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首页 > 解决方案 > 使用 Selection 的 RangeElements 获取 Google Doc 中的所有嵌套文本元素

google-apps-script - 使用 Selection 的 RangeElements 获取 Google Doc 中的所有嵌套文本元素

问题描述

在与上述类似的文档中，我可以使用以下代码获取所有段落：

var paras = body.getParagraphs();

请注意，上面的代码不仅返回顶级段落，还返回ListItems、Tables 等内部的所有子级段落。

我怎样才能在选定的范围内做同样的事情？以下代码仅返回顶级元素。

const selection = DocumentApp.getActiveDocument().getSelection();
var rangeElements = selection.getRangeElements();

例如，上表包含 9 个非空段落，如果它们在选择中，我想一一处理。

我想要实现的类似于通过尽可能保留格式、表格、列表项等来翻译选择中的文本。

标签： google-apps-scriptgoogle-docs

解决方案

.getRangeElements()返回RangeElements数组。范围元素是一个包装对象，用于帮助我们处理部分选择。我们可以调用.getElement()这个数组中的每一项来获取Element 对象，它是一个非常通用的对象，几乎可以表示 Google Doc 的任何部分。Elements有一个.getType()返回ElementType枚举的方法；而且有很多！

让我们使用我们目前所知道的来看看 Google Doc 中可能的类型（我创建了一个类似于你的（img）作为示例）：

function selectionHasWhichTypes() {
  var doc = DocumentApp.getActiveDocument();
  var selection = doc.getSelection();
  var rangeElems = selection.getRangeElements();

  rangeElems.forEach(function(elem){
    var elem = elem.getElement();

    Logger.log(elem.getType());
  });
}

//Logger OUTPUT:
PARAGRAPH
PARAGRAPH
PARAGRAPH
PARAGRAPH
PARAGRAPH
LIST_ITEM
LIST_ITEM
LIST_ITEM
PARAGRAPH
PARAGRAPH
PARAGRAPH
TABLE
PARAGRAPH

啊哈！看起来我们现在只需要处理PARAGRAPH、LIST_ITEM和TABLE ElementTypes ，但是让我们也记住他们的孩子（我们会发现这些是可以有孩子的 5 个中的 3 个）。这听起来像是一个递归函数的工作，它将不断挖掘子元素，直到我们找到并处理它们。

所以让我们试试吧。下一部分可能看起来令人困惑，但本质上是寻找一个元素，检查它是否有子元素，然后查看它们是否有子元素，等等。我们还想检查我们是否也有新的ElementTypes 来处理......

function selectionHasWhichTypes() {
  var doc = DocumentApp.getActiveDocument();
  var selection = doc.getSelection();
  var rangeElems = selection.getRangeElements();

  rangeElems.forEach(function(elem){
    var elem = elem.getElement();

    elemsHaveWhatChildElems(elem, elem.getType());

  });
}

function elemsHaveWhatChildElems(elem, typeChain){
  var elemType = elem.getType();
  if(elemType == "TABLE" || elemType == "LIST_ITEM" || elemType == "PARAGRAPH"){ //Lets see if element is one of our basic 3. If so they could have children.
    var numChildren = elem.getNumChildren(); //How many children are there?
    if(numChildren > 0){
      for(var i = 0; i < numChildren; i++){ //Let's go through them.
        var child = elem.getChild(i);
        elemsHaveWhatChildElems(child, typeChain + "." + child.getType()); //Recursion step to look for more children.
      }
    }else{
       Logger.log(typeChain); //Let's log the chain of Parent to Child elements.
    }
  }else{
    Logger.log("*" + typeChain); //Let's mark the new elemTypeChains we have not seen.
  }
}

//Logger OUTPUT:
*PARAGRAPH.TEXT
PARAGRAPH
*PARAGRAPH.HORIZONTAL_RULE
PARAGRAPH
*PARAGRAPH.TEXT
*LIST_ITEM.TEXT
*LIST_ITEM.TEXT
*LIST_ITEM.TEXT
PARAGRAPH
*PARAGRAPH.TEXT
PARAGRAPH
*TABLE.TABLE_ROW
*TABLE.TABLE_ROW
PARAGRAPH

好的，所以日志的每一行都是元素链及其子元素。我们有一些新的元素类型（HORIZONTAL_RULE、TABLE_ROW和TEXT）。如果链只有 aParagraph并且没有子级，则由“PARAGRAPH”表示。我们可以忽略它，因为它是一个空行。我们也可以忽略HORIZONTAL_RULE，因为这显然不包含文本。

如果我们得到一个 TEXT 元素，这意味着我们可以执行我们的功能（即，对于 OP，它将是一个翻译），就像我们对 LIST_ITEMs 和 PARAGRAPHs 所做的那样。但是，我们仍然需要处理TableRow对象（它的日志是这样的：） TABLE.TABLE_ROW。这类似于我们的主要 3 个元素，可以与我们的if(elemType == "TABLE" || elemType == "LIST_ITEM" || elemType == "PARAGRAPH")which 更改为if(elemType == "TABLE" || elemType == "LIST_ITEM" || elemType == "PARAGRAPH" || elemType == "TABLE_ROW").

