jquery ajax 传递参数？

Question

function a(other){var show=1;var page=1;$.ajax({url:url,type:"post",dataType:"json",data:{show:show,page:page}//这里other传递的参数怎么添加进来？});}function test(){a("a:1,b:2");}

Answer 1

楼主修改一下传参的方式就可以了，试着这样修改

12345678910111213141516

function a(other){    var show=1;     var page=1;     $.ajax({        url:url,         type:"post",         dataType:"json",         data:$.extend({show:show,page:page},other)    });}test({a:1,b:2});//字符串修改成对象传入

Answer 2

type:"POST",
url:url,
data:{'other':other},

Answer 3

function a(other){
var show=1;

var page=1;

$.ajax({
url:url,

type:"post",

dataType:"json",

data:{show:show,page:page,other:other}//这样

});
}
function test(){
a("a:1,b:2");

}