我需要从file.txt中查找带有大量名称的行,例如clientLogin=a@yahoo.com,clientLogin=b@gmail.com.
file.txt有垃圾邮件,邮件是email=a@yahoo.com email=b@gmail.com.我需要过滤掉这些
一旦我得到这些行,我需要grep gmail和yahoo并获得他们的计数
List l = new ArrayList{a@yahoo.com,b@gmail.com}
def gmail = ['sh','-c','grep "clientLogin="$l.get(0) file.txt' | grep gmail | wc -l ]
def yahoo = ['sh','grep "clientLogin="$l.get(1) file.txt' | grep yahoo| wc -l ]
这不起作用.如何动态替换$l.get(1)值？
问题是${l.get(0)}必须在“”内,
即：
def gmail = ['sh','grep "clientLogin=${l.get(0)}" file.txt' | grep gmail | wc -l ]
所以它看起来像：
def gmail = ['sh','grep "clientLogin=a@yahoo.com" file.txt' | grep gmail | wc -l ]
但是clientLogin = ${l.get(0)}不会产生结果.我不确定我哪里出错了.
感谢您的建议,但它不会产生结果,至少在我尝试时.
file.txt有很多垃圾和类似的模式：
Into the domain clientLogin=a@yahoo.com exit on 12/01/2008 etc..
因此,我做到了
def ex = ['sh','grep "domain clientLogin=$client" file.txt'| grep "something more" | wc -l]
这样我可以按照自己的意愿链接grep并最终降落在我需要的数量上.
如果我使用的话,我不确定是否可以连接greps
def ex = ['grep',"$client",'file.txt']
感谢您的输入.
解决方法
你已经在使用groovy,使用正则表达式给你答案吗？
def file = new File("file.txt")
file.delete() // clear out old version for multiple runs
file << """
foobar clientLogin=a@yahoo.com baz quux # should match a@yahoo.com
foobar email=a@yahoo.com baz quux
foobar email=b@gmail.com bal zoom
foobar clientLogin=a@yahoo.com baz quux # should match a@yahoo.com
foobar clientLogin=b@gmail.com bal zoom # should match b@gmail.com
foobar email=b@gmail.com bal zoom
"""
def emailList = ["a@yahoo.com","b@gmail.com"]
def emailListGroup = emailList.join('|')
def pattern = /(?m)^.*clientLogin=($emailListGroup).*$/
def resultMap = [:]
(file.text =~ pattern).each { fullLine,email ->
resultMap[email] = resultMap[email] ? resultMap[email] + 1 : 1
}
assert resultMap["a@yahoo.com"] == 2
assert resultMap["b@gmail.com"] == 1
这对我来说比尝试使用流程并使用它更加清晰,而且它只会选择你正在寻找的“clientLogin =(email)”的确切行.
