多次复制幻灯片 - Google 幻灯片

Question

尝试多次复制 3 张（可能更多，具体取决于演示文稿）幻灯片。

搜索后找到了一个接近的结果： Google Slide API-我如何多次复制一张幻灯片并每次都创建一个唯一的 object id ，但是不允许第三张幻灯片。

到目前为止的代码如下：

function duplicateSlide() {
  var presentationId = "1J14grlx5pMrcsglMvmDVB22vTf89Qt0JUaq394Rltfc"; // Please set the Slides ID.
  var pageId1 = "g1399daed6c0_0_89"; // Please set the page ID of the source slide.
  var pageId2 = "g1399daed6c0_0_9"; // Please set the page ID of the source slide.
  var pageId3 = "g1399daed6c0_0_15"; // Please set the page ID of the source slide.
  
  
  var newID1 = ["First_1", "First_2", "First_3", "First_4"]; // Please set the unique ID here."
  var newID2 = ["Second_1A", "Second_2B", "Second_3C", "Second_4D"]; // Please set the unique ID here."
  var newID3 = ["Third_1A1", "Third_2B2", "Third_3C3", "Third_4D4"]; // Please set the unique ID here."

  var requests = newID1.reverse().map(function(id) {
    var obj = {};
    obj[pageId1] = id;
    return {duplicateObject: {objectId: pageId1, objectIds: obj}};
  });
  var requests = newID2.reverse().map(function(id) {
    var obj = {};
    obj[pageId2] = id;
    return {duplicateObject: {objectId: pageId2, objectIds: obj}};
  });
  var requests = newID3.reverse().map(function(id) {
    var obj = {};
    obj[pageId3] = id;
    return {duplicateObject: {objectId: pageId3 , objectIds: obj}};
  });
//  requests.push({deleteObject: {objectId: pageId}});
  Slides.Presentations.batchUpdate({'requests': requests}, presentationId);
}

谢谢

Answer 1

在您的脚本中，仅使用了第三个requests 。 因为第一个requests被第二个requests覆盖，第二个requests被第三个requests覆盖。 如果要包括所有请求，那么下面的修改呢？

从：

var requests = newID1.reverse().map(function(id) {
  var obj = {};
  obj[pageId1] = id;
  return {duplicateObject: {objectId: pageId1, objectIds: obj}};
});
var requests = newID2.reverse().map(function(id) {
  var obj = {};
  obj[pageId2] = id;
  return {duplicateObject: {objectId: pageId2, objectIds: obj}};
});
var requests = newID3.reverse().map(function(id) {
  var obj = {};
  obj[pageId3] = id;
  return {duplicateObject: {objectId: pageId3 , objectIds: obj}};
});

至：

var requests1 = newID1.reverse().map(function (id) {
  var obj = {};
  obj[pageId1] = id;
  return { duplicateObject: { objectId: pageId1, objectIds: obj } };
});
var requests2 = newID2.reverse().map(function (id) {
  var obj = {};
  obj[pageId2] = id;
  return { duplicateObject: { objectId: pageId2, objectIds: obj } };
});
var requests3 = newID3.reverse().map(function (id) {
  var obj = {};
  obj[pageId3] = id;
  return { duplicateObject: { objectId: pageId3, objectIds: obj } };
});
var requests = [...requests1, ...requests2, ...requests3];