一页的Ajax加载div从另一页调用

Question

I am facing with some glitches regarding the ajax load
OK, so the query is lets say I have page a (picture attached)

the ajax loads the div into the "RESULT" div on clicking the specific id's
here is the ajax script & the html i have come so far. this loads perfectly on the Result div on the page a itself.

<a href="#" id="one" /><br>
<a href="#" id="two" /><br>
<a href="#" id="three" /><br>
<div id="result"  class="functions"></div>

    $.ajaxSetup({
        cache: false
    });
    var ajax_load = "<img class='loading' src='img/load.gif' alt='loading...' />";


    var loadUrl = "content.php";
    $("#one").click(function () {
    $("#result").html(ajax_load).load(loadUrl);
    });

    var load2Url = "content1.php";
    $("#two").click(function () {
    $("#result").html(ajax_load).load(loadUrl);
    });

    var load3Url = "content2.php";
    $("#three").click(function () {
        $("#result").html(ajax_load).load(load2Url);
    });

thing is i need to load the same contents on the result div on page b from another page using the same id calls, lets say page b.

the concept is, on page b, the id is clicked, it LOADS page a first & loads the content on "RESULT" div. both pages are on the same server
Would really appreciate any assistance. thank you.

Answer 1

One way to handle it will be to pass the dynamic item as a request parameter like in page b use the url /pagea.html?section=one etc

Then on page load

$(function(){
    var location = window.location.href;
    var page = location.substring(location.indexOf('=') + 1);
    $('#' + page).trigger('click');
})

Answer 2

You got your loadUrl 's in the wrong pace they should be inside of the function otherwise it's value is the last value it was set to which is content2.php

jsFiddle Demo

$.ajaxSetup({
    cache: false
});

var ajax_load = "<img class='loading' src='img/load.gif' alt='loading...' />";

$("#one").click(function () {
    var loadUrl = "content.php";
    $("#result").html(ajax_load).load(loadUrl);
});

$("#two").click(function () {
    var loadUrl = "content1.php";
    $("#result").html(ajax_load).load(loadUrl);
});

$("#three").click(function () {
    var loadUrl = "content2.php";
    $("#result").html(ajax_load).load(loadUrl);
});