SPFX - Listview command - Get current folder

Question

I have created a SPFX Listview command and I need to get the current URL of the folder that I am in.

Does anyone know if this is possible? I can't find any documentation regarding this on sp-pnp-js github or other sources.

Any help is much appreciated :)

Answer 1

SPFx has some utility classes to help you get the folder url. Its not available in pageContextInfo or any other page property. However, you can split it from the query string's Id parameter.

Add the below import statement:

import { Log, UrlQueryParameterCollection } from '@microsoft/sp-core-library';

And in your execution logic, use it as:

@override
  public onExecute(event: IListViewCommandSetExecuteEventParameters): void {         

      var queryParameters = new UrlQueryParameterCollection(window.location.href);
      var current_folder_path= url_query_param.getValue("Id") || url_query_param.getValue("RootFolder");

      if (queryParameters.getValue("Id")) {
          var folderUrl = decodeURIComponent(current_folder_path);
          Dialog.alert(`${folderUrl}`);
      }        
  }

End result - in the Shared Document library, have created a folder named Test:

Answer 2

It is pretty easy. Because you know which item is selected, you know url of the file. event.selectedRows[0] holds a lot of information about the selected item. Or you can use query string.

@override
public onExecute(event: IListViewCommandSetExecuteEventParameters): void {
  switch (event.itemId) {
    case 'COMMAND_1':
      {
        var folderUrl: string;

        // Get file url of first selected item
        if (event.selectedRows.length > 0) {
          const fileUrl: string = event.selectedRows[0].getValueByName("FileRef");

          // Trim the file name and you have server relative folder url
          folderUrl = fileUrl.substring(0, fileUrl.lastIndexOf('/'));
        } else {
          // Get url form query string
          folderUrl = this.getFolderUrlFromQueryString();
        }

        Dialog.alert(folderUrl);
        break;
      }
    default:
      throw new Error('Unknown command');
  }
}

private getFolderUrlFromQueryString(): string {
  const id = this.getQueryStringParameter("Id");
  const rootFolder = this.getQueryStringParameter("RootFolder");

  var url = id != null ? id : rootFolder;
  if (url === undefined) {
    // in case of root folder
    url = this.context.pageContext.list.serverRelativeUrl;
  }

  return decodeURIComponent(url);
}

private getQueryStringParameter(paramToRetrieve: string) {
  if (document.URL.indexOf("?") == -1)
    return null;

  var params = document.URL.split("?")[1].split("&");
  var strParams = "";
  for (var i = 0; i < params.length; i = i + 1) {
    var singleParam = params[i].split("=");
    if (singleParam[0].toLowerCase() == paramToRetrieve.toLowerCase())
      return singleParam[1];
  }
}

Answer 3

I assume you mean a folder once inside a list or library. Here's how I'd approach it...

var folderUrl = decodeURIComponent(window.location.href.split("?")[1].split("RootFolder=")[1].split("&FolderCTID")[0]);

var folderServerRelativeUrl = this.context.pageContext.web.absoluteUrl + folderUrl;

folderUrl gets the entire query parameter, then gets everything after "RootFolder=" but before "&FolderCTID", then decodes the %2F characters into "/".

Then using the page context concatenate the webs' absoluteUrl and the folderUrl.

Answer 4

As you mentioned that you want the folder URL without selecting any items or folders then you can write your code under the "onListViewUpdated" function. If the returned URL is empty then your outside of the folder.

  @override
  public onListViewUpdated(event: IListViewCommandSetListViewUpdatedParameters): void {
    Dialog.alert(this.getParameterByName('id'));    
  }

  private getParameterByName(name) {
    var url = window.location.href;
    name = name.replace(/[\[\]]/g, "\\$&");
    var regex = new RegExp("[?&]" + name + "(=([^&#]*)|&|#|$)"),
      results = regex.exec(url);
    if (!results) return null;
    if (!results[2]) return '';
    return decodeURIComponent(results[2].replace(/\+/g, " "));
  }