获取文档中的所有链接

Question

假设 Google Docs/Drive 中的“普通文档”（例如段落、列表、表格）包含分散在整个内容中的外部链接，您如何使用 Google Apps 脚本编译存在的链接列表？

具体来说，我想通过在每个 url 中搜索 oldText 并在每个 url 中将其替换为 newText 来更新文档中所有损坏的链接，但不是文本。

我认为开发文档的replacing text 部分不是我需要的——我需要扫描文档的每个元素吗？我可以只使用editAsText 并使用 html 正则表达式吗？示例将不胜感激。

Answer 1

这只是最痛苦的！ 代码可用as part of a gist。

^{_{是的，我不会拼写。}}

getAllLinks

这是一个实用函数，用于扫描文档中的所有 LinkUrls，并以数组的形式返回它们。

/**
 * Get an array of all LinkUrls in the document. The function is
 * recursive, and if no element is provided, it will default to
 * the active document's Body element.
 *
 * @param {Element} element The document element to operate on. 
 * .
 * @returns {Array}         Array of objects, vis
 *                              {element,
 *                               startOffset,
 *                               endOffsetInclusive, 
 *                               url}
 */
function getAllLinks(element) {
  var links = [];
  element = element || DocumentApp.getActiveDocument().getBody();
  
  if (element.getType() === DocumentApp.ElementType.TEXT) {
    var textObj = element.editAsText();
    var text = element.getText();
    var inUrl = false;
    for (var ch=0; ch < text.length; ch++) {
      var url = textObj.getLinkUrl(ch);
      if (url != null) {
        if (!inUrl) {
          // We are now!
          inUrl = true;
          var curUrl = {};
          curUrl.element = element;
          curUrl.url = String( url ); // grab a copy
          curUrl.startOffset = ch;
        }
        else {
          curUrl.endOffsetInclusive = ch;
        }          
      }
      else {
        if (inUrl) {
          // Not any more, we're not.
          inUrl = false;
          links.push(curUrl);  // add to links
          curUrl = {};
        }
      }
    }
    if (inUrl) {
      // in case the link ends on the same char that the element does
      links.push(curUrl); 
    }
  }
  else {
    var numChildren = element.getNumChildren();
    for (var i=0; i<numChildren; i++) {
      links = links.concat(getAllLinks(element.getChild(i)));
    }
  }

  return links;
}

findAndReplaceLinks

此实用程序基于 getAllLinks 进行查找和替换功能。

/**
 * Replace all or part of UrlLinks in the document.
 *
 * @param {String} searchPattern    the regex pattern to search for 
 * @param {String} replacement      the text to use as replacement
 *
 * @returns {Number}                number of Urls changed 
 */
function findAndReplaceLinks(searchPattern,replacement) {
  var links = getAllLinks();
  var numChanged = 0;
  
  for (var l=0; l<links.length; l++) {
    var link = links[l];
    if (link.url.match(searchPattern)) {
      // This link needs to be changed
      var newUrl = link.url.replace(searchPattern,replacement);
      link.element.setLinkUrl(link.startOffset, link.endOffsetInclusive, newUrl);
      numChanged++
    }
  }
  return numChanged;
}

演示用户界面

为了演示这些实用程序的使用，这里有几个 UI 扩展：

function onOpen() {
  // Add a menu with some items, some separators, and a sub-menu.
  DocumentApp.getUi().createMenu('Utils')
      .addItem('List Links', 'sidebarLinks')
      .addItem('Replace Link Text', 'searchReplaceLinks')
      .addToUi();
}

function searchReplaceLinks() {
  var ui = DocumentApp.getUi();
  var app = UiApp.createApplication()
                 .setWidth(250)
                 .setHeight(100)
                 .setTitle('Change Url text');
  var form = app.createFormPanel();
  var flow = app.createFlowPanel();
  flow.add(app.createLabel("Find: "));
  flow.add(app.createTextBox().setName("searchPattern"));
  flow.add(app.createLabel("Replace: "));
  flow.add(app.createTextBox().setName("replacement"));
  var handler = app.createServerHandler('myClickHandler');
  flow.add(app.createSubmitButton("Submit").addClickHandler(handler));
  form.add(flow);
  app.add(form);
  ui.showDialog(app);
}

// ClickHandler to close dialog
function myClickHandler(e) {
  var app = UiApp.getActiveApplication();

  app.close();
  return app;
}

function doPost(e) {
  var numChanged = findAndReplaceLinks(e.parameter.searchPattern,e.parameter.replacement);
  var ui = DocumentApp.getUi();
  var app = UiApp.createApplication();
  
  sidebarLinks(); // Update list

  var result = DocumentApp.getUi().alert(
      'Results',
      "Changed "+numChanged+" urls.",
      DocumentApp.getUi().ButtonSet.OK);
}


