mysql问题，不能修改也不能插入,求高手解决

Question

<?php
$vod_name=$_POST['vod_name'];
$vod_cid=$_POST['vod_cid'];
$vod_area=$_POST['vod_area'];
$vod_year=$_POST['vod_year'];
$vod_actor=$_POST['vod_actor'];
$vod_continu=$_POST['vod_continu'];
$vod_pic=$_POST['vod_pic'];
$vod_play=$_POST['vod_play'];
$vod_url=$_POST['vod_url'];
$vod_language=$_POST['vod_language'];
$vod_content=$_POST['vod_content'];
$vod_server=$_POST['vod_server'];
$vod_director=$_POST['vod_director'];
$vod_letter=$_POST['vod_letter'];
$vod_addtime = strtotime($_POST['vod_addtime']);
$conn=mysql_connect('localhost','root','haojunqing') or die('连接失败');
mysql_select_db('ppvod',$conn) or die('选择数据库失败'); 
$s_sql="select vod_name from ff_vod where vod_name='$vod_name'";
$s_result=mysql_query($s_sql);
$s_row=mysql_fetch_array($s_result);
if($s_row){
$update_sql="update ff_vod set vod_year='$vod_year',vod_actor='$vod_actor',vod_pic='$vod_pic',vod_play='$vod_play',vod_url='$vod_url',vod_addtime='$vod_addtime',vod_server='$vod_server' WHERE vod_name='$vod_name'";
$update_result=mysql_query($updare_sql);
if(mysql_affected_rows($conn)!=0)
echo "修改".$vod_name."成功";
else
echo "修改".$vod_name."失败";
}
else
{
$insert_sql="insert into ff_vod(vod_cid,vod_name,vod_actor,vod_director,vod_content,vod_pic,vod_area,vod_language,vod_year,vod_continu,vod_addtime,vod_play,vod_server,vod_url) values('$vod_cid','$vod_name','$vod_actor','$vod_director','$vod_content','$vod_pic','$vod_area','$vod_language','$vod_year','$vod_continu','$vod_addtime','$vod_play','$vod_server','$vod_url')";
$insert_result=mysql_query($insert_sql) or die('添加失败');
if(mysql_affected_rows($conn)!=0)
echo "添加".$vod_name."成功";
}

?>

Answer 1

"update ff_vod set vod_year='".$vod_year."',....这样写应该就可以了

Answer 2

代码最后加一句 echo mysql_error();
你现在处于项目调试阶段 mysql_query($insert_sql) or die('添加失败'); 这样的返回的错误信息太笼统了，很不方便调试。

Answer 3

$update_sql="update ff_vod set vod_year=$vod_year,vod_actor=$vod_actor,vod_pic=$vod_pic,vod_play=$vod_play,vod_url=$vod_url,vod_addtime=$vod_addtime,vod_server=$vod_server WHERE vod_name=$vod_name";

Answer 4

打开 错误提示。另外vod_name='$vod_name' 是否存在这个update 的条件?

Answer 5

这句$s_sql="select vod_name from ff_vod where vod_name='$vod_name'";
应该是这样的$s_sql="select vod_name from ff_vod where vod_name="$vod_name;

Answer 6

因为在php中单引号，双引号是有区别的，单引号里面的变量做字符串解析，双引号里面的变量还是原来变量的值

Answer 7

打印出sql语句，在mysql里执行一下 是否成功。。 

Answer 8

输出sql打印出来看看就知道了。