使用 Iterable 進行數字字符串排序<component>在 groovy</component>

Question

我想從Iterable<Component> components中對版本進行排序。 Нow 當我在控制台中打印時，它會顯示以下結果：

artifact 1.0.1
artifact 1.0.10
artifact 1.0.11
artifact 1.0.12
artifcat 1.0.2
artifcat 1.0.3
artifcat 1.0.4

這是我的代碼

import org.sonatype.nexus.repository.storage.Component
import org.sonatype.nexus.repository.storage.Query
import org.sonatype.nexus.repository.storage.StorageFacet

def repoName = "artifact"

log.info("delete components for repository: " + repoName)


def repo = repository.repositoryManager.get(repoName)
def tx = repo.facet(StorageFacet).txSupplier().get()
try {
tx.begin()
    Iterable<Component> components = tx.findComponents(Query.builder()
      .where('version < ').param('1.1.0')
      .build(), [repo])
    tx.commit()
    
    for(Component c : components) {
        log.info("Name " + c.name() + " Version" + c.version())
    }
} catch (Exception e) {
    log.warn("Transaction failed {}", e.toString())
    tx.rollback()
} finally {
    tx.close()
}

Answer 1

可以按版本進行一些簡單的排序，如下所示：

def components = []
'''\
artifact 1.2.1
artifact 1.0.1
artifact 1.0.10
artifact 2.0.10
artifact 1.0.11
artifact 1.0.12
artifact 1.4.12
artifcat 1.0.2
artifcat 1.0.3
artifcat 1.0.4'''.splitEachLine( ' ' ){ name, version ->
  components << [ name:name, version:version ]
}

// augment each component with numeric represenation of version
components.each{
  it.versionNumeric = it.version.split( /\./ )*.toInteger().reverse().withIndex().sum{ num, pow -> 100.power( pow ) * num }
}

components.sort{ it.versionNumeric }*.version.join '\n'

印刷

1.0.1
1.0.2
1.0.3
1.0.4
1.0.10
1.0.11
1.0.12
1.2.1
1.4.12
2.0.10

Answer 2

另一個例子：

def components = '''\
artifact 1.2.1
artifact 1.0.1
artifact 1.0.10
artifact 2.0.10
artifact 10.2
artifact 1.0.11
artifact 1.0.12
artifact 1.4.12
artifcat 1.0.2
artifcat 1.0.3
artifcat 1.0.4
artifact 1.0.4.2'''.readLines()*.tokenize(' ').collect { name, version ->
  [name: name, version: version]
}

def sorted = components.sort { a, b ->
  def f = { it.version.tokenize('.')*.toInteger() }
  [f(a), f(b)].transpose().findResult { ai, bi -> 
    ai <=> bi ?: null 
  } ?: a.version <=> b.version
}

sorted.each { c -> 
    println c.version
}

打印：

─➤ groovy solution.groovy
1.0.1
1.0.2
1.0.3
1.0.4
1.0.4.2
1.0.10
1.0.11
1.0.12
1.2.1
1.4.12
2.0.10
10.2

請注意，此解決方案還（至少或多或少）處理具有不同數量元素的版本，例如10.2和1.0.4.2 。