Question

Hello am trying to count how many rows are in location column where there names include any of the names which are in the predefined $uk_array.

mysql prints out that I don’t have any value in my column “there are ()” that matches a name from the array although there is one “Dublin, Ireland”.

Here is the code:

    <?php
    include'connect.php';
    $uk_array = array('Liverpool','London','London UK','UK','London',
    'Dublin, Ireland','Manchester','Norwich','United Kingdom','Norwich','Duplin','England','ENGLAND',
    'united kingdom');

    $string = '';

    foreach($uk_array as $term=>$i) {
        $string = $string."location LIKE '%".$i."%' AND";
    }
    $string = substr($string, 0, -5);
    $query="SELECT COUNT(location) as location FROM tweets WHERE $string";
    $result = mysql_query($query) or die(mysql_error());
    while($row = mysql_fetch_array($result)){
        echo "There are ". $row['location']."";

    }


    ?>

Answer 1

You should be doing

$string = $string."location LIKE '%".$i."%' OR";

Switch the AND operator with the OR operator, otherwise the field has to contain all of the values in uk_array.

Also, you have a typo – Duplin?

Edit: More errors

foreach($uk_array as $term=>$i) {
    if($term) $string .= ' OR '; // only append OR if it's not the first one
    $string .= "location LIKE '%".$i."%'";
}