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小弟我这段代码有什么错啊mysql_query不起作用也不报错的

WBOY
WBOYOriginal
2016-06-13 10:12:071089browse

我这段代码有什么错啊,mysql_query不起作用也不报错的

include("in.php");
if (empty($_POST['YBT']))
{
mysql_close();
echo "<script>alert('非法提交!请重新提交!');history.back();</script>";
exit;
}
$YLX='展会资讯';
$exe="insert into yzx (YLX,YNR,YBT,YDT) values ('$YLX','".phpasp2($_POST["YNR"])."','".htmlencode2($_POST['YBT'])."','".time()."')";
$result=mysql_query($exe);
mysql_close();
echo "<script>alert('添加成功!');location.href='newslist.php';</script>";
exit;
?>

------解决方案--------------------
$result=mysql_query($exe) or die(mysql_error());

如有错误,请贴出。
------解决方案--------------------
那你可没设计好表,你想要自动增值,可是表却没有设置YID为自增

探讨

Field 'YID' doesn't have a default value. 可是我想要的事YID自动增值的呀

------解决方案--------------------
PHP code
//查看sql语句是否正确echo "sql-->>{$exe}<br>";if(mysql_affected_rows() > 0){ //若执行成功  echo "<script>alert('添加成功!');location.href='newslist.php';</script>";}else{  mysql_close();  echo "<script>alert('添加失败!');return;</script>";  exit;}mysql_close();<div class="clear">
                 
              
              
        
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