我刚刚在Form / Google电子表格中添加了一个脚本。它从表单中获取响应URL并将其推送到响应电子表格中的列。我想将URL链接到一个按钮(在HTML中,我当然会用编辑响应URL来锚定我的图像,但现在我有点困惑,因为我不是一个超级经验丰富的脚本编辑器)。如何将它集成到我的脚本中?:
function assignEditUrls() {
var form = FormApp.openById('1-Sxpvd9jktE-SVXV0_dfp018xwcIoa3aXMA_fdff9W8');
//enter form ID here
var sheet = SpreadsheetApp.getActiveSpreadsheet().getSheetByName('Form Responses 1');
//Change the sheet name as appropriate
var data = sheet.getDataRange().getValues();
var urlCol = 5; // column number where URL's should be populated; A = 1, B = 2 etc
var responses = form.getResponses();
var timestamps = [], urls = [], resultUrls = [];
for (var i = 0; i < responses.length; i++) {
timestamps.push(responses[i].getTimestamp().setMilliseconds(0));
urls.push(responses[i].getEditResponseUrl());
}
for (var j = 1; j < data.length; j++) {
resultUrls.push([urls[timestamps.indexOf(data[j][0].setMilliseconds(0))]]);
}
sheet.getRange(2, urlCol, resultUrls.length).setValues(resultUrls);
}
答案 0 :(得分:1)
无法以编程方式将按钮或图像添加到电子表格中 你可以做的是在这些单元格中添加url作为fomula =超链接(“url”,yoururl),使它看起来更漂亮。