这是我在这个论坛上提出的先前问题的延续。我通过添加 else 条件对代码进行了一些更改。我想要解决的是在页面加载时显示当前状态。当前代码仅在单击按钮时显示状态。如何在页面加载时显示当前状态，并在单击按钮时更新状态。CODE.GSfunction doGet(e) { return HtmlService.createHtmlOutputFromFile('Index');}function checkStatus(notify) { var employee = "John Peter"; var ss = SpreadsheetApp.getActiveSpreadsheet(); var mainSheet = ss.getSheetByName("MAIN"); var data = mainSheet.getDataRange().getValues(); for (var j = 0; j < data.length; j++){ var row = data[j]; var mainSheet2 = row[4]; var mainSheet3 = row[0]; var status = (mainSheet2 =="IN" && mainSheet3 == employee) ; if (status == true){ var notify = employee +" You Are In" return notify; } else{ var status = (mainSheet2 =="OUT" && mainSheet3 == employee) ; if (status == true){ var notify2 = employee +" You Are Out" return notify2; }} }}索引.html<body> <div> <button onclick="onStatus()">Check Status</button> <font color='Green' id="status" onbeforeprint="onStatus" ></font> </div> <script> function onStatus() { google.script.run.withSuccessHandler(updateStatus) // Send the backend result to updateStatus() .checkStatus(); // Call the backend function } function updateStatus(notify) { document.getElementById('status').innerHTML= notify; } </script> </body>
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眼眸繁星
TA贡献1873条经验 获得超9个赞
您可以在页面加载时调用此函数，如下所示。在正文标记结束之前添加此脚本。
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js">
</script>
<script>
// The code in this function runs when the page is loaded.
$(function() {
google.script.run.withSuccessHandler(updateStatus)
.checkStatus();
});
</script>
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