Question

I have a mysql query which is checking three fields. If any of them are a 1 I want it to display the text I have following. The problem I’m having is that it is only showing the first time a variable is set, it will not output the other lines. My code is

while ($row =mysql_fetch_array($cust_check)){
    if ($row['emailmatch'] == "1"){
        $output_error = 'Your email is in our database<br />';  
    }
    if ($row['phonematch'] == "1"){
        $output_error .= 'Your phone number is in our database<br />';
    }
    if ($row['addressmatch'] == "1"){
        $output_error .= 'Your address is in our database<br />';
    }
}

Then I just want $output_error to show all that have a value of 1. Can someone help? Thanks

Answer 1

If your email matches, you are resetting your $output_error variable.

$output_error = '';
while ($row =db_fetch_array($cust_check)){
    if ($row['emailmatch'] == "1"){
        $output_error .= 'Your email is in our database<br />'; 
    }
    if ($row['phonematch'] == "1"){
        $output_error .= 'Your phone number is in our database<br />'; 
    }
    if ($row['addressmatch'] == "1"){
        $output_error .= 'Your address is in our database<br />';  
    }
}