PHP加上if判断语句后如何做JS跳转？

Question

chk.php <?php 
               $yer=$_GET['yer'];
   $mon=$_GET['mon'];
   $day=$_GET['day'];
               $link=@mysql_connect("localhost","root","root") or die ("数据库链接失败".mysql_error());
               mysql_select_db("bnk",$link);
               mysql_query("set names gb2312"); 
               $sql=mysql_query("select * from bnk_stm  where YER='".$yer. "' and MON='".$mon. "'and DAY='".$day."' and RAT_S=1 order by rand()  limit 1");
               $row=mysql_fetch_array($sql);
   if($row==false) {
      echo "跳转到没有资料的页面";
  }
  else{
  echo "跳转到有资料的页面";
  
   ?>    <?php 
}
mysql_close($link);
 ?>上面的代码 判断值为空或不空时如何做JS跳转？就是check过后直接跳转到另一个页面

Answer 1

echo '<script language="javascript" type="text/javascript">           window.location.href="要跳转的页面"</script>'

Answer 2

if($row==false) {
        echo '<script>document.location.href="php.php";</script>'
 }
 else{
        echo '<script>document.location.href="php.php";</script>'
}

Answer 3

echo
 '<script language="javascript"type="text/javascript">
window.location.href="../STAMP_CENTER/index.php?STAMP=$row[STAMP]</script>';地址后我想让STAMP等于这个字段的值，应该怎样用引号？

Answer 4

echo "<script language=\"javascript\"type=\"text/javascript\">
window.location.href=\"../STAMP_CENTER/index.php?STAMP=".$row[STAMP]."</script>";

Answer 5

echo '<script language="javascript"type="text/javascript">
window.location.href="../STAMP_CENTER/index.php?STAMP='.$row['STAMP'].'</script>';

Answer 6

function alert_go($msg,$url){
echo('<script language="JavaScript">');
//echo("alert('$msg');");
echo("location.href='$url';");
echo('</script>');
exit;
}
alert_go($url);

Answer 7

function alert_go($msg,$url){
    echo('<script language="JavaScript">');
    echo("alert('$msg');");
    echo("location.href='$url';");
    echo('</script>');
    exit;
}
alert_go('记录不能为空',$url);//提示后自动跳转到指定的页面去