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Home Backend Development PHP Tutorial 求圣人检查下面的代码错在哪里

求圣人检查下面的代码错在哪里

Jun 13, 2016 am 11:48 AM
content insert mysql nbsp

求高人检查下面的代码错在哪里。
代码如下,添加后提示添加失败。

<br /><meta http-equiv='Content-Type' content='text/html; charset=utf-8' /> <br /><?php<br />phpinfo();<br />//这是一个信息增、删、改操作处理页面<br /><br />//一、导入配置文件。<br />		require("dbconfig.php");<br />//二、链接MYSQL、并选择数据库。<br />		$link = @mysql_connect(HOST,USER,PASS) or die("数据库链接错误!");<br />		mysql_select_db(DBNAME,$link);<br />//三、根据action的值,来判断所操作,执行对应的代码。<br /> switch ($_GET["action"]){<br />	 case "add": //执行添加<br />	 //1、获取要添加的信息,并补充其它信息<br />	 $title = $_POST["title"];<br />	 $keywords = $_POST["keywords"];<br />	 $author = $_POST["author"];<br />	 $content = $_POST["content"];<br />	 $addtime = time();<br />	 //2、添加信息过过滤(省略)<br />	 //3、拼装添加SQL语句,并执行添加操作<br />		$sql = "insert into news valuse(null,'{$title}','{$keywords}','{$author}','{$addtime}','{$content}')";<br />//      echo $sql;  //这一步打印正常。<br />		mysql_query($sql,$link);<br />	 //4、判断是否添加成功,<br />	 	$id = mysql_insert_id($link);//判断刚刚添加的信息ID值值<br />		echo $id;<br />		break;<br />		if ($id>0){<br />			echo"<h3 id="添加成功">添加成功</h3>";<br />			}else{<br />			echo"<h3 id="添加失败">添加失败</h3>";<br />				}<br />	 break;<br />	 <br />	 case "del": //执行删除<br />	 <br />	 break;<br />	 <br />	 case "update": //执行修改<br />	 <br />	 break;<br />	 <br />	 }<br />//四、关闭数据库<br />mysql_close($link);<br />?><br />
Copy after login



------解决方案--------------------
第一段:insert?into?news?valuse
第二段:insert into news values
一个是valuse另一个是valuse

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