我在WordPress中有一個AJAX函數,它調用一個PHP函數來返回數據庫中的瞬態記錄的值。爲什麼AJAX json腳本返回額外的0(零)
當我使用jQuery調用函數時,我收到結果,但它總是有一個額外的0(零)附加到值。
這裏是我的jQuery函數:
(function($) {
$(document).ready(function() {
var AdvancedDashboardWidget = function(element, options)
{
var ele = $(element);
var settings = $.extend({
action: '',
service: '',
countof: '',
query: '',
callback:''
}, options || {});
this.count=0;
var url='';
switch(settings.service)
{
case 'facebook':
if(settings.countof=='likes' || settings.countof=='talks')
{
ajaxCall(action,ele,settings);
}
}
};
var ajaxCall = function(action,ele,settings){
opts = {
url: ajaxurl, // ajaxurl is defined by WordPress and points to /wp-admin/admin-ajax.php
type: 'POST',
async: true,
cache: false,
dataType: 'json',
data:{
action: settings.action // Tell WordPress how to handle this ajax request
},
success:function(response) {
//alert(response);
ele.html(response);
return;
},
error: function(xhr,textStatus,e) { // This can be expanded to provide more information
alert(e);
//alert('There was an error');
return;
}
};
$.ajax(opts);
};
$.fn.advanceddashboardwidget = function(options)
{
return this.each(function()
{
var element = $(this);
// Return early if this element already has a plugin instance
if (element.data('advanceddashboardwidget')) return;
// pass options to plugin constructor
var advanceddashboardwidget = new AdvancedDashboardWidget(this, options);
// Store plugin object in this element's data
element.data('advanceddashboardwidget', advanceddashboardwidget);
});
};
});
})(jQuery);
但是也有參與,這是主要的jQuery的功能與WordPress通信,並返回PHP函數的值更輔助功能。
的問題是,如果該值例如返回爲「99
」,它將爲「990
」
這裏是jQuery是調用PHP函數返回:
/**
* Get Facebook Likes
*/
public function get_facebook_likes(){
echo 99;
}
如果我將以上內容更改爲return 99;
,我會收到純文本0
你添加一個數字時,你應該增加一個數量爲數字的字符串,無論是在PHP端或使用Javascript結束? – JayC
只是爲了調試我有PHP函數「return 99;」。它仍然返回990 – Jason