PHP - Moved: Check For Mysql Results

This topic has been moved to PHP Installation & Configuration.

http://www.phpfreaks.com/forums/index.php?topic=326567.0

I haven't coded in PHP for a while and I'm trying to refresh my memory on some things. =( I was hoping someone here could help me with a problem. I have a script I made to post articles to my database, one of the columns includes a number. 1 (yes) and 0 (no). Now i'm using this to keep track of what articles have been approved. When I edit an article I want to include a check box. If my database field contains 1 its checked... if its 0 then its not checked. how would I go about doing that?

Been trying to find an answer and keep running around in circles...
I have: ( left test echos in there)      

        if ($r['left_house_number'] != '') {
            /* Now see what they are supposed to get */
            $left_customer_query = "SELECT * FROM customer WHERE house_number ='".$r['left_house_number']."' AND street = '".$r['left_house_street']."' AND route_number = '".$_SESSION['route_number']."'";
// echo "SELECT * FROM customer WHERE house_number ='".$r['left_house_number']."' AND street = '".$r['left_house_street']."' AND route_number = '".$_SESSION['route_number']."'";
            $left_customer_result = mysql_query($left_customer_query) or die(mysql_error());
                $left_customer_result_count = mysql_num_rows($left_customer_result);
echo $r['left_house_number'].' =lhn1';
            while ($lrow = mysql_fetch_array($left_customer_result)) {
echo $r['left_house_number'].' =lhn2';
                /* See if there is any customer if not display blank row */
echo $left_customer_result_count.' = count ';
                if ($left_customer_result_count < 1) {
echo $r['left_house_number'].' =lhn3';
                    echo '<span class="delivery_nothing">'.$r['left_house_number'].'&nbsp;</span>&nbsp;1<span class="delivery_nothing">'.$r['left_house_street'].'&nbsp;';
                } else {
                    /* Have a customer, now check to see if they are supposed to get something */
                    if ($lrow['status'] != "AC" && $lrow['status'] != "VH") {
                        /* Customer is stopped or on vacation so show house number and street not bolded */
                        echo '<span class="delivery_nothing">'.$r['left_house_number'].'</span>2<span class="delivery_nothing">'.$r['left_house_street'];
echo $r['left_house_number'].' =lhn4';
                    } else {
                        /* Now check for correct delivery day and return data */
                        if (Show_Product($lrow['product']) == '1') {
echo $r['left_house_number'].' =lhn5';
                            echo '<span class="delivery_bold">'.$lrow['house_number'].'</span><span class="delivery_bold">'.$lrow['placement'].'</span><span class="delivery_bold">'.$lrow['street'].'</span>';
                            echo Product_Formater($lrow['product']);
                        }

I have tried...
if ($left_customer_result_count < 1) {
if ($left_customer_result_count == 0) {
if ($left_customer_result_count == '') {
if ($left_customer_result_count === 0) {

And I can't get it to work.
the count "test echo" is either blank or 1

If I echo the sql statement with values I know aren't in the db and paste it into phpmyadmin:
SELECT * FROM customer WHERE house_number = '3333' AND street = 'Chili Avenue' AND route_number = '68-24-630'

I get:
"MySQL returned an empty result set"

And if I change it to something that is in the db:
SELECT * FROM customer WHERE house_number = '3019' AND street = 'Chili Avenue' AND route_number = '68-24-630'

I get:
"Showing rows 0 - 0 (1 total, Query took 0.0012 sec)"


So what do I have to do to get this to work right?
What does it actually output when there are no results returned?
thanks,

I'm working on a directory component and have kind of a tricky thing I need to do.

The list of results pop through ajax on the left side along with it's google map number tag, and the google map pops on the right side. To make it even trickier the results are popped through a radius search by proximity.

The problem is if there are multiple results with the same address you end up having all of the number tags on the map buried under the last tag with that address, plus it makes for kind of a long list.

What I need to do is run a check to see if the previous result is the same address, and if yes not pop that result in a new table cell, rather stick it in a jquery drop down div so that all of the following results with the same address are stuffed in the jquery drop down in the same table cell under the first result with that address.

In a perfect world all of the listings with the same addresses would be submitted in order so I could just do a check for the previous itemid's address, but of course that's not the case. I thought about running a query to grab ALL of the addresses and then check the item's address against it to see if it matches and if yes echo the jquery div instead of the <td>, but obviously that won't work because that will cause the first item with that address to be put in the div also. Plus I don't know if that's a very efficient way to go about it anyway.

Here's kind of the basic thing I'm working with:
Code: [Select]
$query = "SELECT data_txt FROM jos_sobi2_fields_data WHERE fieldid = 76 AND itemid = $mySobi->id";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
$myaddress = $row['data_txt'];
}
echo "<td $style>";

Any ideas?

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=312470.0