Javascript 等待异步响应时显示加载程序

我不知道在等待javascript编写的函数响应时如何显示加载程序

我有一个

，其中包含响应：

<input type="button" name="btn-s" value="Search">
<div align="center" id="div_feed_data">
</div>

现在我想放置一个加载gif而不是按钮，直到响应准备好，但我不知道怎么做，因为我已经找到了纯ajax代码的方法，我在js方面不如英语好，所以我需要帮助，谢谢大家

见：

见：

---

创建Ajax请求时，它将经历以下状态

0: request not initialized 
1: server connection established
2: request received 
3: processing request 
4: request finished and response is ready

当您触发.send（）方法时，将处理这些状态。因此，您尝试显示加载程序gif一旦调用Ajax调用，当就绪状态为4时，您可以隐藏加载程序。

伪代码

function yourFunction() {
   //call the show loader function here
   if(xmlReq.readyState == 4 && smlReq.status == 200){
        //call the hide loader function here
   }
}

function yourFunction() {
   if(xmlReq.readyState == 1 ){
        //call the show loader function here
        //console.log('connected');
   }

   if(xmlReq.readyState == 2 ){
        //call the show loader function here
        //console.log('Request Received');
   }

    if(xmlReq.readyState == 3 ){
        //call the show loader function here
        //console.log('Processing');
   }

   if(xmlReq.readyState == 4 && xmlReq.status == 200){
        //call the hide loader function here
   }
}

注意：您甚至可以基于readystate值显示每个状态的消息

伪代码

function yourFunction() {
   //call the show loader function here
   if(xmlReq.readyState == 4 && smlReq.status == 200){
        //call the hide loader function here
   }
}

function yourFunction() {
   if(xmlReq.readyState == 1 ){
        //call the show loader function here
        //console.log('connected');
   }

   if(xmlReq.readyState == 2 ){
        //call the show loader function here
        //console.log('Request Received');
   }

    if(xmlReq.readyState == 3 ){
        //call the show loader function here
        //console.log('Processing');
   }

   if(xmlReq.readyState == 4 && xmlReq.status == 200){
        //call the hide loader function here
   }
}

---

创建Ajax请求时，它将经历以下状态

0: request not initialized 
1: server connection established
2: request received 
3: processing request 
4: request finished and response is ready

当您触发.send（）方法时，将处理这些状态。因此，您尝试显示加载程序gif一旦调用Ajax调用，当就绪状态为4时，您可以隐藏加载程序。

伪代码

function yourFunction() {
   //call the show loader function here
   if(xmlReq.readyState == 4 && smlReq.status == 200){
        //call the hide loader function here
   }
}

function yourFunction() {
   if(xmlReq.readyState == 1 ){
        //call the show loader function here
        //console.log('connected');
   }

   if(xmlReq.readyState == 2 ){
        //call the show loader function here
        //console.log('Request Received');
   }

    if(xmlReq.readyState == 3 ){
        //call the show loader function here
        //console.log('Processing');
   }

   if(xmlReq.readyState == 4 && xmlReq.status == 200){
        //call the hide loader function here
   }
}

注意：您甚至可以基于readystate值显示每个状态的消息

伪代码

function yourFunction() {
   //call the show loader function here
   if(xmlReq.readyState == 4 && smlReq.status == 200){
        //call the hide loader function here
   }
}

function yourFunction() {
   if(xmlReq.readyState == 1 ){
        //call the show loader function here
        //console.log('connected');
   }

   if(xmlReq.readyState == 2 ){
        //call the show loader function here
        //console.log('Request Received');
   }

    if(xmlReq.readyState == 3 ){
        //call the show loader function here
        //console.log('Processing');
   }

   if(xmlReq.readyState == 4 && xmlReq.status == 200){
        //call the hide loader function here
   }
}

---

这是更准确的，所以我认为它也可以帮助更好的其他人，正因为如此，我选择这个答案是正确的。但是Machalot的答案也是正确的。这更准确，所以我认为它也能帮助更好的其他人，正因为如此，我选择了这个答案作为正确答案。但马切洛特的观点也是正确的。