jquery+ajax+php实现搜索功能

转自：http://blog.csdn.net/less_cold/article/details/52353620

html代码

 "Content-Type" content="text/html"; charset="utf-8">
 搜索
 "jquery-3.0.0/jquery-3.0.0.min.js">
 "pagebar.css" rel="stylesheet" type="text/css" />
 "stylesheet" href="bootstrap.css" type="text/css"/>
 "bootstrap.min.js">

 "en">
 class="container">
 "text" id="gover_search_key" placeholder="请输入内容进行搜索"/>
 "submit" οnclick="search()">搜索

  

"container" class="rows">  

  

"text/javascript">  

function search(){  

    var keyword=$("#gover_search_key").val();  

    $.ajax({  

        type:"GET",  

        url:'search.php',   

        dataType:"json",   

        data:{'keyword':keyword},  

        success: function (data) {  

        //alert("test");  

                var html='';  

                var path1=eval(data)[0];  

                inum=eval(data)[4][0];  

                for(i=0;i

                  html = html+'+eval(data)[0][i]+'"style="height:200px,width:200px"/>

'+eval(data)[2][i]+'

';  

                  }  

                 $("#container").html(html);  

                      

        }  

    }  

    )  

}  

  

  

php代码

 header('Content-type:text/json; charset=utf-8');
 @ $db = mysqli_connect('localhost','root','');
 mysqli_query($db,"set names utf8;");
 mysqli_select_db($db,'mpicture');//找到数据库mpicture
 if(mysqli_connect_errno()){
  echo "Error:Could not connect to mysqli database.";
  exit;
 }
 $keyword=$_GET['keyword'];
 $q="SELECT count(*) FROM picture where title like '%$keyword%'";
 $result1 = mysqli_query($db,$q);
 $row1 = mysqli_fetch_array($result1);
 $allnum=$row1[0];
 //echo $allnum;
 $k="SELECT * FROM picture where title like '%$keyword%';";//对表picture进行选择
 $result = mysqli_query($db,$k);//执行$k的mysql语句，并赋给result
 $rownum = mysqli_num_rows($result);//获取result的数据数量
 $picture=array(array());
 for($i=0;$i<$rownum;$i++){
     $row = mysqli_fetch_assoc($result);//获取result的一条数据
     $picture[0][$i]= $row['path1'];//获得缩略图的路径
     $picture[1][$i]= $row['path2'];//获得全图的路径
     $picture[2][$i]= $row['title'];//获得文字描述
     $picture[3][$i]= $row['Id'];//获得id
 }
 $picture[4][0]=$allnum;
 echo json_encode($picture);
 ?>