/**
 * Shows a custom HTML user interface in a sidebar in the Google Docs editor.
 */
function sidebarLinks() {
  var links = getAllLinks();
  var sidebar = HtmlService
          .createHtmlOutput()
          .setTitle('URL Links')
          .setWidth(350 /* pixels */);

  // Display list of links, url only.
  for (var l=0; l<links.length; l++) {
    var link = links[l];
    sidebar.append('<p>'+link.url);
  }
  
  DocumentApp.getUi().showSidebar(sidebar);
}

Answer 2

我为您的第一个问题提供了另一个更简短的答案，关于迭代文档正文中的 all 链接。这个指导性代码返回当前文档正文中的一个平面链接数组，其中每个链接由一个对象表示，其条目指向文本元素 (text)、包含它的段落元素或列表项元素 (@987654322 @)、链接出现的文本中的偏移索引 (startOffset) 和 URL 本身 (url)。希望您会发现它很容易满足您自己的需求。

它使用getTextAttributeIndices() 方法而不是遍历文本的每个字符，因此预计比以前写的答案要快得多。

编辑：自从最初发布这个答案以来，我修改了几次函数。它现在还 (1) 包括每个链接的 endOffsetInclusive 属性（请注意，对于延伸到文本元素末尾的链接，它可以是 null - 在这种情况下，可以使用 link.text.length-1 代替）； (2) 在文档的所有部分中查找链接，不仅是正文，并且 (3) 包括 section 和 isFirstPageSection 属性以指示链接所在的位置； (4) 接受参数mergeAdjacent，当设置为真时，将只返回一个链接条目，用于链接到同一 URL 的连续文本段（例如，如果部分文本是样式与另一部分不同）。

为了在所有部分下包含链接，引入了一个新的实用函数 iterateSections()。

/**
 * Returns a flat array of links which appear in the active document's body. 
 * Each link is represented by a simple Javascript object with the following 
 * keys:
 *   - "section": {ContainerElement} the document section in which the link is
 *     found. 
 *   - "isFirstPageSection": {Boolean} whether the given section is a first-page
 *     header/footer section.
 *   - "paragraph": {ContainerElement} contains a reference to the Paragraph 
 *     or ListItem element in which the link is found.
 *   - "text": the Text element in which the link is found.
 *   - "startOffset": {Number} the position (offset) in the link text begins.
 *   - "endOffsetInclusive": the position of the last character of the link
 *      text, or null if the link extends to the end of the text element.
 *   - "url": the URL of the link.
 *
 * @param {boolean} mergeAdjacent Whether consecutive links which carry 
 *     different attributes (for any reason) should be returned as a single 
 *     entry.
 * 
 * @returns {Array} the aforementioned flat array of links.
 */
function getAllLinks(mergeAdjacent) {
  var links = [];

  var doc = DocumentApp.getActiveDocument();


  iterateSections(doc, function(section, sectionIndex, isFirstPageSection) {
    if (!("getParagraphs" in section)) {
      // as we're using some undocumented API, adding this to avoid cryptic
      // messages upon possible API changes.
      throw new Error("An API change has caused this script to stop " + 
                      "working.\n" +
                      "Section #" + sectionIndex + " of type " + 
                      section.getType() + " has no .getParagraphs() method. " +
        "Stopping script.");
    }

    section.getParagraphs().forEach(function(par) { 
      // skip empty paragraphs
      if (par.getNumChildren() == 0) {
        return;
      }

      // go over all text elements in paragraph / list-item
      for (var el=par.getChild(0); el!=null; el=el.getNextSibling()) {
        if (el.getType() != DocumentApp.ElementType.TEXT) {
          continue;
        }

        // go over all styling segments in text element
        var attributeIndices = el.getTextAttributeIndices();
        var lastLink = null;
        attributeIndices.forEach(function(startOffset, i, attributeIndices) { 
          var url = el.getLinkUrl(startOffset);

          if (url != null) {
            // we hit a link
            var endOffsetInclusive = (i+1 < attributeIndices.length? 
                                      attributeIndices[i+1]-1 : null);

