Ajax - How refresh <DIV> after submit
Asked Answered
S

5

7

How can I refresh just part of the page ("DIV") after my application releases a submit? I'm use JQuery with plugin ajaxForm. I set my target with "divResult", but the page repeat your content inside the "divResult". Sources:

   <script>      
       $(document).ready(function() {      
           $("#formSearch").submit(function() {      
                var options = {    
                  target:"#divResult",
                  url: "http://localhost:8081/sniper/estabelecimento/pesquisar.action"  
                }      
               $(this).ajaxSubmit(options);      
               return false;      
          });      
      })
   </script>

Page

 <s:form id="formSearch" theme="simple" class="formulario" method="POST">      
 ...      

 <input id="btTest" type="submit" value="test" >      

 ...      

                 <div id="divResult" class="quadro_conteudo" >      
                     <table id="tableResult" class="tablesorter">      
                         <thead>      
                             <tr>      
                                 <th style="text-align:center;">      
                                     <input id="checkTodos" type="checkbox" title="Marca/Desmarcar todos" />      
                                 </th>      
                                 <th scope="col">Name</th>      
                                 <th scope="col">Phone</th>      
                             </tr>      
                         </thead>      

                         <tbody>      
                             <s:iterator value="entityList">      
                                 <s:url id="urlEditar" action="editar"><s:param name="id" value="%{id}"/></s:url>      
                                <tr>      
                                    <td style="text-align:center;"><s:checkbox id="checkSelecionado" name="selecionados" theme="simple" fieldValue="%{id}"></s:checkbox></td>      
                                    <td> <s:a href="%{urlEditar}"><s:property value="name"/></s:a></td>      
                                    <td> <s:a href="%{urlEditar}"><s:property value="phone"/></s:a></td>      
                                </tr>      
                             </s:iterator>      
                         </tbody>      
                     </table>      

                     <div id="pager" class="pager">      
                         <form>      
                             <img src="<%=request.getContextPath()%>/plugins/jquery/tablesorter/addons/pager/icons/first.png" class="first"/>      
                             <img src="<%=request.getContextPath()%>/plugins/jquery/tablesorter/addons/pager/icons/prev.png" class="prev"/>      
                             <input type="text" class="pagedisplay"/>      
                             <img src="<%=request.getContextPath()%>/plugins/jquery/tablesorter/addons/pager/icons/next.png" class="next"/>      
                             <img src="<%=request.getContextPath()%>/plugins/jquery/tablesorter/addons/pager/icons/last.png" class="last"/>      
                             <select class="pagesize">      
                                 <option selected="selected" value="10">10</option>      
                                 <option value="20">20</option>      
                                 <option value="30">30</option>      
                                 <option value="40">40</option>      
                                 <option value="<s:property value="totalRegistros"/>">todos</option>      
                             </select>      
                             <s:label>Total de registros: <s:property value="totalRegistros"/></s:label>      
                         </form>      
                     </div>      
                     <br/>      
             </div> 

Thanks.

Skep answered 15/5, 2009 at 14:9 Comment(0)
S
8

To solve this using jquery I would try this;

$(document).ready(function() {
    $("#formSearch").submit(function() {
        var options = {
            /* target:"#divResult", */

            success: function(html) {
                $("#divResult").replaceWith($('#divResult', $(html)));
            },

            url: "http://localhost:8081/sniper/estabelecimento/pesquisar.action"
        }

        $(this).ajaxSubmit(options);
        return false;
    });
});

alternatively, you could get the server to return just the html that needs to be inserted into the div rather than the rest of the html document.

I don't really know the TableSorter plugin but I do know that you will need to reinitialize your TableSorter plugin each time you reload the element. so add a line to your success function that targets your table such as

success: function(html) {
    var resultDiv = $("#divResult").replaceWith($('#divResult',     $(html)));

    $('table.tablesorter', resultDiv).TableSorter();
}
Shaveling answered 15/5, 2009 at 14:32 Comment(0)
I
3

Your problem is on the server side: you have to make a page that returns only the div you want, and then change the 'url' to match that.

Currently you're loading the full page with the AJAX call, which is why it's returning the whole page.

Inescutcheon answered 15/5, 2009 at 14:31 Comment(0)
R
0

You can use most of the normal jquery ajax function parameters with ajaxSubmit. Just pass in a success function.

$('#formSearch').ajaxSubmit({success: function(){ /* refresh div */ }); 

See here for a more elaborate example.

Resonate answered 15/5, 2009 at 14:11 Comment(1)
How use normal JQuery in this case?Skep
A
0

If you just want to refresh your div with the same static content that was in it before it was overridden by your post results, you can maybe try something like:

$("#Container").html($("#Container").html());

of course be careful that the inner HTML has not changed, or alternatively, you can store the innerHTML in a variable in the document.ready() function, then load it whenever you need to. Sorry, written in haste.

Ancell answered 16/4, 2010 at 21:40 Comment(0)
B
0

Using jQuery, sometimes when I perform a .get ajax call to update a database, I reload/refresh another part of the page using a simple jQuery .load statement in the success callback function to replace the contents of a div.

$("#div_to_refresh").load("url_of_current_page.html #div_to_refresh")

I'd like to note that this is not the most efficient way to reload a small portion of data, and thus I would only use it on a low traffic web app, but it is simple to code.

Baler answered 4/8, 2014 at 23:54 Comment(0)

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