Question

我知道URL可以按预期的方式工作，就像我将其记录到控制台一样，这很好。但是，当readyState == 4和status == 200时，我无法获得“好消息”登录到控制台。我尝试删除readState，但仍然无法记录。我尝试记录状态，并且只会以0值触发一次。这是我第一次使用Ajax，因此可以提供任何帮助。

function setupRequest(){
  var bttn = document.querySelector('#send');

  bttn.addEventListener('click', sendData)
}

setupRequest();

function sendData () {

console.log('ran')
  var url = 'localhost/bev/drinks.php';

  var data = document.getElementById('input').value;
  url += '?' + 'alcohol=' + data;

console.log(url)

 var request = new XMLHttpRequest();
 request.onreadystatechange = function () {
  if (this.readyState == 4 && this.status == 200) {
    console.log('good news')
            console.log(this.responseText)
        } else {
    console.log(this.status)
        }
    }

 request.open('GET', url, true);
 request.send;

  console.log('sent')

}

Answer 1

您需要实际致电send()。每当您说request.send;

时，您什么也没做

function setupRequest() {
  var bttn = document.querySelector('#send');

  bttn.addEventListener('click', sendData)
}

setupRequest();

function sendData() {

  console.log('ran')
  var url = 'https://jsonplaceholder.typicode.com/posts';

  var data = document.getElementById('input').value;
  //url += '?' + 'alcohol=' + data;

  console.log(url)

  var request = new XMLHttpRequest();
  request.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      console.log('good news')
      console.log(this.responseText)
    } else {
      console.log(this.status)
    }
  }

  request.open('GET', url, true);

  // You wrote (without parentheses):
  ///////////////////
  // request.send; //
  ///////////////////

  // You need to write
  request.send();

  console.log('sent')

}

<button type="button" id="send">Btn</button>
<input type="text" id="input">

Ajax不会在readystatechange上触发

1 个答案: