Question

怎么用Thinkphp+jquery实现ajax的提交，接收处理和返回

Answer 1

$.ajax({
type: "POST",
url: "/index.php/Home/Variable/add_variables.html",
data: {"set_id":st_id,"t_id":t_id,"varList":varList,"source_title":source_title},
dataType: "JSON",
success: function(ret){
console.log(ret);
addVarFlag=true;
if(ret.code==200){
for(var i=0;i<checkBoxs.length;i++){
if(!$(checkBoxs[i]).hasClass("checkAll")){
var cbox=$(checkBoxs[i]).parents("tr").remove();
}
}
fillColVarList(ret.data)
}else{
AlertBox("添加失败！");
}
}
});