[SOLVED] Using an array with a db query

[spillage]

Sorry to be a noobie pain in the butt but having problems trying to work this one out.

 

if $cust=$_POST['all']{
while($row = mysql_fetch_array($result))
   {
echo $row['printer_type']."  | ".$row['Cust_Email']," | ".$row['Friends_Email'];
    echo "<br />";
   }}
   
else if $cust=$_POST['cust']
   {
while($row = mysql_fetch_array($result))
   {
echo $row['printer_type']."  | ".$row['Cust_Email']," | ";
    echo "<br />";
   }}

 

$cust is from a drop down menu which is working as I can echo the result and get the corrct output

and though the above would work but have searched around the correct answer without joy and starting

to really struggling.

 

If I am (and probally am) being a thick idiot please let me know and Ill spent a while longer searching for the

answer as I quite often have my brain in the wrong gear.

 

Cheers,

 

Spill

[Barand]

I may be missing something but is there a question in there?

[spillage]

sorry

 

all that happens is the second script runs. So if $cust=$_POST['all'] only first two rows are printed.

 

although if I also add at the beforehand

 

echo $cust then 'all' is printed???

 

Cheers.

[Barand]

What is your SQL query?

 

That's going to control the results

[spillage]

is as follows

 

$result = mysql_query("SELECT * FROM submit WHERE printer_type LIKE '%$printer%'");

 

if this is used with

while($row = mysql_fetch_array($result))
   {
echo $row['printer_type']."  | ".$row['Cust_Email'],";
    echo "<br />";
}

this works fine regardless of which $row I select to be printed. But am trying to select the results printed.

 

Spill

[sasa]

try if ($cust == 'all') etc.

[spillage]

Thanks very much sasa.

 

Works a treat.

 

Spill