Hi,

       I think you should read a beginner tutorial about SQL. there are just
four things Insert, Delete, Update and select, that need to be known for
working with database in normal conditions, after that you can move to
create, drop, alter or much more like joins etc.

http://www.lmpx.com/mysql_ref.php


Zareef Ahmed



----- Original Message ----- 
From: "Anasta" <[EMAIL PROTECTED]>
To: <php-general@lists.php.net>
Sent: Sunday, December 18, 2005 7:41 AM
Subject: [PHP] insert to DB


> It just wont insert new data into DB--any ideas as i have gone over and
> over.
>
>
>
> <?
> mysql_connect("localhost","anasta","silteren");
>
> mysql_select_db("mytipperv1");
>
> if(!isset($cmd))
> {
>    $result = mysql_query("select * from leaderboard order by ID");
>
>    while($r=mysql_fetch_array($result))
>
>    {
>       //grab the title and the rating of the entry
>       $id=$r["ID"];//take out the id
>    $Name=$r["Name"];//take out the Name
>    $Rating=$r["Rating"];//take out the Rating
>
>
>     echo "$Name <a href='edit.php?cmd=edit&ID=$id'> Edit</a><br>";
>
>
>     }
> }
> ?>
> <?
> if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
> {
>    if (!isset($_POST["submit"]))
>    {
>       $id = $_GET["ID"];
>       $sql = "SELECT * FROM leaderboard WHERE ID=$id";
>       $result = mysql_query($sql);
>       $myrow = mysql_fetch_array($result);
>       ?>
>
>       <form action="../craigs_DB/edit.php" method="post">
>       <input type=hidden name="ID" value="<?php echo $myrow["ID"] ?>">
>
>       Name: <INPUT TYPE="TEXT" NAME="Name"  VALUE="<?php echo
$myrow["Name"]
> ?>" SIZE=15 maxlength="15"><br><br>
>       Rating: <INPUT TYPE="TEXT" NAME="Rating"  VALUE="<?php echo
> $myrow["Rating"] ?>" SIZE=3 maxlength="3"><br><br>
>       <input type="hidden" name="cmd" value="edit">
>       <input type="submit" name="submit" value="submit">
>       </form>
>
> <? } ?>
> <?
>    if ($_POST["$submit"])
>    {
>       $Name = $_POST["Name"];
>    $Rating = $_POST["Rating"];
>
>    $sql = "UPDATE leaderboard SET Name='$Name', Rating='$Rating' WHERE
> ID=$id";
>       //replace news with your table name above
>       $result = mysql_query($sql);
>       echo "Thank you! Information updated.";
>  }
> }
> ?>
>
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>
>
>

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