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Display results from two colums in PHP & MySQL

I have two columns in my database 'type' and 'type2'

I am send a url paramater over along the lines of <a href="type.php?id=Land Investments">

This works fine and 'Land Investments' is sent over in the URL.

I need to display results if 'Land Investments' appears in either the 'type' or 'type2' column. How would I code the query?

The code I currently use which only cjecks the 'type' column is:

$colname_Recordset1 = "-1";
if (isset($_GET['id'])) {
  $colname_Recordset1 = (get_magic_quotes_gpc()) ? $_GET['id'] : addslashes($_GET['id']);
}
mysql_select_db($database_PPA, $PPA);
$query_Recordset1 = sprintf("SELECT id, name, price, location, `currency`, `description`, type, type2, country, bedrooms, features, img1 FROM property WHERE type = %s", GetSQLValueString($colname_Recordset1, "text"));
$Recordset1 = mysql_query($query_Recordset1, $PPA) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
 if($totalRows_Recordset1 == '0') {
      header('Location: noresults.php');
}
0
BrighteyesDesign
Asked:
BrighteyesDesign
1 Solution
 
plshrkCommented:
WHERE (type = %s OR type2 = %s)
0
 
BrighteyesDesignAuthor Commented:
Thanks for that.

I get an error:

Warning: sprintf() [function.sprintf]: Too few arguments in /home2/brightf7/public_html/personalproperty/type.php on line 89
Query was empty

When using that though?

$query_Recordset1 = sprintf("SELECT id, name, price, location, `currency`, `description`, type, type2, offer, country, bedrooms, features, img1 FROM property WHERE (type = %s OR type2 = %s)"
0
 
manhdnCommented:
SELECT * FROM table WHERE type1 LIKE ‘%$word%’ OR type2 LIKE ‘%$word%’
0
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BrighteyesDesignAuthor Commented:
Thanks,

That code...

"SELECT id, name, price, location, `currency`, `description`, type, type2, offer, country, bedrooms, features, img1 FROM property WHERE type LIKE '%$word%' OR type2 LIKE '%$word%'"

...just shows all properties, it doesn't filter anyhting.
0
 
Lukasz ChmielewskiCommented:
Try this, there are two arguments required after the query

$query_Recordset1 = sprintf("SELECT id, name, price, location, `currency`, `description`, type, type2, country, bedrooms, features, img1 FROM property WHERE type = %s OR type2 = %s", GetSQLValueString($colname_Recordset1, "text"), GetSQLValueString($colname_Recordset1, "text"));

Open in new window

0
 
BrighteyesDesignAuthor Commented:
Yes, that does the trick!

Thanks for that
0
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