下一页/上一页的 AJAX 请求帮助 [英] AJAX Request Help for next/previous page

问题描述

如何使用 AJAX 获取下一页/上一页.

如何在浏览器中调用:

page.php?page=1-1

或

page.php?page=1

返回只是文本.

应该以这种格式加载页面:

1-1或者1

当用户单击下一页/上一页按钮时，我如何将该页码传递给 ajax 调用并显示结果.

另外，我如何跟踪用户当前正在查看的页面?以及如何为页面设置最大最小值，例如我有 100 页不要调用第 101 页

http://jsfiddle.net/2b8gR/5/

HTML

<input id="loadPages" name="loadPages" type="button" value="Previous"/><div id="displayResults" name="displayResults">

JS(这不起作用)

$("#loadPages").click(function(){$.ajax({网址:'page.php'，数据:{'页面':'1-1'}，错误:函数(){警报('错误')；},成功:函数(返回数据){警报(返回数据)；$('#displayResults').append(returnData);}});});

解决方案

尝试这样的事情...保留一个名为 currentPage 的全局变量，并相应地调整页码.

现场演示 http://jsfiddle.net/Jaybles/MawSB/

HTML

<input id="prev" type="button" value="Previous"/><div id="displayResults" name="displayResults">当前页面:1</div>

JS

var currentPage=1;加载当前页面()；$("#next, #prev").click(function(){当前页面 =($(this).attr('id')=='next') ?currentPage + 1 : currentPage - 1;if (currentPage==0)//检查最小值当前页=1；else if (currentPage==101)//检查最大值当前页=100；别的加载当前页面()；});函数 loadCurrentPage(){$('input').attr('disabled','disabled');//禁用按钮//显示加载图像$('#displayResults').html('<img src="http://blog-well.com/wp-content/uploads/2007/06/indicator-big-2.gif"/>');$.ajax({url: '/echo/html/',数据:'html=当前页面:'+ currentPage+'&delay=1',类型:'POST'，成功:功能(数据){$('输入').attr('禁用','');//重新启用按钮$('#displayResults').html(data);//更新div}});}

那么你的php页面就可以访问$_REQUEST['page'];并相应地返回数据.

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

http://jsfiddle.net/2b8gR/5/

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);
        }
    });
});

解决方案

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.

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