Home > Backend Development > PHP Tutorial > ajax search prompt_PHP tutorial

ajax search prompt_PHP tutorial

WBOY
Release: 2016-07-13 17:37:12
Original
899 people have browsed it

数据库表: 复制内容到剪贴板
代码:

CREATE TABLE `xqbar`.`suggest` (
`id` INT NOT NULL AUTO_INCREMENT ,
`title` VARCHAR( 100 ) NOT NULL ,
`hits` INT NOT NULL DEFAULT 0,
PRIMARY KEY ( `id` )
) ENGINE = InnoDB

insert into suggest(title,hits)values(xqbar.com,100);
insert into suggest(title,hits)values(www.xqbar.com,410);
insert into suggest(title,hits)values(http://xqbar.com,700);
insert into suggest(title,hits)values(mail:xqbar.com,200);
insert into suggest(title,hits)values(ftp:xqbar.com,100);
insert into suggest(title,hits)values(http://www.xqbar.com,70)search.php
(关于php我也是接触不久,下面的php如果罗嗦还望高手指点)
返回的信息字符串要为 xxx1|xxx2$200|100 前后对应 复制内容到剪贴板
代码:

if($_GET["action"]!=){
#获取用户输入的关键字
$keyword=$_GET["keyword"];
#过滤关键字
$keyword=str_replace("[","[[]",$keyword);
$keyword=str_replace("&","[&]",$keyword);
$keyword=str_replace("%","[%]",$keyword);
$keyword=str_replace("^","[^]",$keyword);
#链接数据库
$conn=mysql_connect("localhost","xqbar","xqbaradmin");
#选择数据库
@mysql_select_db("xqbar") or die(sorry);
mysql_query(set names utf-8);
#查询语句
$sql="select title,hits from suggest where title like %".$keyword."% order by hits desc limit 10";
$result=mysql_query($sql);
#循环得到查询结果,返回字符串
#格式为 结果1|结果2$结果1点击次数|结果2点击次数
if($result){
$i=1;$title=;$hits=;
while($row=mysql_fetch_array($result,MYSQL_BOTH))
{
$title=$title.$row[title];
$hits=$hits.$row[hits];
if($i {
$title=$title."|";
$hits=$hits."|";
}
$i++;
}
}
mysql_close();
}
?>
js代码 复制内容到剪贴板
代码:


引入prototye.js有朋友说这个库太大,或者把,不习惯的朋友可以使用jquery.js框架或者直接创建ajax对象,这个我就不想说了,这里直接引用prototye.js框架

创建层和显示查询结果的js代码

Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template