            // check if this and the last found link are continuous
            if (mergeAdjacent && lastLink != null && lastLink.url == url && 
                  lastLink.endOffsetInclusive == startOffset - 1) {
              // this and the previous style segment are continuous
              lastLink.endOffsetInclusive = endOffsetInclusive;
              return;
            }

            lastLink = {
              "section": section,
              "isFirstPageSection": isFirstPageSection,
              "paragraph": par,
              "textEl": el,
              "startOffset": startOffset,
              "endOffsetInclusive": endOffsetInclusive,
              "url": url
            };

            links.push(lastLink);
          }        
        });
      }
    });
  });


  return links;
}

/**
 * Calls the given function for each section of the document (body, header, 
 * etc.). Sections are children of the DocumentElement object.
 *
 * @param {Document} doc The Document object (such as the one obtained via
 *     a call to DocumentApp.getActiveDocument()) with the sections to iterate
 *     over.
 * @param {Function} func A callback function which will be called, for each
 *     section, with the following arguments (in order):
 *       - {ContainerElement} section - the section element
 *       - {Number} sectionIndex - the child index of the section, such that
 *         doc.getBody().getParent().getChild(sectionIndex) == section.
 *       - {Boolean} isFirstPageSection - whether the section is a first-page
 *         header/footer section.
 */
function iterateSections(doc, func) {
  // get the DocumentElement interface to iterate over all sections
  // this bit is undocumented API
  var docEl = doc.getBody().getParent();

  var regularHeaderSectionIndex = (doc.getHeader() == null? -1 : 
                                   docEl.getChildIndex(doc.getHeader()));
  var regularFooterSectionIndex = (doc.getFooter() == null? -1 : 
                                   docEl.getChildIndex(doc.getFooter()));

  for (var i=0; i<docEl.getNumChildren(); ++i) {
    var section = docEl.getChild(i);

    var sectionType = section.getType();
    var uniqueSectionName;
    var isFirstPageSection = (
      i != regularHeaderSectionIndex &&
      i != regularFooterSectionIndex && 
      (sectionType == DocumentApp.ElementType.HEADER_SECTION ||
       sectionType == DocumentApp.ElementType.FOOTER_SECTION));

    func(section, i, isFirstPageSection);
  }
}

Answer 3

我在玩耍并合并了@Mogsdad 的answer——这是真正复杂的版本：

var _ = Underscorejs.load(); // loaded via http://googleappsdeveloper.blogspot.com/2012/11/using-open-source-libraries-in-apps.html, rolled my own
var ui = DocumentApp.getUi();

// #region --------------------- Utilities -----------------------------

var gDocsHelper = (function(P, un) {
  // heavily based on answer https://stackoverflow.com/a/18731628/1037948

  var updatedLinkText = function(link, offset) {
    return function() { return 'Text: ' + link.getText().substring(offset,100) + ((link.getText().length-offset) > 100 ? '...' : ''); }
  }

  P.updateLink = function updateLink(link, oldText, newText, start, end) {
    var oldLink = link.getLinkUrl(start);

    if(0 > oldLink.indexOf(oldText)) return false;

    var newLink = oldLink.replace(new RegExp(oldText, 'g'), newText);
    link.setLinkUrl(start || 0, (end || oldLink.length), newLink);
    log(true, "Updating Link: ", oldLink, newLink, start, end, updatedLinkText(link, start) );

    return { old: oldLink, "new": newLink, getText: updatedLinkText(link, start) };
  };

  // moving this reused block out to 'private' fn
  var updateLinkResult = function(text, oldText, newText, link, urls, sidebar, updateResult) {
    // and may as well update the link while we're here
    if(false !== (updateResult = P.updateLink(text, oldText, newText, link.start, link.end))) {
       sidebar.append('<li>' + updateResult['old'] + ' &rarr; ' + updateResult['new'] + ' at ' + updateResult['getText']() + '</li>'); 
    }

    urls.push(link.url); // so multiple links get added to list
  };

  P.updateLinksMenu = function() {
    // https://developers.google.com/apps-script/reference/base/prompt-response
    var oldText = ui.prompt('Old link text to replace').getResponseText();
    var newText = ui.prompt('New link text to replace with').getResponseText();

    log('Replacing: ' + oldText + ', ' + newText);
    var sidebar = gDocUiHelper.createSidebar('Update All Links', '<h3>Replacing</h3><p><code>' + oldText + '</code> &rarr; <code>' + newText + '</code></p><hr /><ol>');

    // current doc available to script
    var doc = DocumentApp.getActiveDocument().getBody();//.getActiveSection();

    // Search until a link is found
    var links = P.findAllElementsFor(doc, function(text) {
      var i = -1, n = text.getText().length, link = false, url, urls = [], updateResult;

