Hiya,

Thanks for the quick reply. I used the PHP manual example and it connects
to the database successfully but cannot select the database.

I'm not sure why this is? I've looked hard at it and I cannot see where I have
gone wrong.

Thanks.

GF.
  ----- Original Message ----- 
  From: David Jackson 
  To: [EMAIL PROTECTED] 
  Cc: [EMAIL PROTECTED] 
  Sent: Monday, December 31, 2001 1:48 AM
  Subject: Re: [PHP] MySQL problem


  Here's the example from the PHP manual:
  The tutorial here are very helpfull:
  http://www.melonfire.com/community/columns/trog/

  -- David

  <?php
  // Connecting, selecting database
  $link = mysql_connect("mysql_host", "mysql_login", "mysql_password")
      or die("Could not connect");
  print "Connected successfully";
  mysql_select_db("my_database")
      or die("Could not select database");

  // Performing SQL query
  $query = "SELECT * FROM my_table";
  $result = mysql_query($query)
      or die("Query failed");

  // Printing results in HTML
  print "<table>\n";
  while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
      print "\t<tr>\n";
      foreach ($line as $col_value) {
          print "\t\t<td>$col_value</td>\n";
      }
      print "\t</tr>\n";
  }
  print "</table>\n";

  // Closing connection
  mysql_close($link);
  ?>




  > Hello,
  >
  > I am extremely new to MySQL and have never managed to get working
  > smoothly with PHP before. I am trying really hard to understand
  > how to work it, and am almost there.
  >
  > I have a problem which I do not know how to resolve and was
  > wondering if anybody could help me. I have no idea what is wrong
  > with the code and why I am getting the error message;
  >
  > Warning: Supplied argument is not a valid MySQL result resource in
  > C:\apache\htdocs\sams\chapter10\results.php on line 47
  >
  > I am currently using a book to aid me with MySQL, and this is an
  > example from the book. It does not seem to work and I have no idea what
  > I may have done wrong to obtain this warning.
  >
  > I have changed my login and password to question marks.
  >
  > <?
  >
  >  if (!$searchtype || !$searchterm)
  >
  >  {
  >     echo "You have not entered search details.  Please go back and try
  > again.";
  >
  >     exit;
  >
  >  }
  >
  >
  >  $searchtype = addslashes($searchtype);
  >
  >  $searchterm = addslashes($searchterm);
  >
  >  @ $db = mysql_pconnect("mesh", "bookorama", "bookorama123");
  >
  >  if (!$db)
  >
  >  {
  >     echo "Error: Could not connect to database.  Please try again
  >     later.";
  >
  >     exit;
  >
  >  }
  >
  >  mysql_select_db("booktest");
  >
  >  $query = "select * from booktest where ".$searchtype." like
  > '%".$searchterm."%'";
  >
  >  $result = mysql_query($query);
  >
  >  $num_results = mysql_num_rows($result);
  >
  >  echo "<p>Number of books found: ".$num_results."</p>";
  >
  >  for ($i=0; $i <$num_results; $i++)
  >
  >  {
  >
  >     $row = mysql_fetch_array($result);
  >
  >     echo "<p><strong>".($i+1).". Title: ";
  >
  >     echo stripslashes($row["title"]);
  >
  >     echo "</strong><br>Author: ";
  >
  >     echo stripslashes($row["author"]);
  >
  >     echo "<br>ISBN: ";
  >
  >     echo stripslashes($row["isbn"]);
  >
  >     echo "<br>Price: ";
  >
  >     echo stripslashes($row["price"]);
  >
  >     echo "</p>";
  >
  >  }
  >
  > ?>
  >
  > The problem seems to be around the lines of code;
  >
  > $result = mysql_query($query);
  >
  > $num_results = mysql_num_rows($result);
  >
  > Any assistance is appreciated.
  >
  > Yours,
  >
  > GF.
  >
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