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查询问题
[code=PHP][/code]
include("conn.php");
$ip=$_SERVER['REMOTE_ADDR'];
$result=mysql_query("select * form ip where iip=$ip",$conn )or die(mysql_error());
if($result){
echo "<script> alert('失败');window.location.href='index.html'</script>";
}else{
mysql_query("insert into ip(id,ip) values(null,'$ip')",$conn)or die(mysql_error());
}
?>
我这样写为什么不对呀?奇怪啦,我要判断客户端的ip是否存在ip表中如果不存在就插入,存在就输出失败跳转到首页,可是怎么也调试不对呀?
------解决方案--------------------
$result=mysql_query("select * form ip where iip='$ip'",$conn )or die(mysql_error());
if($result){
------解决方案--------------------
[Quote=引用:]
[code=PHP][/code]
include("conn.php");
$ip=$_SERVER['REMOTE_ADDR'];
$result=mysql_query("select * form ip where iip=$ip",$conn )or die(mysql_error());
if($result){
echo "<script> alert('失败……<br />[/Quote]<br />楼主确定是iip不是ip?<br />$result=mysql_query("select * form ip where ip='".$ip."'",$conn )or die(mysql_error());
<br /><font color='#e78608'>------解决方案--------------------<br />
$result=mysql_query("select * from ip where iip='$ip'",$conn )or die(mysql_error()); //from写错了 <div class="clear"></script>