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ajax处理php返回json数据例子_PHP教程

时间:2021-07-01 10:21:17 帮助过:20人阅读

/-----------test.html---------------/

function ajax_init()
{
var ajax=false;
try {
ajax = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
ajax = new ActiveXObject("Microsoft.XMLHTTP");
} catch (E) {
ajax = false;
}
}
if (!ajax && typeof XMLHttpRequest!=undefined) {
ajax = new XMLHttpRequest();
}
return ajax;
}
function post_data()
{
var url = "/server/xml.php";

var show_check = document.getElementById("show_check");
var show_flag = document.getElementById("show_flag");

var ajax = ajax_init();
ajax.open("GET", url, true);

//var img = "";
ajax.onreadystatechange = function()
{
if (ajax.readyState == 4 && ajax.status == 200)
{
var ret = ajax.responseText;
var show_info = eval(ret);
show_check.innerHTML = show_info[0].no1; // bsd
show_flag.innerHTML = show_info[0].no2; // wuddy
//alert("responseXMLs value: " + info[0].firstChild.data);
}
/*else
{
show.innerHTML = img;
}*/
}
ajax.send(null);
}

/*------------json.php-----------------*/
/* 这边如果写成 $info = array(no1=>bsd, no2=>wuddy, no3=>xie); */
/* 同样要把html文件中必成 var show_info = eval([ + ret + ]); 都是返回一个对象 */
$info = array(array(no1=>bsd, no2=>wuddy, no3=>xie));
$userinfo = json_encode($info);
echo $userinfo;
?>

www.bkjia.comtruehttp://www.bkjia.com/PHPjc/486585.htmlTechArticle/-----------test.html---------------/ label onclick="javascript:post_data();"click /label function ajax_init() { var ajax=false; try { ajax = new ActiveXObject("Msxml2.XMLHTTP"); }...

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