简单的PHP+MySQL+AJAX留言板实例

对初学AJAX的同学会有帮助的。 ----------ajax.js--------------- var request = false; try { request = new XMLHttpRequest(); } catch (trymicrosoft) { try { request = new ActiveXObject("Msxml2.XMLHTTP"); } catch (othermicrosoft) { try { request = new ActiveXObject("Microsoft.XMLHTTP"); } catch (failed) { request = false; } } } if (!request) alert("Error initializing XMLHttpRequest!"); function getCustomerInfo() { document.getElementById("output").innerHTML = "Loading..."; document.getElementById("submit").disabled = true; var name = document.getElementById("name").value; var text = document.getElementById("text").value; var key = document.getElementById("key").value; var answer = document.getElementById("answer").value; var poststr = "name="+name+"&text="+text+"&key="+key+"&answer="+answer; var url = "doit.php"; request.open("POST", url, true); request.setRequestHeader("Content-Type","application/x-www-form-urlencoded"); request.send(poststr); request.onreadystatechange = updatePage; } function updatePage() { if (request.readyState == 4) if (request.status == 200) document.getElementById("output").innerHTML = request.responseText; else if (request.status == 404) alert("目标地址不存在"); else alert("警告: 错误代码 " + request.status); } function getInfo() { document.getElementById("output1").innerHTML = "Loading..."; var url = "show.php"; var d = new Date().getTime(); request.open("GET", url+"?d="+d.toString(), true); request.onreadystatechange = updatePage1; request.send(); } function updatePage1() { if (request.readyState == 4) if (request.status == 200) document.getElementById("output1").innerHTML = request.responseText; else if (request.status == 404) alert("目标地址不存在"); else alert("警告: 错误代码 " + request.status); }