简单的ajax分页疑问,该如何处理

WBOY

发布： 2016-06-13 13:33:55

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888 人浏览过

简单的ajax分页疑问

HTML code

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<title>ajax分页</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<meta name="description" content="">
<meta name="keywords" content="">
<script type="text/javascript">
function ajax() {
    var ajax = false;
    if(window.XMLHttpRequest) {
        ajax = new XMLHttpRequest();
    } else {
        ajax = new ActiveXObject("Microsoft.XMLHTTP");
    }
    return ajax;
}
window.onload = function check(node) {
    var parameter = "page=" + node;
    var nokia = ajax();
    //alert(nokia);return;
    nokia.open('POST',"process.php",true);
    nokia.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
    nokia.send(parameter);
    nokia.onreadystatechange = function () {
        if(nokia.readyState==4 && nokia.status==200) {
            document.getElementById('span1').innerHTML = this.responseText;
        }
    }
}
</script>

<style type="text/css">
</style>

    
        <span id="span1"></span>

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PHP code

<!--

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--><?php $page = isset($_POST['page'])?$_POST['page']+0:1;
if($page==0) {
$page =1;
}
$conn = mysql_connect('localhost','root','111111');
mysql_select_db('msg');
$sql = 'select count(*) from news';
$info = mysql_query($sql,$conn);
$msg = mysql_fetch_row($info);
$counts = $msg[0]; 
$perpage = 5; 
$pages = ceil($counts/$perpage); 
$start = $page - (5-1)/2;
$end = $page + (5-1)/2;
$start = $start<1?1:$start;
$end = ($start+5-1)>$pages?$pages:($start+5-1);
$end = $end>$pages?$pages:$end;
$start = ($end-5+1)' . $i . ' ';
}
echo $link;

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运行上面的html文件可以显示 1 2 3 4 5
可一单击超链接时，firefox的debug显示 check is not defined
不是已经innerHTML进来了么，而且奇怪的是那个$page如果不加零的话显示的object element，望朋友解答！！！

------解决方案--------------------

这样呢
function check(node) {
var parameter = "page=" + node;
var nokia = ajax();
//alert(nokia);return;
nokia.open('POST',"process.php",true);
nokia.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
nokia.send(parameter);
nokia.onreadystatechange = function () {
if(nokia.readyState==4 && nokia.status==200) {
document.getElementById('span1').innerHTML = this.responseText;
}
}
}
window.onload = check(node);

------解决方案--------------------

JScript code

window.onload = function ()
{
   check(1);
}
function check(node) {
    var parameter = "page=" + node;
    var nokia = ajax();
    //alert(nokia);return;
    nokia.open('POST',"process.php",true);
    nokia.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
    nokia.send(parameter);
    nokia.onreadystatechange = function () {
        if(nokia.readyState==4 && nokia.status==200) {
            document.getElementById('span1').innerHTML = this.responseText;
        }
    }
}
<br><font color="#e78608">------解决方案-------------------- <div class="clear">
                 
              
              
        
            </div></font>

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