javascript - Javascript在出错后继续运行代码

Question

我正在对一些值进行测试，看看哪个给出了正确的输出。

每次更改代码时都需要重新部署到服务器，这会影响服务器上的并发用户。

我想尽量减少这种情况。

即使发现错误，是否可以继续运行代码？

Javascript：

var schedule = document.getElementById("<%=ddlExecutionSchedule.ClientID%>").value;
console.log("schedule is " + schedule);  //returns UNDEFINED

var schedule2 = document.getElementById("<%=ddlExecutionSchedule.ClientID%>").length;
console.log("schedule2 is " + schedule2); //returns UNDEFINED

var schedule3 = document.getElementById('ddlExecutionSchedule'); 
console.log("schedule3 is " + schedule3); //returns HTML OBJECT

var schedule4 = document.getElementById('ddlExecutionSchedule').value;
console.log("schedule4 is " + schedule4); //returns UNDEFINED

var schedule5 = document.getElementById('ddlExecutionSchedule').length;
console.log("schedule5 is " + schedule5); //returns UNDEFINED

var schedule6 = document.getElementById(<%=ddlExecutionSchedule.ClientID%>);
console.log("schedule6 is " + schedule6); //returns NULL

var options = schedule.getElementsByTagName('input');
//^ = The line above caught an error of <Uncaught TypeError: 
      Cannot read property 'getElementsByTagName' of undefined

var options2 = schedule2.getElementsByTagName('input');
//^ = To test

...

感谢您的帮助。

标签： javascriptsql-serverdeployment

Answer 1

try并catch会有所帮助：

var schedule1 = document.getElementById("id1").length;
console.log("schedule1 is " + schedule1); //returns UNDEFINED

var schedule2 = document.getElementById("id1");
console.log("schedule2 is " + schedule2);

try {
  var options1 = schedule1.getElementsByTagName('input');
  console.log('options1: ' + options1);
} catch (err) {
  console.log('error: '+ err.message);
}
var options2 = schedule2.getElementsByTagName('input');
console.log('options2: ' + options2);

<label id="id1"><input></label>