[RESOLVED] php + mysql question

Something odd is going on here. The following code does not return $site_id when it should:

Code:

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                        $query = 'select `site_id` from `sites` where `url`='$url'';
                        $result = mysql_query($query);
                        echo mysql_error();
                        $row = mysql_fetch_row($result);
                        if ($site_id != "")
                                siteScreen($site_id, $message);
                        else
                                showsites($message);

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I excute the same query from PhpMyAdmin, and it works fine:

Code:

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select `site_id` from `sites` where `url`='http://www.yahoo.com/'

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If its incorrect, then how do I get the $site_id based on a given $url string?

Code:

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print $result;

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... prints: Resource id #11

None of the mysql_* functions automatically populate variables for you. You access fields returned in a dataset by using the index notation, either by numerical index, or, if you use mysql_fetch_assoc(), the field name.

Also, you should check whether you need to call mysql_error() or not.

PHP Code:

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$result = mysql_query('select `site_id` from `sites` where `url`='$url'');
if (!is_resource($result)) echo mysql_error();

if (mysql_num_rows($result) > 0) {
  $row = mysql_fetch_assoc($result);
  if ($row['site_id'] != '')
    siteScreen($site_id, $message);
  else
    showsites($message);
}
else {
  // Not found, place handler code here
} 


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Doesnt work either :(

PHP Code:

---

$result = mysql_query('select `site_id` from `sites` where `url`=\'$url\''); 
if (!is_resource($result)) { echo mysql_error(); }
if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    if ($row['site_id'] != '') {
        siteScreen($row['site_id'], $message);
        }
    else {
        showsites($message);
        }
    }
else { 
    showsites($message);
    }
print "<hr>".$row['site_id']; 


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What happens then?

Also, on an unrelated note, don't use print or echo to output HTML code, unless it's just for debugging. There's no point otherwise.

Have you tried using print_r to display the array of results that is being returned from your query?

PHP Code:

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print "<hr>";
print_r ($row);
print "<hr>";
print_r ($result); 


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$row does not print anything.
$result prints Resource id #11

What could that be?

Thanks guys, I got this working:

PHP Code:

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            $result = mysql_query("select site_ID from sites where url='$url'");
            echo mysql_error();
            $row = mysql_fetch_row($result);
            $site_id = $row[0];

            print "<hr>".$site_id; 


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