这给了我们链中的另一个新元素；TableCell（类似：的日志TABLE.TABLE_ROW.TABLE_CELL），我们可以再次将其添加到 if 语句中：if(elemType == "TABLE" || elemType == "LIST_ITEM" || elemType == "PARAGRAPH" || elemType == "TABLE_ROW" || elemType == "TABLE_CELL")

是时候看看我们处理 Table ElementTypes 时会发生什么了。

function selectionHasWhichtypeChains() {
  var doc = DocumentApp.getActiveDocument();
  var selection = doc.getSelection();
  var rangeElems = selection.getRangeElements();

  rangeElems.forEach(function(elem){
    var elem = elem.getElement();

    elemsHaveWhatChildElems(elem, elem.getType());

  });
}

function elemsHaveWhatChildElems(elem, typeChain){
  var elemType = elem.getType();
  if(elemType == "TABLE" || elemType == "LIST_ITEM" || elemType == "PARAGRAPH" || elemType == "TABLE_ROW" || elemType == "TABLE_CELL"){ //Lets see if element is one of our basic 5 if so they could have children.
    var numChildren = elem.getNumChildren(); //How many children are there?
    if(numChildren > 0){
      for(var i = 0; i < numChildren; i++){ //Let's go through them.
        var child = elem.getChild(i);
        elemsHaveWhatChildElems(child, typeChain + "." + child.getType()); //Recursion step to look for more children.
      }
    }else{
       Logger.log(typeChain); //Let's log the chain of Parent to Child elements.
    }
  }else{
    Logger.log("*" + typeChain); //Let's mark the new elemTypeChains we have not seen.
  }
}

//Logger OUTPUT:
*PARAGRAPH.TEXT
PARAGRAPH
*PARAGRAPH.HORIZONTAL_RULE
PARAGRAPH
*PARAGRAPH.TEXT
*LIST_ITEM.TEXT
*LIST_ITEM.TEXT
*LIST_ITEM.TEXT
PARAGRAPH
*PARAGRAPH.TEXT
PARAGRAPH
*TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
*TABLE.TABLE_ROW.TABLE_CELL.TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
*TABLE.TABLE_ROW.TABLE_CELL.TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
*TABLE.TABLE_ROW.TABLE_CELL.TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
*TABLE.TABLE_ROW.TABLE_CELL.TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH
*TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
*TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.HORIZONTAL_RULE
*TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
*TABLE.TABLE_ROW.TABLE_CELL.PARAGRAPH.TEXT
PARAGRAPH

这很棒！我们已经深入到每个父元素的深处，并且达到了文本元素或空白段落！从这里我们可以稍微修改我们的代码以添加我们想要执行的功能，同时保持文档的结构：

function myFunction() {
  var doc = DocumentApp.getActiveDocument();
  var selection = doc.getSelection();
  var rangeElems = selection.getRangeElements(); //Get main Elements of selection

  rangeElems.forEach(function(elem){ //Let's rn through each to find ALL of their children.
    var elem = elem.getElement(); //We have an ElementType. Let's get the full element.
    getNestedTextElements(elem, elem.getType()); //Time to go down the rabbit hole.
  });
}

function getNestedTextElements(elem, typeChain){
  var elemType = elem.getType();
  if(elemType == "TABLE" || elemType == "LIST_ITEM" || elemType == "PARAGRAPH" || elemType == "TABLE_ROW" || elemType == "TABLE_CELL"){ //Lets see if element is one of our basic 5, if so they could have children.
    var numChildren = elem.getNumChildren(); //How many children are there?
    if(numChildren > 0){
      for(var i = 0; i < numChildren; i++){ //Let's go through them.
        var child = elem.getChild(i);
        getNestedTextElements(child, typeChain + "." + child.getType()); //Recursion step to look for more children.
      }
    }
  }else if(elemType == "TEXT"){
    //THIS IS WHERE WE CAN PERFORM OUR OPERATIONS ON THE TEXT ELEMENT
    var text = elem.getText();


  }else{
    Logger.log("*" + typeChain); //Let's log the new elem we dont deal with now - for future proofing.
  }
}

繁荣！完毕。我知道这是一篇很长的文章，但我已将解决方案的每个部分分解为多个部分，以帮助新的 Apps 脚本编码人员了解选择的结构（我猜是文档正文）以及在结构时如何修改它非常复杂（许多嵌套元素）。我真的希望这会有所帮助。如果有人看到可以改进的部分，请告诉我。

作为对 OP 的说明：请注意，这不一定处理元素的部分选择，但可以通过稍微修改第一个函数以检查isPartial()RangeElement来轻松处理。

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