我目前正在编写一个脚本,该脚本从 google 表格中提取数据并将其转换为 html 表。但我的谷歌表格中有链接,所以我希望这些链接可以在一个看起来不错但不知道如何做到这一点的按钮中获得。我现在有这个loob
for (var i = 0; i < entry.length; i++) {
html += '<tr>';
html += '<td>' + entry[i]['gsx$komponente']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$name']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$hersteller']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$preis']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$link']['$t'] + '</td>';
html += '</tr>';
}
html += '</table>';
这entry[i]['gsx$link']['$t']给了我链接,我只是无法让它在按钮内工作如果您知道如何解决这个问题,请帮助我
这是完整的代码
<!DOCTYPE html>
<meta charset="ISO-8859-1">
<html>
<body>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js">
</script>
<script>
// ID of the Google Spreadsheet
var spreadsheetID = "1Xx0qGY_5Ic1KNB7m8Lu5mZqHE4XQzauvcugTVUGwgqk";
// Make sure it is public or set to Anyone with link can view
var url = "https://spreadsheets.google.com/feeds/list/" + spreadsheetID + "/od6/public/values?alt=json";
// make JSON call to Google Data API
$.getJSON(url, function(data) {
// set global html variable
var html = '';
// build table headings
html += '<table>';
html += '<tr>';
html += '<th>Komponente</th>';
html += '<th>Name</th>';
html += '<th>Hersteller</th>';
html += '<th>Preis</th>';
html += '<th>Link</th>';
html += '</tr>';
// loop to build html output for each row
var entry = data.feed.entry;
/**
** Change to descending order
** for (var i = entry.length - 1; i >= 0; i -= 1) {
*/
for (var i = 0; i < entry.length; i++) {
html += '<tr>';
html += '<td>' + entry[i]['gsx$komponente']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$name']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$hersteller']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$preis']['$t'] + '</td>';
html += '<td>' + entry[i]['gsx$link']['$t'] + '</td>';
html += '</tr>';
}
弑天下
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