Hallo allemaal

Ik ben bezig met een scriptje waarmee ik bepaalde records kan toevoegen, wijzigen en verwijderen.

Het probleem is dat als ik een rij selecteer dan krijg ik niet die specifieke rij maar gelijk alle rijen te zien uit de database. Ik heb bijvoorbeeld op de index pagina tabel 1 en tabel 2, ik klik dan op tabel 1 om hem te wijzigen en ik wordt doorgeschakeld naar wijzigen.php. Nu moet ik eigenlijk alleen tabel 1 krijgen maar ik krijg ook tabel 2 en dat is dus het probleem.

Hieronder het script:

<?

require('database.php');

$query = "SELECT * FROM info";
$sql = mysql_query("$query");
while ($fill = mysql_fetch_array($sql)) {
$id = $fill['id_rel'];
$ftp = $fill['ftp'];
$ftp_user = $fill['ftp_user'];
$ftp_pass = $fill['ftp_pass'];
$database = $fill['db'];
$db_user = $fill['db_user'];
$db_pass = $fill['db_pass'];
$admin_shop = $fill['admin_shop'];
$shop_user = $fill['shop_user'];
$shop_pass = $fill['shop_pass'];
$admin_website = $fill['admin_website'];
$admin_user = $fill['admin_user'];
$admin_pass = $fill['admin_pass'];

?>


<form>
<table width="291" border="0" cellpadding="0" cellspacing="1" bordercolor="#999999">
<tr>
<td width="138"><span class="style1">FTP</span></td>
<td width="150"><input type="text" name="FTP" value="<? echo "$ftp"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">FTP User </span></td>
<td><input type="text" name="FTP_USER" value="<? echo "$ftp_user"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">FTP Password</span></td>
<td><input type="text" name="FTP_PASS" value="<? echo "$ftp_pass"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Database</span></td>
<td><input type="text" name="DB" value="<? echo "$database"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Db User</span></td>
<td><input type="text" name="DB_USER" value="<? echo "$db_user"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Db Password</span></td>
<td><input type="text" name="DB_PASS" value="<? echo "$db_pass"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Admin Winkel </span></td>
<td><input type="text" name="ADMIN_SHOP" value="<? echo "$admin_shop"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Winkel User</span></td>
<td><input type="text" name="SHOP_USER" value="<? echo "$shop_user"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Winkel Password</span></td>
<td><input type="text" name="SHOP_PASS" value="<? echo "$shop_pass"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Admin website</span></td>
<td><input type="text" name="ADMIN_WEBSITE" value="<? echo "$admin_website"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Admin User </span></td>
<td><input type="text" name="ADMIN_USER" value="<? echo "$admin_user"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style1">Admin Password</span></td>
<td><input type="text" name="ADMIN_PASS" value="<? echo "$admin_pass"; ?>" size="25" ></td>
</tr>
<tr>
<td><span class="style2"><? echo "$id" ?></span></td>
<td><input type="submit" name="Opslaan" value="save" style=" WIDTH:120PX; "></td>
</tr>
</table>
</form>

<?
}

if(isset($Opslaan))
{

mysql_query("update info SET ftp='$FTP', ftp_user='$FTP_USER', ftp_pass='$FTP_PASS', db='$DB', db_user='$DB_USER', db_pass='$DB_PASS', admin_shop='$ADMIN_SHOP', shop_user='$SHOP_USER', shop_pass='$SHOP_PASS', admin_website='$ADMIN_WEBSITE', admin_user='$ADMIN_USER', admin_pass='$ADMIN_PASS' where id_rel='$ID_REL' ");

?>

<script>

location.href="index.php";
</script>

<?
}
?>

wie o wie kan me helpen

alvast bedankt
Ferdi schreef op 04.01.2006 23:32
deze doet het ook niet

$query = "
SELECT *
FROM info
WHERE id='" . $_GET['id'] "'";

krijg de volgende melding
Parse error: parse error, unexpected T_CONSTANT_ENCAPSED_STRING in


Punt vergeten:

$query = "
SELECT *
FROM info
WHERE id='" . $_GET['id'] . "'";
Goed punt, SanThe
Jan Koehoorn schreef op 05.01.2006 01:18
Goed punt, SanThe


Geen punt, Jan.
maar nu krijg ik gelijk de volgende foutmelding

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
dit is de code van index.html

include("database.php");

$query = "SELECT * FROM info";
$sql = mysql_query("$query");
while ($fill = mysql_fetch_array($sql)) {
$id = $fill['id_rel'];
$ftp = $fill['ftp'];
$ftp_user = $fill['ftp_user'];
$ftp_pass = $fill['ftp_pass'];
$database = $fill['db'];
$db_user = $fill['db_user'];
$db_pass = $fill['db_pass'];
$admin_shop = $fill['admin_shop'];
$shop_user = $fill['shop_user'];
$shop_pass = $fill['shop_pass'];
$admin_website = $fill['admin_website'];
$admin_user = $fill['admin_user'];
$admin_pass = $fill['admin_pass'];

?>

<style type="text/css">
<!--
.style1 {
font-family: Verdana, Arial, Helvetica, sans-serif;
font-size: 12px;
}
.style2 {
font-family: Verdana, Arial, Helvetica, sans-serif;
font-weight: bold;
font-size: 12px;
}
A:link {
COLOR: #336699; TEXT-DECORATION: none
}
A:visited {
COLOR: #336699; TEXT-DECORATION: none
}
A:hover {
COLOR: #318fc8; TEXT-DECORATION: underline
}

-->
</style>

<p class="style2">Informer</p>
<table width="346" border="1" cellpadding="0" cellspacing="0" bordercolor="#CCCCCC">
<tr>
<td width="145" class="style1">ID</td>
<td width="195" class="style1"><? echo "$id"."<BR>"; ?> </td>
</tr>
<tr>
<td class="style1">FTP</td>
<td class="style1"><? echo "$ftp"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">FTP User</td>
<td class="style1"><? echo "$ftp_user"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">FTP Password </td>
<td class="style1"><? echo "$ftp_pass"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Database</td>
<td class="style1"><? echo "$database"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Db User</td>
<td class="style1"><? echo "$db_user"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Db Password </td>
<td class="style1"><? echo "$db_pass"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Admin Winkel </td>
<td class="style1"><? echo "$admin_shop"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Winkel User</td>
<td class="style1"><? echo "$shop_user"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Winkel Password </td>
<td class="style1"><? echo "$shop_pass"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Admin website</td>
<td class="style1"><? echo "$admin_website"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Admin User </td>
<td class="style1"><? echo "$admin_user"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">Admin Password</td>
<td class="style1"><? echo "$admin_pass"."<BR>"; ?></td>
</tr>
<tr>
<td class="style1">&nbsp;</td>
<td class="style1"><form name="form1" method="post" action="index.php?MYSQL=DEL">
<input type="hidden" name="ID_REL" value="<? echo "$id"; ?> ">
<input type="image" src="recycle_bin_icon.jpg" alt="verwijderen" align="middle">
<a href="update2.php" class="style1" onclick="return popitup('update2.php')">website wijzigen</a>
</form>



</td>
</tr>
</table>

hieronder komt nog iets maar is niet van toepassing
$sql = mysql_query("$query");

Quotes horen daar niet.

$sql = mysql_query($query);
jongens bedankt voor jullie hulp, ik heb het eindelijk voor elkaar gekregen.

thnx (Y)

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