      // note: the following only gets the FIRST link in the text -- while(i < n && !(url = text.getLinkUrl(i++)));

      // scan the text element for links
      while(++i < n) {

        // getLinkUrl will continue to get a link while INSIDE the stupid link, so only do this once
        if(url = text.getLinkUrl(i)) {
          if(false === link) {
            link = { start: i, end: -1, url: url };
            // log(true, 'Type: ' + text.getType(), 'Link: ' + url, function() { return 'Text: ' + text.getText().substring(i,100) + ((n-i) > 100 ? '...' : '')});
          }
          else {
            link.end = i; // keep updating the end position until we leave
          }
        }
        // just left the link -- reset link tracking
        else if(false !== link) {
          // and may as well update the link while we're here
          updateLinkResult(text, oldText, newText, link, urls, sidebar);
          link = false; // reset "counter"
        }

      }

      // once we've reached the end of the text, must also check to see if the last thing we found was a link
      if(false !== link) updateLinkResult(text, oldText, newText, link, urls, sidebar);

      return urls;
    });

    sidebar.append('</ol><p><strong>' + links.length + ' links reviewed</strong></p>');
    gDocUiHelper.attachSidebar(sidebar);

    log(links);
  };

  P.findAllElementsFor = function(el, test) {
    // generic utility function to recursively find all elements; heavily based on https://stackoverflow.com/a/18731628/1037948

    var results = [], searchResult = null, i, result;
    // https://developers.google.com/apps-script/reference/document/body#findElement(ElementType)
    while (searchResult = el.findElement(DocumentApp.ElementType.TEXT, searchResult)) {
      var t = searchResult.getElement().editAsText(); // .asParagraph()

      // check to add to list
      if(test && (result = test(t))) {
        if( _.isArray(result) ) results = results.concat(result); // could be big? http://jsperf.com/self-concatenation/
        else results.push(result);
      }
    }
    // recurse children if not plain text item
    if(el.getType() !== DocumentApp.ElementType.TEXT) {
      i = el.getNumChildren();

      var result;
      while(--i > 0) {
        result = P.findAllElementsFor(el.getChild(i));
        if(result && result.length > 0) results = results.concat(result);
      }
    }

    return results;
  };

  return P;  
})({});

// really? it can't handle object properties?
function gDocsUpdateLinksMenu() {
  gDocsHelper.updateLinksMenu();
}

gDocUiHelper.addMenu('Zaus', [ ['Update links', 'gDocsUpdateLinksMenu'] ]);

// #endregion --------------------- Utilities -----------------------------

---

为了完整起见，我在下面包含了用于创建菜单、侧边栏等的“额外”实用程序类：

var log = function() {
  // return false;

  var args = Array.prototype.slice.call(arguments);

  // allowing functions delegates execution so we can save some non-debug cycles if code left in?

  if(args[0] === true) Logger.log(_.map(args, function(v) { return _.isFunction(v) ? v() : v; }).join('; '));
  else
    _.each(args, function(v) {
      Logger.log(_.isFunction(v) ? v() : v);
    });
}

// #region --------------------- Menu -----------------------------

var gDocUiHelper = (function(P, un) {

  P.addMenuToSheet = function addMenu(spreadsheet, title, items) {
    var menu = ui.createMenu(title);
    // make sure menu items are correct format
    _.each(items, function(v,k) {
      var err = [];

      // provided in format [ [name, fn],... ] instead
      if( _.isArray(v) ) {
        if ( v.length === 2 ) {
          menu.addItem(v[0], v[1]);
        }
        else {
          err.push('Menu item ' + k + ' missing name or function: ' + v.join(';'))
        }
      }
      else {
        if( !v.name ) err.push('Menu item ' + k + ' lacks name');
        if( !v.functionName ) err.push('Menu item ' + k + ' lacks function');

        if(!err.length) menu.addItem(v.name, v.functionName);
      }

      if(err.length) {
        log(err);
        ui.alert(err.join('; '));
      }

    });

    menu.addToUi();
  };

  // list of things to hook into
  var initializers = {};

  P.addMenu = function(menuTitle, menuItems) {
    if(initializers[menuTitle] === un) {
      initializers[menuTitle] = [];
    }
    initializers[menuTitle] = initializers[menuTitle].concat(menuItems);
  };

  P.createSidebar = function(title, content, options) {
    var sidebar = HtmlService
    .createHtmlOutput()
    .setTitle(title)
    .setWidth( (options && options.width) ? width : 350 /* pixels */);

    sidebar.append(content);

    if(options && options.on) DocumentApp.getUi().showSidebar(sidebar);
    // else { sidebar.attach = function() { DocumentApp.getUi().showSidebar(this); }; } // should really attach to prototype...

