Can Not Perform The New Script

[yangkai9999]

Hello,bellow are my test PHP codes.the first one is used to get a pull down list from exist mysql table and pass a selected value to a post variable and perfrom the next PHP code. when click go button. it seems work (not sure, but no error message come out when I run the code).the second code is used to present the values by using the difference method to transfer. The problem is that can not go to the next page and stay at the current page.Where is the wrong?Thanks,code 1:<?phpmysql_connect("localhost","root","*****");mysql_select_db("mytest") or die( "Unable to select database/ this is my note");echo "<form action='./update2.php' method='post'>"; echo "Please choose a patient for update:  "; echo "<select id='num' Onchange='changeLocation(this)' name=getid>"; echo "<option selected >"; $i=0; $query="SELECT p_id from mytest.p2 order by P_ID;"; $results = mysql_query($query) or die('no such table exist ' . mysql_error());; while ($row=mysql_fetch_row($results)) { if ($i == 0) { $i++; echo "<option value='p_id=$row[0]'>$row[0] </option>"; } else { echo "<option value='p_id=$row[0]'>$row[0]</option>"; echo "$row[0]"; } } echo "</select>"; echo "<input type='submit' name='go' value='go'>";echo "</form>";++$i;?>second one:<?php$p_id = $row[0];$p_id2= $_get($getid);$p_id3= $_post($getid);echo $p_id ;echo $p_id2 ;echo $p_id3 ;?>

[justsomeguy]

Add this to the top of all of your pages:

ini_set('display_errors', 1);error_reporting(E_ALL);

On the second page $row does not exist, $row only exists on the first page. Also, you need to use $_GET and $_POST, not $_get and $_post. You can use print_r to see the contents of an array:print_r($_GET);