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Javascript 无法更改image.src属性+;函数执行,但不返回值_Javascript_Google Apps Script_Google Sheets - Fatal编程技术网

Javascript 无法更改image.src属性+;函数执行,但不返回值

Javascript 无法更改image.src属性+;函数执行,但不返回值,javascript,google-apps-script,google-sheets,Javascript,Google Apps Script,Google Sheets,请查看IndexQ3文件中的SPOT 1注释。这就是我的问题所在。当用户单击下一个问题按钮时xxx() 执行并运行myFunc_3。后者有一些问题,我无法检测到,即使它是一个5行函数。我知道它是通过在电子表格中输入一个新值来执行的。我知道它有问题,因为它返回“undefined” 问题:我做错了什么 IndexQ3文件: <!DOCTYPE html> <html> <head> <base target="_top"> <

请查看IndexQ3文件中的SPOT 1注释。这就是我的问题所在。当用户单击下一个问题按钮时
xxx()
执行并运行
myFunc_3
。后者有一些问题,我无法检测到,即使它是一个5行函数。我知道它是通过在电子表格中输入一个新值来执行的。我知道它有问题,因为它返回“undefined”

问题:我做错了什么

IndexQ3文件:

<!DOCTYPE html>
<html>
  <head> 
    <base target="_top">
  </head>
  <body>
      <script>
          function saveResponse() {return document.getElementById('text').value;}
          function returnText () {return "New Text!"} //function that changes the xx element
          //function returnSrc () {var aaaa = '"' + returnCellValue(6) + '"'} //this approach is not working

          function xxx() //function that is run when user clics button Next Question
          {
            document.getElementById("xx").innerHTML = returnText(); //Works

            //SPOT 1 
            //the next line of code works partly - the myFunc_3() does get and write the value correctly
            //but the variable uu is undefined despite "return b"
            var uu = google.script.run.myFunc_3(); //
            if(typeof uu == typeof "asdf") {uu = "1"} else {uu = "2"} //src value will be changed to 2 because of undefined

            //document.getElementById("img").src = "https://www.google.com/images/srpr/logo3w.png"
            document.getElementById("img").src = uu; //the scr value comes out as undefined
          }

      </script>
  <div style="width: 100%; display: table;">
    <div style="display: table-row">
      <div style="display: table-cell;"> <!-- width: 400px -->
          <p id="xx"> Question:</p>    
          <br>
          <p style="font-size: 14pt"><?!= returnCellValue(2) ?></p>
          <br>
            <textarea id="text" rows="10" cols="30" style="font-size: 14pt">Enter the response here..</textarea>
           <br>
            <input type="button" value="Save Response" onclick='google.script.run.myFunc_3(saveResponse())' />  
            <input type="button" value="Next Question" onclick='google.script.run.nextQuestion(); xxx()' />
          </div>     
        <div style="display: table-cell;">
          <br>
          <img id="img" src="<?!= returnCellValue(6) ?>" alt="No Image" width="650">
          </div>
    </div>
</div>
  </body>
</html>
大牛,谢谢你

如果还有其他人在为此挣扎,以下是我所做的。在函数xxx()中,我添加了
google.script.run.withSuccessHandler(onSuccess.myFunc_3()
。然后在正文中添加脚本标记

function onSuccess(sss) {
             var uu = sss;
             document.getElementById("img").src = sss;   
           }

“[该函数]是异步的,不直接返回;但是,[它]可以将值作为传递给成功处理程序的参数返回给客户机”请看,Daniel,我不知道如何将您的注释标记为最佳答案,因此我们开始:这是最佳答案!
function onSuccess(sss) {
             var uu = sss;
             document.getElementById("img").src = sss;   
           }