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我正嘗試從laravel blade模板中的json響應創建分頁鏈接。我知道它可以創建使用PHP,但我怎麼能做到這一點從JSON響應。如何創建laravel blade中的json數據的分頁鏈接
這裏是我的控制器方法:
function getContact()
{
return $contacts = Contact::where(array('is_active'=>1))
->paginate(2);
}
,這裏是我的jQuery Ajax代碼顯示在表中的數據...
$.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);
}
});
和執行console.log是 -
JSON響應{"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":"[email protected]","work_email"
:"[email protected]","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":"[email protected]","work_email":"[email protected]","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代碼顯示數據到表中,但我怎麼能顯示從分頁
使用jquery。
感謝@Ponce的回答,我增加的控制器編碼,但它返回數據[ '數據']爲{} –
我編輯你的答案,讓其他人得到這個線程的完美解決方案,現在這個解決方案完全適合我。 –