Help with 'ELSE' statement

igolo · Apr 11, 2002

The following code searches a database for a store location, then returns the IP info. How do I get the code to ask if I would like to enter a new location only if the search location is not in the database?
I have the form (addstor.php) that will allow me to add a new location.

<?php
if ($searchstring)
{
$sql="SELECT * FROM StorIPScheme WHERE $searchtype LIKE '%$searchstring%' ORDER BY StoreName ASC";
$db = mysql_connect("-------", "mforney", "-------&quot

;
mysql_select_db("CalicoStores",$db);
$result = mysql_query($sql,$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD>Store Number<TD>Store Name<TD>Remote IP<TD>Remote WAN
<TD>Default Gateway<TD>DLCI#</TR>";
while($myrow = mysql_fetch_array($result)) {
echo "<TR><TD>".$myrow["StoreNo"]."<TD>".$myrow["StoreName"]."<TD>".$myrow["RemoteIP"].
"<TD>".$myrow["RemoteWAN"]."<TD>".$myrow["Gateway"]."<TD>".$myrow["DLCI_No"];
echo "<TD><a href=\"view1.php?id=".$myrow["id"]."\">View</a>";
}
echo "</TABLE>";
}
?>

iTST · Apr 11, 2002

<?php
if ($searchstring)
{
$sql="SELECT * FROM StorIPScheme WHERE $searchtype LIKE '%$searchstring%' ORDER BY StoreName ASC";
$db = mysql_connect("-------", "mforney", "-------&quot

;
mysql_select_db("CalicoStores",$db);
$result = mysql_query($sql,$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD>Store Number<TD>Store Name<TD>Remote IP<TD>Remote WAN
<TD>Default Gateway<TD>DLCI#</TR>";
if ($myrow = mysql_fetch_array($result)) {
echo "<TR><TD>".$myrow["StoreNo"]."<TD>".$myrow["StoreName"]."<TD>".$myrow["RemoteIP"].
"<TD>".$myrow["RemoteWAN"]."<TD>".$myrow["Gateway"]."<TD>".$myrow["DLCI_No"];
echo "<TD><a href=\"view1.php?id=".$myrow["id"]."\">View</a>";
while($myrow = mysql_fetch_array($result)) {
echo "<TR><TD>".$myrow["StoreNo"]."<TD>".$myrow["StoreName"]."<TD>".$myrow["RemoteIP"].
"<TD>".$myrow["RemoteWAN"]."<TD>".$myrow["Gateway"]."<TD>".$myrow["DLCI_No"];
echo "<TD><a href=\"view1.php?id=".$myrow["id"]."\">View</a>";
}
} else {
_____________past add form here____________-
}
echo "</TABLE>";
}
?>

Just hacked away, but the attempt should be clear. cu, Sascha

http://www.itst.org/

danielhozac · Apr 11, 2002

Or you could use the mysql_num_rows($result) to see if the query returned anything.
Like this
<?php
if ($searchstring)
{
$sql="SELECT * FROM StorIPScheme WHERE $searchtype LIKE '%$searchstring%' ORDER BY StoreName ASC";
$db = mysql_connect("-------", "mforney", "-------&quot

;
mysql_select_db("CalicoStores",$db);
$result = mysql_query($sql,$db);
echo "<TABLE BORDER=2>";
echo"<TR><TD>Store Number<TD>Store Name<TD>Remote IP<TD>Remote WAN
<TD>Default Gateway<TD>DLCI#</TR>";
if (mysql_num_rows($result) > 0)
{
while($myrow = mysql_fetch_array($result)) {
echo "<TR><TD>".$myrow["StoreNo"]."<TD>".$myrow["StoreName"]."<TD>".$myrow["RemoteIP"].
"<TD>".$myrow["RemoteWAN"]."<TD>".$myrow["Gateway"]."<TD>".$myrow["DLCI_No"];
echo "<TD><a href=\"view1.php?id=".$myrow["id"]."\">View</a>";
}
}
else
{
echo "Form";
}
echo "</TABLE>";
}
?> //Daniel

igolo · Apr 18, 2002

Thanks guys

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Help with 'ELSE' statement

igolo

IS-IT--Management

iTST

Technical User

danielhozac

Programmer

igolo

IS-IT--Management

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