php mysql search engine

[Reaper0167]

php mysql search engine displays the blob image like this....what is going on?

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[jeger003]

you should post your code so we can see whats going on

[genericnumber1]

You likely need to send the image header...

 

Place

header('Content-type:image/jpeg');

at the top

 

or whatever type of image it is (above if it's jpeg, replace jpeg with png if it's a png, etc).

[Reaper0167]

I will post the code here in a little while. The images are bmp, png, jpg and gif. I was told that you cannot display all those types on one page, but I find that hard to believe. I will try the header now.

[Reaper0167]

by the way, this is not my script, i was just wondering how good it worked so I could get an idea of search engines. I tried the header code at the top of the page and all it shows is a little box with an x in the middle. And all the other content on the page was missing.

<?php
header('Content-type:image/x-png');
include "upload_db_info.php";

  // Get the search variable from URL
  $var = @$_GET['q'] ;
  $trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10; 

// check for an empty string and display a message.
if ($trimmed == "")
  {
  echo "<p>Please enter a search...</p>";
  exit;
  }

// check for a search parameter
if (!isset($var))
  {
  echo "<p>We dont seem to have a search parameter!</p>";
  exit;
  }

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("----","-----","-------"); //(host, username, password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("-------") or die("Unable to select database"); //select which database we're using

// Build SQL Query  
$query = "select * from $table_name where name like \"%$trimmed%\"  
  order by name"; // EDIT HERE and specify your table and field names for the SQL query

$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);

// If we have no results, offer a google search as an alternative

if ($numrows == 0)
  {
  echo "<h4>Results</h4>";
  echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>";

// google
echo "<p><a href=\"http://www.google.com/search?q=" 
  . $trimmed . "\" target=\"_blank\" title=\"Look up 
  " . $trimmed . " on Google\">Click here</a> to try the 
  search on google</p>";
  }

// next determine if s has been passed to script, if not use 0
  if (empty($s)) {
  $s=0;
  }

// get results
  $query .= " limit $s,$limit";
  $result = mysql_query($query) or die("Couldn't execute query");

// display what the person searched for
echo "<p>You searched for: "" . $var . ""</p>";

// begin to show results set
echo "Results";
$count = 1 + $s ;

// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["content"];

  echo "$count.) $title" ;
  $count++ ;
  }

$currPage = (($s/$limit) + 1);

//break before paging
  echo "<br />";

  // next we need to do the links to other results
  if ($s>=1) { // bypass PREV link if s is 0
  $prevs=($s-$limit);
  print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< 
  Prev 10</a>&nbsp ";
  }

// calculate number of pages needing links
  $pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder from division

  if ($numrows%$limit) {
  // has remainder so add one page
  $pages++;
  }

// check to see if last page
  if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

  // not last page so give NEXT link
  $news=$s+$limit;

  echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>";
  }

$a = $s + ($limit) ;
  if ($a > $numrows) { $a = $numrows ; }
  $b = $s + 1 ;
  echo "<p>Showing results $b to $a of $numrows</p>";
  
?>

[genericnumber1]

You have to give the images their own page which you then link to with the html img tag, that's what people meant by "you can't have all of the images on one page"

[Reaper0167]

Can that be done on the fly(by itself). People will be uploading there picture, picture name, and description all day, every day. So when they upload, it is automatically given a page of its own. I'm still a little confused.

[Reaper0167]

bump ttt

[Reaper0167]

i found this little bit of code

<?php
<html> 
2 <head> 
3   <title>My page title</title> 
4 </head> 
5 <body> 
6 
7  <img height="125" width="125" alt="Flower" src="flower01.jpg" /> 
8 
9 </body> 
10</html>
?>

Each picture will need a page like this, and then echo out the php page on my search results page????? How can this be done with all the images in the database?

[genericnumber1]

This is the reason it's not very efficient to store images in a database. You could create a separate page that retrieves the image from the database based upon a GET id.

 

Eg.

<img src="image.php?id=12345" />

[Reaper0167]

There are different types of files in the db.... gif, bmp, png, etc. So would this line be changed or removed?

header('Content-type:image/x-png');

 

<img src="image.php?id=12345" />    <---|

Could you explain the line above please---^

[genericnumber1]

If you don't really understand much about storing image files in a database, you should do some googling. Explaining every detail would be spending a lot of time repeating what hundreds of other people have already said elsewhere.

[Reaper0167]

storing them I do know about. What I want to do I can't find anywhere. So thanks.

[gizmola]

I don't think you understand what you've been told.  A browser understands http protocol.  In reference to html, the browser supports the rendering of html documents.  Documents can have embedded elements like images, but the base html for those images is the img tag.  The img tag needs to reference a source for the image --- namely a seperate url that indicates where the image can be located in web space.  The client is responsible for integrating all the elements referenced in the html page and rendering them.

 

So -- no you can't just read a blob out the database and display that.  What you *can* do, is have a script that reads the image blob data from the database and returns that with the appropriate Mime header, as genericnumber1 suggested.  You can then use the call to that script in a page. 

 

So--- you need to write a script that returns ONE image.  The simplest manner would be using the get param as he illustrated.  Let's say that the script is called:

 

image.php.

 

It would accept an image id, or something that can be used to find the data in the mysql database.

 

You'd call it with :  http://yoursite.com/image.php?id=5

 

Working properly, this should render the image in your browser.

 

 

Once this is working, it's easy enough to have a page with as many images as you need:

 

Image one!

 

Image two!

 

etc.

 

 

[Reaper0167]

So are you saying that every time someone uploads a picture, i need to edit that php file? I hope not.

[genericnumber1]

That's not what he's saying.

[gizmola]

So are you saying that every time someone uploads a picture, i need to edit that php file? I hope not.

 

No. The script example I provided assumed a generic script.  We don't have any information from you about your database structure, so the best we can do is hypothesize.  Often people use AUTO_INCREMENT to provide a numeric primary key on the table.  My example assumed such a mechanism.  I could easily write a script that would query the image table, and provide a list of the file names and the images.  This would be "generic" in that the next time someone uploaded an image it would be reflected in the list.

 

We don't know what the application is or what you are trying to do.  You haven't provided any code to look at.  We're just guessing at this point.