我有一个字符串数据，如下所示：
a<- "\n Update Your Profile to Dissolve This Message\nSocial Media Learning and behaviour\n Uploaded on May 3, 2020 at 10:56 in Research\n View Forum\n \n"
为此，我必须使用以下代码提取字符串“ Social Media Learning and behaviour”：
gsub(" Uploaded on .* ", "", gsub("\n Update Your Profile to Dissolve This Message\n", "",a))
这给了我如下输出
"Social Media Learning and behaviour\n\n"
我无法匹配确切的模式。没有“ \ n \ n”的情况下提取“社交媒体学习和行为”的确切模式是什么？
答案 0 :(得分：1)
您可以捕获组中的上一行，并匹配包含Uploaded的下一行：
(.*)\r?\n[^\S\r\n]+Uploaded on
a<- "\n Update Your Profile to Dissolve This Message\nSocial Media Learning and behaviour\n Uploaded on May 3, 2020 at 10:56 in Research\n View Forum\n \n"
stringr::str_match(a, "(.*)\\r?\\n[^\\S\\r\\n]+Uploaded on")
答案 1 :(得分：0)
您可以提取SELECT
Employee_conferenceDay
From Employee
Where ID_Employee IN (
Select
ID_Employee
From Conference
Where ID_conference = 206247
);
和"Update Your Profile to Dissolve This Message"之间的部分
"Uploaded on"您还可以使用sub(".*Update Your Profile to Dissolve This Message\n(.*)\n\\s+Uploaded on.*", "\\1", a)
#[1] "Social Media Learning and behaviour"
中的str_match
stringr