Question

我错过了什么吗？该按钮应该基于无线电选择启动各个站点

＆＃13;

  
function navigateSite(){
  var getUrl = document.querySelector('input[name = "myRadio"]:checked').value
  console.log(getUrl);
}

＆＃13;

 <INPUT TYPE="radio" id="Orange" name="myRadio" value="www.google.com"> Orange
 <INPUT TYPE="radio" id="Apple" name="myRadio" value="www.mozilla.com"> Apple
 <INPUT TYPE="radio" id="Mango" name="myRadio" value="www.microsoft.com"> Mango
 <button onclick="navigateSite()">Select</button>

＆＃13;

Answer 1

弹出窗口仅在代码段之外工作

＆＃13;

  
function navigateSite(){
  var getUrl = document.querySelector('input[name = "myRadio"]:checked').value
  console.log(getUrl);
  window.open(getUrl, getUrl, "width=600,height=500");
}

＆＃13;

 <INPUT TYPE="radio" id="Orange" name="myRadio" value="www.google.com"> Orange
 <INPUT TYPE="radio" id="Apple" name="myRadio" value="www.mozilla.com"> Apple
 <INPUT TYPE="radio" id="Mango" name="myRadio" value="www.microsoft.com"> Mango
 <button onclick="navigateSite()">Select</button>

＆＃13;

带有无线电选择的Javascript无法正常工作

1 个答案: