedPopStatus}  |  Run Keyword And Return Status  |  Page Should Contain Element  |  &{locator_logged_cancel_btn}[${PLATFORM}]

    \    Run Keyword If  |  ${loggedPopStatus}  |  Click Element  |  &{locator_logged_cancel_btn}[${PLATFORM}]

如何在机器人框架中始终满足条件的情况下循环？

0 个答案: