我正试图从 const deleteNotification$ = (id, { ajax }) => ajax({ url: `api/delete/{id}` });
// showYesNo is an action to eventually show a dialog using this approach https://stackoverflow.com/a/35641680/235659
const showYesNo = payload => ({
type: SHOW_MODAL,
modalType: MODAL_TYPE_YES_NO,
modalProps: { ...payload },
});
const deleteNotificationEpic = (action$, store, dependencies) => {
let uid = dependencies.uid; // dependencies.uid is added here to allow passing the uid during unit test.
return merge(
// Show confirmation dialog.
action$.pipe(
ofType(NOTIFICATION_DELETE_REQUEST),
map(action => {
uid = shortid.generate();
return showYesNo({
message: 'NOTIFICATION_DELETE_CONFIRMATION',
payload: {
notificationId: action.notificationId,
uid,
},
})
}),
),
// Deletes the notification if the user clicks on Yes
action$.pipe(
ofType(MODAL_YES_CLICKED),
filter(({ payload }) => payload.uid === uid),
mergeMap(({ payload }) =>
deleteNotification$(payload.notificationId, dependencies).pipe(
mergeMap(() => of(deleteNotificationSuccess())),
catchError(error => of(deleteNotificationSuccess(error))),
),
),
),
);
};
命令中获取USB名称（MYUSB）。目前，我有以下输出。我想获取以blkid开头的USB名称。所以输出只是LABEL。我怎样才能做到这一点？
MYUSB答案 0 :(得分：1)
尝试blkid | grep -o -P '(?<=LABEL=")([^"]*)'
这应该将blkid的输出传递给grep，它只返回与正则表达式匹配的字符串部分。
应该注意的是，如果blkid的输出中出现多个标签，则会输出所有标签，每个标签都在一个新行上。因此，要仅获取最后一个标签，请使用：
blkid | grep -o -P '(?<=LABEL=")([^"]*)' | tail -1
正则表达式解释：
(?<=LABEL=") - 断言字符串中当前位置前面的内容为LABEL="。([^"]*) - 匹配并捕获零个或多个不是"的字符（引号之间的部分）。