【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】: