foreach ?

[harkly]

I have a code that does a search on zip codes and returns ones that are in a specified range. This is done in an array. I am very very new to arrays, with this being my real first one!

 

What I need to do is take all the zip codes that it returns, use them to find & pull info from other tables and echo it out.

 

Can someone point me in the right direction of how to do this?

 

So it will go something like this::

 

from table user find all zips from within a 10 miles radius of 90210

select the userID

then select all info from table background

 

[harkly]

my code I have

 

$result = mysql_query("SELECT userID FROM user WHERE zip IN (". $zips_in_range .")");

while ($row = mysql_fetch_assoc($result) ) {

foreach ($row as $userID)
{

    extract($row, EXTR_PREFIX_SAME, "userID");

echo "This is the userID - $userID <br>";

    echo "Select * FROM photos WHERE userID = $userID <BR>";
    
    $query = ("Select * FROM photos WHERE userID = $userID ");
      while ($r=mysql_fetch_array($query))
        {
          $userID=$r["userID"]; 
          $photo_1=$r["photo_1"];

          echo " $userID & $photo_1";
        }
}
}

 

my results

 

This is the userID - kelly

Select * FROM photos WHERE userID = kelly

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/i/d/2/id2294/html/singles/results.php on line 32

This is the userID - adfggg

Select * FROM photos WHERE userID = adfggg

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/i/d/2/id2294/html/singles/results.php on line 32

This is the userID - neer

Select * FROM photos WHERE userID = neer

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/i/d/2/id2294/html/singles/results.php on line 32

 

 

[vmicchia]

$query = ("Select * FROM photos WHERE userID = $userID ");
$result = mysql_query($query);
      while ($r=mysql_fetch_array($result))
        {
          $userID=$r["userID"]; 
          $photo_1=$r["photo_1"];

          echo " $userID & $photo_1";
        }

I think that change should take care of it.

 

[harkly]

Thanks this worked great!

I just had to add '' to the $userID in this line

$query = ("Select * FROM photos WHERE userID = $userID ");