给定 Google Docs/Drive 中的“普通文档”（例如段落、列表、表格），其中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，而不是文本。
我不认为开发文档的替换文本部分是我需要的——我需要扫描文档的每个元素吗？我可以只editAsText并使用 html 正则表达式吗？示例将不胜感激。
问问题
15295 次
7 回答
19
这只是最痛苦的！代码作为 gist 的一部分提供。
是的，我不会拼音。
获取所有链接
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;
}
查找并替换链接
该实用程序建立在getAllLinks执行查找和替换功能的基础上。
/**
* 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);
}
于 2013-09-11T02:15:17.250 回答
8
我为您的第一个问题提供了另一个更简短的答案，涉及遍历文档正文中的所有链接。此指令性代码返回当前文档正文中的平面链接数组，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 ( paragraph)、偏移索引在链接出现的文本 ( startOffset) 和 URL 本身 ( url) 中。希望您会发现它很容易满足您自己的需求。
它使用该getTextAttributeIndices()方法而不是遍历文本的每个字符，因此预计比以前编写的答案执行得更快。
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endOffsetInclusive每个链接的属性（请注意，它可以null用于延伸到文本元素末尾的链接 - 在这种情况下可以使用link.text.length-1）；(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 true 时，将只返回一个链接条目，用于链接到同一 URL 的连续文本（例如，如果部分文本的样式不同于另一部分）。
为了在所有部分下包含链接，iterateSections()引入了一个新的实用函数 , 。
/**
* 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);
}
}
于 2016-11-21T21:52:41.703 回答
1
我在玩耍并结合了@Mogsdad 的答案——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'] + ' → ' + 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> → <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，我在下面包括了用于创建菜单、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();
}
于 2013-09-13T21:11:10.833 回答
1
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发现它有用。
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();
}
于 2017-08-05T18:56:02.110 回答
0
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
于 2019-04-04T12:40:05.973 回答
-1
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）并且每次使用 getLinkUrl() 来获取 oldText 如果不是 null ， setLinkUrl(newText) .... 保持显示的文本不变
于 2013-09-10T21:53:23.477 回答
-1
这是一种无需编写脚本即可实现相同目标的快速而肮脏的方法：
- 在 Google Docs 中，将文档保存为 RTF 格式。
- 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。
- 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，它的优点是可以以一种简单的形式完全公开所有超链接、格式和有关文档的所有其他内容编辑和应用正则表达式工具。
于 2019-12-19T17:58:16.727 回答