PHP While loop with mysql not adding to variable

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

Change emailmatch to also append, what happens if $output_error phonematch equals 1, addressmatch equals 1, and then lastly emailmatch equals 1? $output_error would only contain the one error message for emailmatch? — Ryan Kempt
– Ryan Kempt, Commented Nov 1, 2012 at 22:47
Actually, why don't you just change your SQL statement to select only the ones that equal 1 and then just display all of them instead of using your (possibly) many if statements? — Ryan Kempt
– Ryan Kempt, Commented Nov 1, 2012 at 22:49

AndrewR · Accepted Answer · 2012-11-01 22:46:39Z

4

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 />';  
    }
}

answered Nov 1, 2012 at 22:46

AndrewR

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