Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/283.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Javascript ajax PHP MySQL查询_Javascript_Php_Mysql_Ajax - Fatal编程技术网

Javascript ajax PHP MySQL查询

Javascript ajax PHP MySQL查询,javascript,php,mysql,ajax,Javascript,Php,Mysql,Ajax,我需要ajax调用的帮助,但我是ajax的新手,我不知道该怎么做 我有以下PHP代码(phonecall.PHP): 每当有新的内容发布到表中时,我想自动实时更新页面 这是我的非工作页面: <!DOCTYPE html> <html lang="en"> <head> <meta charset="utf-8"> <title>Phone calls</title> </head> <body

我需要ajax调用的帮助,但我是ajax的新手,我不知道该怎么做

我有以下PHP代码(phonecall.PHP):


每当有新的内容发布到表中时,我想自动实时更新页面

这是我的非工作页面:

<!DOCTYPE html> 
<html lang="en"> 
<head> 
<meta charset="utf-8"> 
<title>Phone calls</title> 
</head>
<body>
<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code
function ajaxFunction() {
  var ajaxRequest;

  try {
      // Opera 8.0+, Firefox, Safari
      ajaxRequest = new XMLHttpRequest();
  } catch (e) {
      // Internet Explorer Browsers
      try {
        ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
      } catch (e) {
          try {
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
          } catch (e) {
            // Something went wrong
            alert("Your browser broke!");
            return false;
          }
      }
  }

  ajaxRequest.onreadystatechange = function(){
  var ajaxDisplay = document.getElementById('call');
  ajaxDisplay.innerHTML = ajaxRequest.responseText;
}

setInterval(function() { //Broken
    ajaxRequest.open();  //Not sure what to put here.
}, 1000);
}
//-->
</script>
</body>
</html>

电话
根据XMLHttpRequest规范,您的ajaxRequest.open()方法接受3个参数:

  • 请求的方法(POST、GET等)
  • 您将请求发送到的文件
  • 请求是否异步
因此:

将向yourfile.php生成一个异步GET请求

您还缺少ajaxRequest().send(),它实际上会将您的请求发送到服务器


关于这一点有很多需要了解的,所以我建议用谷歌搜索一下,因为你似乎缺乏基本知识。

什么不起作用?据我所知,你从来没有调用过
ajaxFunction()
正确的,我不确定我需要做什么才能让它工作。我只需要显示PHP页面上已有的内容。我试图让它与setInterval函数一起工作。我该怎么办?老实说,我对ajax请求一无所知。您应该在循环中调用
ajaxFunction()
<!DOCTYPE html> 
<html lang="en"> 
<head> 
<meta charset="utf-8"> 
<title>Phone calls</title> 
</head>
<body>
<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code
function ajaxFunction() {
  var ajaxRequest;

  try {
      // Opera 8.0+, Firefox, Safari
      ajaxRequest = new XMLHttpRequest();
  } catch (e) {
      // Internet Explorer Browsers
      try {
        ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
      } catch (e) {
          try {
            ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
          } catch (e) {
            // Something went wrong
            alert("Your browser broke!");
            return false;
          }
      }
  }

  ajaxRequest.onreadystatechange = function(){
  var ajaxDisplay = document.getElementById('call');
  ajaxDisplay.innerHTML = ajaxRequest.responseText;
}

setInterval(function() { //Broken
    ajaxRequest.open();  //Not sure what to put here.
}, 1000);
}
//-->
</script>
</body>
</html>
ajaxRequest().open('GET','yourfile.php',true);