【问题标题】:how to create pagination links from json data in laravel blade如何从 laravel 刀片中的 json 数据创建分页链接
【发布时间】:2016-04-06 10:23:47
【问题描述】:

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,但是我怎样才能从 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);
}

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);
           }
       });

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}]}

我可以使用上面的 jquery 代码将数据显示到表格中,但是如何显示来自"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, 的分页

使用 jquery。

【问题讨论】:

    标签: php jquery json laravel


    【解决方案1】:

    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()
                                    )
                              );
    }
    

    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
           }
       });
    

    【讨论】:

    • 感谢@Ponce 的回答,我添加了控制器代码,但它返回 data['data'] 作为 {}
    • 我编辑你的答案,以便其他人得到这个线程的完美解决方案,现在这个解决方案非常适合我。
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