    return sidebar;
  };

  P.attachSidebar = function(sidebar) {
    DocumentApp.getUi().showSidebar(sidebar);
  };


  P.onOpen = function() {
    var spreadsheet = SpreadsheetApp.getActive();
    log(initializers);
    _.each(initializers, function(v,k) {
      P.addMenuToSheet(spreadsheet, k, v);
    });
  };

  return P;
})({});

// #endregion --------------------- Menu -----------------------------

/**
 * A special function that runs when the spreadsheet is open, used to add a
 * custom menu to the spreadsheet.
 */
function onOpen() {
  gDocUiHelper.onOpen();
}

Answer 4

在让 Mogsdad 的解决方案发挥作用时遇到了一些麻烦。具体来说，它错过了结束其父元素的链接，因此没有尾随的非链接字符来终止它。我已经实现了一些解决这个问题并返回标准范围元素的东西。在这里分享，以防有人觉得有用。

function getAllLinks(element) {
  var rangeBuilder = DocumentApp.getActiveDocument().newRange();

  // Parse the text iteratively to find the start and end indices for each link
  if (element.getType() === DocumentApp.ElementType.TEXT) {
    var links = [];
    var string = element.getText();
    var previousUrl = null; // The URL of the previous character 
    var currentLink = null; // The latest link being built
    for (var charIndex = 0; charIndex < string.length; charIndex++) {
      var currentUrl = element.getLinkUrl(charIndex);
      // New URL means create a new link
      if (currentUrl !== null && previousUrl !== currentUrl) {
        if (currentLink !== null) links.push(currentLink);
        currentLink = {};
        currentLink.url = String(currentUrl);
        currentLink.startOffset = charIndex;
      }
      // In a URL means extend the end of the current link
      if (currentUrl !== null) {
        currentLink.endOffsetInclusive = charIndex;
      }
      // Not in a URL means close and push the link if ready
      if (currentUrl === null) {
        if (currentLink !== null) links.push(currentLink);
        currentLink = null;
      }
      // End the loop and go again
      previousUrl = currentUrl;
    }
    // Handle the end case when final character is a link
    if (currentLink !== null) links.push(currentLink);
    // Convert the links into a range before returning
    links.forEach(function(link) {
      rangeBuilder.addElement(element, link.startOffset, link.endOffsetInclusive);
    });
  }

  // If not a text element then recursively get links from child elements
  else if (element.getNumChildren) {
    for (var i = 0; i < element.getNumChildren(); i++) {
      rangeBuilder.addRange(getAllLinks(element.getChild(i)));
    }
  }

  return rangeBuilder.build();
}

Answer 5

此 Excel 宏列出了 Word 文档中的链接。您需要先将数据复制到 Word 文档中。

Sub getLinks()
Dim wApp As Word.Application, wDoc As Word.Document
Dim i As Integer, r As Range
Const filePath = "C:\test\test.docx"
Set wApp = CreateObject("Word.Application")
'wApp.Visible = True
Set wDoc = wApp.Documents.Open(filePath)
Set r = Range("A1")
For i = 1 To wDoc.Hyperlinks.Count
    r = wDoc.Hyperlinks(i).Address
    Set r = r.Offset(1, 0)
Next i
wApp.Quit
Set wDoc = Nothing
Set wApp = Nothing
End Sub

Answer 6

你是对的......搜索和替换在这里不适用。 使用 setLinkUrl() https://developers.google.com/apps-script/reference/document/container-element#setLinkUrl(String)

基本上你必须递归地遍历元素（元素可以包含元素）并且对于每个 使用 getLinkUrl() 获取 oldText 如果不是 null ， setLinkUrl(newText) .... 保持显示的文本不变

Answer 7

这是一种无需脚本即可实现相同目标的快速而肮脏的方法：

1. 从 Google 文档中，将文档保存为 RTF 格式。
2. 在您选择的编辑器中，编辑 RTF 文件中的链接（在我的例子中，我想修改所有超链接，所以我使用了 Emacs 和 regexp-replace）。完成后保存文件。
3. 创建一个全新的 Google Doc，然后从菜单中选择 File>Open 并打开 RTF 文件。 Docs 会将您编辑的 RTF 文件转换回正确的 Google Doc，从而恢复所有格式。

Google Docs 的 RTF 格式非常完整——我没有注意到在往返过程中有任何保真度损失，它的优点是可以在一个易于编辑和应用正则表达式工具的表单。