如何从laravel刀片中的json数据创建分页链接

use*_*387 5 php jquery json laravel

我正在尝试从 laravel 刀片模板中的 json 响应创建分页链接。我知道它可以简单地使用 php 创建,但是我怎么能从 json 响应中做到这一点。

这是我的控制器方法:

function getContact()
{   
     return $contacts = Contact::where(array('is_active'=>1))
                        ->paginate(2);
}
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这是我的 jquery ajax 代码,用于在表上显示数据...

   $.ajax({
            type:'GET',
            url: '<?php echo URL::to('contactgroup/contact') ?>', 

            dataType: 'json',               
            success: function(data){
                //console.log(data['data'].length);
            var table='<thead><th><input type="checkbox" id="checkAll" name="checkAll[]"/></th><th>Name</th><th>contact No</th></thead>';
            for(var i=0; i<data['data'].length; i++){
                table += '<tr><td><input type="checkbox" name="checkbox[]" class="individualCheckbox" value="'+data['data'][i].id+'"/></td><td>'+data['data'][i].contact_name+'</td><td>'+data['data'][i].primary_contact_no+'</td></tr>';

            }
            $('#contactTable').empty();
            $('#contactTable').append(table);
           }
       });
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console.log 中的 json 响应是-

{"total":3,"per_page":2,"current_page":1,"last_page":2,"next_page_url":"http:\/\/localhost\/smsapi\/public\/contactgroup\/contact\/?page=2","prev_page_url":null,"from":1,"to":2,"data":[{"id":1,"contact_name"
:"M Islam","primary_contact_no":"017********","personal_email":"m@company.com","work_email"
:"m@company.com","personal_phone":"017********","work_phone":"017********","personal_address":"abc","work_address":"ring road","is_active":1,"entry_by":7},{"id":4,"contact_name":"sdsdf","primary_contact_no":"242342","personal_email":"mu@company.com","work_email":"mu@company.com","personal_phone":"12142","work_phone":"fgbf","personal_address":"gfg","work_address":"fgfg"
,"is_active":1,"entry_by":7}]}
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我可以使用上面的 jquery 代码将数据显示到表格中,但是我如何显示分页 "total":3,"per_page":2,"current_page":1,"last_page":2,"next_page_url":"http:\/\/localhost\/smsapi\/public\/contactgroup\/contact\/?page=2","prev_page_url":null,"from":1,"to":2,

使用jQuery。

rfp*_*pdl 5

在你的控制器中

function getContact()
{   
   $contacts = Contact::where(array('is_active'=>1))
                    ->paginate(2);

   //return multiple value in JSON format
   return \Response::JSON(array(
                                   'data'       => $contacts, 
                                   'pagination' => (string) $contacts->links()
                                )
                          );
}
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在你看来

 $.ajax({
        type:'GET',
        url: '<?php echo URL::to('contactgroup/contact') ?>', 

        dataType: 'json',               
        success: function(data){
            //console.log(data['data'].length);
        var table='<thead><th><input type="checkbox" id="checkAll" name="checkAll[]"/></th><th>Name</th><th>contact No</th></thead>';
        for(var i=0; i<data['data'].length; i++){
            table += '<tr><td><input type="checkbox" name="checkbox[]" class="individualCheckbox" value="'+data['data'][i].id+'"/></td><td>'+data['data'][i].contact_name+'</td><td>'+data['data'][i].primary_contact_no+'</td></tr>';

        }
        $('#contactTable').empty();
        $('#contactTable').append(table);
        $('#pagination').html(data['pagination']);   //add this element in your HTML as well
       }
   });
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