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无法通过ajax将数据发送到php_Php_Html_Ajax - Fatal编程技术网

无法通过ajax将数据发送到php

无法通过ajax将数据发送到php,php,html,ajax,Php,Html,Ajax,大家好,我是ajax新手,我正在尝试从php代码中获取数据这是我的ajax代码: function blodvotingview(contentid) { var xmlhttp; xmlhttp=GetXmlHttpObject(); if (xmlhttp==null) { alert ("Your browser does not support XMLHTTP!"); return; } var ur

大家好,我是ajax新手,我正在尝试从php代码中获取数据这是我的ajax代码:

function blodvotingview(contentid)
{
    var xmlhttp;    
    xmlhttp=GetXmlHttpObject();
    if (xmlhttp==null)
    {
        alert ("Your browser does not support XMLHTTP!");
        return;
    }
    var url="index.php";
    url=url+"?hp=1";
    url=url+"&m=blogenvoting";
    url=url+"&contentid="+contentid;
    xmlhttp.onreadystatechange=stateChanged;
    xmlhttp.open("GET",url,true);
    xmlhttp.send(null);
}

function stateChanged()
{
    if (xmlhttp.readyState==4)
    {
        document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
}

function GetXmlHttpObject()
{
    if (window.XMLHttpRequest)
    {
        // code for IE7+, Firefox, Chrome, Opera, Safari
        return new XMLHttpRequest();
    }
    if (window.ActiveXObject)
    {
        // code for IE6, IE5
        return new ActiveXObject("Microsoft.XMLHTTP");
    }
    return null;
}
这是我的html代码{entry_id}是数值参数:

<a href="" onclick="blodvotingview({entry_id});return false;" title="Vote Up">view</a>
<p>Suggestions: <span id="txtHint"></span></p>

建议:

这是我想回应的php代码:

 <?php
showcomment()

function showcomment()
{

    echo "yes";
}
?>


<>但是它不起作用,请帮助我。

如果你考虑使用jQuery和jQueY.SerialEdvin插件,你可以通过这个例子很容易地做到这一点:

$.post('URL', $('#form_id').serialize(), function(r) {
   console.log(r);
});

试试这个

function blodvotingview(contentid)
{

  var xmlhttp;    
  if (str==""){
      document.getElementById("txtHint").innerHTML="";
      return;
  }
  if (window.XMLHttpRequest){// code for IE7+, Firefox, Chrome, Opera, Safari
      xmlhttp=new XMLHttpRequest();
  }else{// code for IE6, IE5
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange=function(){
    if (xmlhttp.readyState==4 && xmlhttp.status==200){
      document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
  xmlhttp.open("GET","index.php?hp=1q="+str,true);
  xmlhttp.send();
}
应该很好,
Wezy

您的php语法无效;首先,在定义函数之前,您正在使用它。其次,对该函数的调用没有分号。@Dez请使用此站点上单词的完整拼写。人们会更认真地对待你。
$.get('URL', function(r) {
   console.log(r);
},'json'); // to parse response as JSON
function blodvotingview(contentid)
{

  var xmlhttp;    
  if (str==""){
      document.getElementById("txtHint").innerHTML="";
      return;
  }
  if (window.XMLHttpRequest){// code for IE7+, Firefox, Chrome, Opera, Safari
      xmlhttp=new XMLHttpRequest();
  }else{// code for IE6, IE5
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange=function(){
    if (xmlhttp.readyState==4 && xmlhttp.status==200){
      document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
  xmlhttp.open("GET","index.php?hp=1q="+str,true);
  xmlhttp.send();
}