lucene Alfresco REST核心API按路径查找节点

7d7tgy0s  于 2022-11-23  发布在  Lucene
关注(0)|答案(2)|浏览(207)

我正在开发一个基于REST API的Alfresco客户端。
但是我还没有找到一个方法,允许用它的路径来检索一个节点。

[ base url: /alfresco/api/-default-/public/alfresco/versions/1 , api version: 1 ]

你能告诉我怎么做吗?

7uhlpewt

7uhlpewt1#

下面的代码将从文件夹/文件路径返回您所期望的noderef。

public static void main(String[] args) throws IOException {
        client = new AlfrescoClient.Builder().connect("http://localhost:8080/alfresco", "admin", "admin").build();
        NodeRepresentation node = findNodeByPath("Sites/testsite/documentLibrary/TestFolder/UploadTest.txt");
        System.out.println(node == null ? "null" : node.getName());
    }

private static NodeRepresentation findNodeByPath(String path) throws IOException {
    if (path.startsWith("/"))
        path = path.substring(1);
    if (path.endsWith("/"))
        path = path.substring(0, path.length() - 1);
    return findNodeByPath("-root-", path);
}

private static NodeRepresentation findNodeByPath(String parentNodeId, String path) throws IOException {
    String[] pathParts = path.split("/");
    String name = pathParts[0];
    String remaining = pathParts.length == 1 ? "" : path.substring(name.length() + 1, path.length());
    List<NodeRepresentation> children = client.getNodesAPI().listNodeChildrenCall(parentNodeId).execute().body()
            .getList();
    for (NodeRepresentation child : children) {
        if (child.getName().equals(name)) {
            return pathParts.length == 1 ? child : findNodeByPath(child.getId(), remaining);
        }
    }
    return null;
}
eqfvzcg8

eqfvzcg82#

你可以尝试使用AFTS和PATH操作符,它应该工作。你基本上是在写一个查询,而不是使用一个特定的API方法来"按路径查找"。例如:

{
  "query": {
    "language": "afts",
    "query": "PATH:'/app:company_home/st:sites/cm:swsdp/cm:documentLibrary/cm:Agency_x0020_Files//*' OR PATH:'/app:company_home/st:sites/cm:swsdp/cm:documentLibrary/cm:Budget_x0020_Files//*'"
  }
}

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