AJAX请求帮助下一/ previous页 [英] AJAX Request Help for next/previous page
问题描述
我如何使用AJAX来获取下一个/ previous页(S)。
如何在浏览器中调用:
page.php?页= 1-1
或
page.php?页= 1
返回仅仅是文本。
应该载入的页面格式为:
1-1 要么 1
当用户点击下一个/ previous页按钮(S)我怎么传递页码Ajax调用并显示结果。
另外我如何跟踪哪些页面的用户正在观看的? 我如何把一个最高分的页面,比如我有100个页面不作要求101页
HTML
<输入ID =loadPagesNAME =loadPages类型=按钮值=下一步/>
<输入ID =loadPagesNAME =loadPages类型=按钮值=previous/>
< DIV ID =displayResultsNAME =displayResults>
< / DIV>
JS(这不工作)
$(#loadPages)。点击(函数(){
$阿贾克斯({
网址:page.php,
数据:{页面:1-1},
错误:函数(){警报(错误); },
成功:函数(returnData){
警报(returnData);
$('#displayResults)追加(returnData)。
}
});
});
尝试这样的事情...随身携带一本名为当前页
全局变量,只是调整相应的页码
现场演示 http://jsfiddle.net/Jaybles/MawSB/一>
HTML
<输入ID =下一个类型=按钮值=下一步/>
<输入ID =preV类型=按钮值=previous/>
< DIV ID =displayResultsNAME =displayResults>当前页:1< / DIV>
JS
VAR当前页= 1;
loadCurrentPage();
$(#接下来,#preV)。点击(函数(){
当前页=
($(本).attr('身份证')=='下一步')?当前是+ 1:当前页 - 1;
如果(当前是== 0)//检查分钟
当前是= 1;
否则,如果(当前页== 101)//检查最大
当前是= 100;
其他
loadCurrentPage();
});
功能loadCurrentPage(){
$('输入')ATTR(禁用,禁用)。 //禁用按钮
//显示加载图像
$('#displayResults)HTML('< IMG SRC =http://blog-well.com/wp-content/uploads/2007/06/indicator-big-2.gif/>') ;
$阿贾克斯({
网址:/回声/ HTML /',
数据:'HTML =当前位置:'+当前页+'和;延迟= 1',
键入:POST,
成功:功能(数据){
$('输入')ATTR('残疾','')。 //重新启用按钮
$('#displayResults')的HTML(数据); //更新事业部
}
});
}
那么你的PHP页面可以访问 $ _ REQUEST ['页'];
键,返回相应数据。
How do I use AJAX to get the Next/Previous Page(s).
How to call in a browser:
page.php?page=1-1
or
page.php?page=1
The return is just text.
Should load pages in this format:
1-1 or 1
When the user clicks the next/previous page button(s) how do I pass that page number to the ajax call and display the results.
Also how do I keep track of what page the user is currently viewing? And how do I put a max min for pages, like I have 100 pages don't make call for page 101
HTML
<input id="loadPages" name="loadPages" type="button" value="Next" />
<input id="loadPages" name="loadPages" type="button" value="Previous" />
<div id="displayResults" name="displayResults">
</div>
JS (This is not working)
$("#loadPages").click(function(){
$.ajax({
url: 'page.php',
data:{'page': '1-1'},
error : function (){ alert('Error'); },
success: function (returnData) {
alert(returnData);
$('#displayResults').append(returnData);
}
});
});
Try something like this... Keep a global variable called currentPage
and simply adjust the page number accordingly.
LIVE DEMO http://jsfiddle.net/Jaybles/MawSB/
HTML
<input id="next" type="button" value="Next" />
<input id="prev" type="button" value="Previous" />
<div id="displayResults" name="displayResults">Current Page: 1</div>
JS
var currentPage=1;
loadCurrentPage();
$("#next, #prev").click(function(){
currentPage =
($(this).attr('id')=='next') ? currentPage + 1 : currentPage - 1;
if (currentPage==0) //Check for min
currentPage=1;
else if (currentPage==101) //Check for max
currentPage=100;
else
loadCurrentPage();
});
function loadCurrentPage(){
$('input').attr('disabled','disabled'); //disable buttons
//show loading image
$('#displayResults').html('<img src="http://blog-well.com/wp-content/uploads/2007/06/indicator-big-2.gif" />');
$.ajax({
url: '/echo/html/',
data: 'html=Current Page: ' + currentPage+'&delay=1',
type: 'POST',
success: function (data) {
$('input').attr('disabled',''); //re-enable buttons
$('#displayResults').html(data); //Update Div
}
});
}
Then your php page can access $_REQUEST['page'];
and return data accordingly.
这篇关于AJAX请求帮助下一/ previous页的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!