ThinkPHP实现ajax仿官网搜索功能实例

//搜索,如果在1不在0?
function search(){?
??? $keyword = $_POST['search'];?
??? $Goods=M('goods');?
? //这里我做的一个模糊查询到名字或者对应的id,主要目的因为我这个系统是?
? //商城系统里面用到直接看产品ID?
??? $map['goods_id|goods_name']? = array('like','%'.$keyword.'%');?
??? // 把查询条件传入查询办法?
??? if($goods=$Goods->where($map)->select())?
???? {?
????????????? $this->ajaxReturn($goods,'查询成功!',1);?
???? }else{?
????????????? $this->ajaxReturn($data,"查询失败,数据不存在!",0);?
?}?
}

$(document).ready(function(){?
?? $(".show_message").hide();?
?? var $search=$('#search_box');?
?? $("#submit_from").click(function(){?
??? if($("#search_box").attr("value")=='')?
??? {?
??????? //alert('请输入文字!');?
??????? $(".show_message").html('错误提示：搜索框文本不能为空!');?
??????? $(".show_message").fadeIn(1000);?
??????? $(".show_message").fadeOut(1000);?
??????? $search.focus();?
??????? //return false;?
??? }else{?
??????? //开始ajax执行数据?
??????? $.ajax({?
??????????? type: "POST",?
??????????? url:"/index.php/Goods/search",?
??????????? data:{?
??????????????? search:$search.val()?
??????????? },?
??????????? dataType: "json",?
??????????? success: function (data) {?
??? if (data.status == 1) {?
??????????? //alert(data.info);?
??????????? var html='';?
??????????????????? $.each(data.data,function(no,items){?????
??????????????????? html+='';?
??????????????????? });?
??????????????????? html+="?
'+items.goods_id+' '+items.goods_name+' '+items.add_time+' '+items.brand+' '+items.price+'";?????
???????????????????? $(".goods-list").html(' ').html(html);?
?????????????????? // alert(html);?
??? }?
??? else if (data.status == 0) {?
??????? $(".show_message").show();?
??????? $(".show_message").html(data.info);?
??????????????????? $(".show_message").fadeOut(3000);?
??? //??? alert(data.info);?
????????? return false;?
??????? }?
????? }?
???????? });?
??? }?
? });?
});