Question

我有树状结构

class FSNode {
    FSEntity data
    def parent
    def children = []

    boolean isRoot() {
        parent == null
    }
}

我像这样实例化

// create filesystem folder
def fsEntityDirectory = new FSEntity(name: 'folder')
def fsDirectoryNode = new FSNode(parent: null, data: fsEntityDirectory)

def fsEntityDirectory2 = new FSEntity(name: 'folder1')
def fsDirectoryNode2 = new FSNode(parent: fsDirectoryNode, data: fsEntityDirectory2)

// create filesystem file
def fsEntity = new FSEntity(name: 'basic.txt', size: 10) // file system entity
def fsEntityNode = new FSNode(data: fsEntity, parent: fsDirectoryNode)

def fsEntity2 = new FSEntity(name: 'file.png', size: 12)
def fsEntityNode2 = new FSNode(data: fsEntity2, parent: fsDirectoryNode2)
fsDirectoryNode2.children << fsEntityNode2

fsDirectoryNode.children << fsDirectoryNode2
fsDirectoryNode.children << fsEntityNode

最后我想得到这样的东西

<filesystem-root>
    <directory name="folder">
        <directory name="folder1">
            <file name="file.png" size="12" />
        </directory>
        <file name="basic.txt" size="10" />
    </directory>
</filesystem-root>

这是我序列化的方式

static def serializeFS(node, output, xml = new MarkupBuilder(output)) {
    if (node.children) {
        xml.directory(name: node.data.name)
        node.children.each {
            xml = serializeFS(it, output, xml)
        }
    } else {
        node.data.size ?
                xml.file(name: node.data.name, size: node.data.size) :
                xml.directory(name: node.data.name, "")
    }

    xml
}

static def wrapFS(output, xml) {
    xml."filesystem-root"(output)
}

但最终我得到了

<directory name='folder' />
<directory name='folder1' />
<file name='file.png' size='12' />
<file name='basic.txt' size='10' />

有什么问题？

Answer 1

您可以尝试以下方式：

def serializeFSNode(FSNode node, MarkupBuilder m) {
    if(node.children) {
        m.directory(name: node.data.name) {
            node.children.each { child ->
                serializeFSNode(child, m)
            }
        }
    }
    else {
        m."${node.data.size ? 'file' : 'directory'}"(name: node.data.name, size: node.data.size)
    }
}

以递归方式从树状结构创建xml

1 个答案: