Missing something but what?!?!

[badboybill4281]

below is my display results from a search for a site that is in the DB, i have already inserted everything including the screenshot dir in the DB and when i try to display the rows they don't show whats missing or worng?

$query = "select * from movies where ".$searchtype." like
	  '%".$searchterm."%'";
      $result = mysql_query($query, $conn);
	  $num_results = mysql_num_rows($result);

  echo "<p>number of sites found: ".$num_results."</p>";

  for ($i=0; $i <$num_results; $i++)
  {
    $row = mysql_fetch_array($result);
	echo "<p><strong>".($i+1).". Site Name: ";
	echo htmlspecialchars( stripslashes($row["site_name"]));
	echo "<strong><br>Site URL: ";
	echo htmlspecialchars( stripslashes($row["<a href='site_tour_url'> Preview </a>"]));
	echo "<br>Join URL: ";
	echo htmlspecialchars( stripslashes($row["<a href='site_join_url'> Join </a>"]));
	echo "<br>Gallery URL: ";
	echo htmlspecialchars( stripslashes($row["<a href='site_gall_url'> Content </a>"]));
	echo "<br>Site Screenshoot: ";
	echo htmlspecialchars( stripslashes($row["<img src='site_screen'>"]));
	echo "</p>";
  }

[nnichols]

echo "<p>number of sites found: ".$num_results."</p>";

Does this show "0" or not at all?

You need to explain what does/doesn't happen a little clearer.

[badboybill4281]

sorry 🙁 i will comment whats going on in the code

$query = "select * from movies where ".$searchtype." like
      '%".$searchterm."%'";
      $result = mysql_query($query, $conn);
      $num_results = mysql_num_rows($result);

  echo "<p>number of sites found: ".$num_results."</p>"; //this shows the actual sites it works

  for ($i=0; $i <$num_results; $i++)
  {
    $row = mysql_fetch_array($result);
    echo "<p><strong>".($i+1).". Site Name: "; 
    echo htmlspecialchars( stripslashes($row["site_name"])); //this works shows the site name fine
    echo "<strong><br>Site URL: ";
    echo htmlspecialchars( stripslashes($row["<a href='site_tour_url'> Preview </a>"])); //this doesn't work should say preview with preveiw as a link to the site url
    echo "<br>Join URL: ";
    echo htmlspecialchars( stripslashes($row["<a href='site_join_url'> Join </a>"])); // same as above but with the join url
    echo "<br>Gallery URL: ";
    echo htmlspecialchars( stripslashes($row["<a href='site_gall_url'> Content </a>"])); //same as the 2 above but with the gallery url
    echo "<br>Site Screenshoot: ";
    echo htmlspecialchars( stripslashes($row["<img src='site_screen'>"])); // same as above but should show an image the currect dir is in this field in the DB
    echo "</p>";
  }

[nnichols]

Can you post the database structure for me to have a look at? Or even better you could zip your php file along with a small sample data dump from your db, that way I can have a proper look at what is going wrong.

Nick

[nnichols]

Sorry I didn't read your code properly. Try this -

echo "<strong><br>Site URL: ";
$url = htmlspecialchars( stripslashes($row['site_tour_url']));
echo "<a href=\"$url\"> Preview </a>";

instead of the way you are doing it now.

Hope it works 😉

[badboybill4281]

that worked thanx 😃 i'm sure i'm going to need alil help on this next part which deals with the same thing but i'm gonna try to do it then i'll run back if i get stuck

[badboybill4281]

also heres the full code

 trim($searchterm);
   if (!$searchtype || !$searchterm)
   {
     echo "You have not entered search details. Please go back and try again.";
	   exit;
	 }
  $searchtype = addslashes($searchtype);
  $searchterm = addslashes($searchterm);

  $conn = mysql_pconnect("localhost", "dbuser", "dbpass");
    if (!$conn)
	{
	  echo "Error: Could not connect to database. Please try again later.";
	  exit;
	  }
  mysql_select_db("DBhere",$conn);
    $query = "select * from table where ".$searchtype." like
	  '%".$searchterm."%'"; //replace table with table to search
      $result = mysql_query($query, $conn);
	  $num_results = mysql_num_rows($result);

  echo "<p>number of sites found: ".$num_results."</p>";

  for ($i=0; $i <$num_results; $i++)
  {
    $row = mysql_fetch_array($result);
	echo "<p><strong>".($i+1).". Site Name: ";
	echo htmlspecialchars( stripslashes($row["site_name"]));
	echo "<strong><br>Site URL: ";
	$url = htmlspecialchars( stripslashes($row['site_tour_url']));
    echo "<a href=\"$url\"> Preview </a>";
	echo "<br>Join URL: ";
	$join = htmlspecialchars( stripslashes($row['site_join_url']));
	echo "<a href=\"$join\"> Join </a>";
	echo "<br>Gallery URL: ";
	$gall = htmlspecialchars( stripslashes($row['site_gall_url']));
	echo "<a href=\"$gall\"> Pictures </a>";
	echo "<br>Site Screenshot: ";
	$screen = htmlspecialchars( stripslashes($row['site_screen']));
	echo "<img src=\"$screen\">";
	echo "</p>";
  }

and the search form is like this

<html>
<head>
<TITLE>Untitled Document</TITLE>
<meta HTTP-EQUIV="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<BODY>
<form action="searchresults.php" method="post">
Choose Search Type:<br>
<select name="searchtype">
<option value="site_name">Site Name
<option value="site_tour_url">Tour URL
<option value="site_join_url">Join URL
<option value="site_gall_url">Gallery URL
<option value="site_screen">Site Screenshot
</select>
<br>
Enter search term:<br>
<input name="searchterm" type=text>
<br>
<input type=submit value="search">
</form>
</BODY>
</HTML>

[remote2]

i think there is something wrong with your sql statement
when i do
printf("select * from movies where ".$searchtype." like'%".$searchterm."%'");

it tells me
Warning: printf(): too few arguments in ...

if i remove the % then everything prints out ok

i'm new to PHP... so i dont know what % means...
i'm assuming you are trying to print ' and using % as an escape
char..

but it seems to work without %

[nnichols]

Hello remote2

He is using "%" because it is a wildcard in mysql, so with the "%" before and after his "$searchterm" it will search for the value of "$searchterm" anywhere in that column.

Hello badboybill4281

I'm glad that works now. I was absolutely wiped out yesterday which is why it took me so long to spot the problem.

Nick 😉

[badboybill4281]

np nick thanx for the help i'm alil stuck on the next part still ::sigh:: maybe you can help on it as well i need to take the information from the search term and display it in 3 cols width and 3 rows in height

top row of each col haveing the site_name in it

and the row below that having the site_screen in each col across

and the bottom row which in each col will be split into 3 cols and from left to right for each col

, site_tour_url,site_join_url,site_gall_url

all like the codee above so it takes the result and displays them as many wide as it needs to up to 3 wide and then returns below those if above 3 to make 4,5,6 and so on, any ideas??? i tried tables but couldn't seem to get it to work :mad:

[nnichols]

I would do it using an array. If you loop through all the results returned by the query and store them in an array you will be able to access any element of the array in any order. It makes it a lot easier for building tables in this way.

Have a go. If you struggle with it post again and I'll see what I can do.

Good luck! 😉