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创建 Menu 不够方便
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#I9KB1
qtdebug
Opened this issue
2016-06-21 17:45
感觉 MenuApi.createMenu() 创建菜单不够方便，希望提供一个结构，MenuApi.createMenu() 直接接受 Menu 的 JSON 字符串创建菜单，例如 ``` public ResultType createMenu(String menuJson) { BeanUtil.requireNonNull(menu, "menu is null"); String url = BASE_API_URL; if (BeanUtil.isNull(menu.getMatchrule())) { //普通菜单 LOG.debug("创建普通菜单....."); url += "cgi-bin/menu/create?access_token=#"; } else { //个性化菜单 LOG.debug("创建个性化菜单....."); url += "cgi-bin/menu/addconditional?access_token=#"; } BaseResponse response = executePost(url, menuJson); return ResultType.get(response.getErrcode()); } ```
感觉 MenuApi.createMenu() 创建菜单不够方便，希望提供一个结构，MenuApi.createMenu() 直接接受 Menu 的 JSON 字符串创建菜单，例如 ``` public ResultType createMenu(String menuJson) { BeanUtil.requireNonNull(menu, "menu is null"); String url = BASE_API_URL; if (BeanUtil.isNull(menu.getMatchrule())) { //普通菜单 LOG.debug("创建普通菜单....."); url += "cgi-bin/menu/create?access_token=#"; } else { //个性化菜单 LOG.debug("创建个性化菜单....."); url += "cgi-bin/menu/addconditional?access_token=#"; } BaseResponse response = executePost(url, menuJson); return ResultType.get(response.getErrcode()); } ```
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