0

我很少坚持使用 XML。XML中的路径sqlfile必须取自使用加载 SQLfiles 的文件夹,dir并且dir1我在构建 XML 时遇到问题:

<databaseChangeLog

<changeSet author="John" id="JRIA" failOnError="true" runAlways="false">
    <sqlFile path="path.sql" relativeToChangelogFile="true" encoding="utf8" />
    <rollback>
        <sqlFile path="rollback/path.sql" relativeToChangelogFile="true" encoding="utf8" />
    </rollback>
</changeSet>

我的例子:

import groovy.io.FileType
import groovy.xml.*


def dir = new File("C:\\Users\\John\\git\\changelogs\\version1\\db")
def dir1 = new File("C:\\Users\\John\\git\\changelogs\\version1\\rollback")

def sw = new StringWriter()
def xml = new groovy.xml.MarkupBuilder(sw)
xml.changeSet(author:"John", ID:"JIRA", failOnError: "True", runAlways: "false"){
    sqlFile(path:"From DIR", relativeToChangelogFile="true")
    rollback(){
        sqlFile(path:"From DIR1", relativeToChangelogFile="true")}
}

如何使用dirdir1以良好的方式生成该 XML?以及如何获取特定的扩展文件(sql)

4

1 回答 1

0

这很简单,只需使用.each

xml.dataBaseChangeLog(){
dir.eachFileRecurse(FileType.FILES) { file ->   
    changeSet(author:"John", ID:"JIRA", failOnError: "True", runAlways: "false")
    sqlFile(path:file, relativeToChangelogFile="true")    
    rollback(){       
        sqlFile(path:file, relativeToChangelogFile="true")
}}}
于 2017-09-06T11:38:41.697 回答