Problem with results not showing up.

mcclellanfsu · February 21, 2007

Hello, I will tell you all right off the bat...I am just starting to learn PHP. I am trying to create just a very basic page that will insert book data into a database. When I run it I get NO ERRORS, but I also get no results. Nothing is inserted into the database.

Here is my code:

<html>
<head>
  <title>Book-O-Rama - New Book Entry</title>
</head>

<body>
  <h1>Book-O-Rama - New Book Entry</h1>

  <form action="insert_book.php" method="post">
    <table border="0">
      <tr>
        <td>ISBN</td>
         <td><input type="text" name="isbn" maxlength="13" size="13"></td>
      </tr>
      <tr>
        <td>Author</td>
        <td> <input type="text" name="author" maxlength="30" size="30"></td>
      </tr>
      <tr>
        <td>Title</td>
        <td> <input type="text" name="title" maxlength="60" size="30"></td>
      </tr>
      <tr>
        <td>Price $</td>
        <td><input type="text" name="price" maxlength="7" size="7"></td>
      </tr>
      <tr>
        <td colspan="2"><input type="submit" value="Register"></td>
      </tr>
    </table>
  </form>
</body>
</html>

and here is the php insert page:

<html>
<head>
  <title>Book-O-Rama Book Entry Results</title>
</head>
<body>
<h1>Book-O-Rama Book Entry Results</h1>
<?php
  // create short variable names
  $isbn=$_POST['isbn'];
  $author=$_POST['author'];
  $title=$_POST['title'];
  $price=$_POST['price'];

  if (!$isbn || !$author || !$title || !$price)
  {
     echo 'You have not entered all the required details.<br />'
          .'Please go back and try again.';
     exit;
  }
  if (!get_magic_quotes_gpc())
  {
    $isbn = addslashes($isbn);
    $author = addslashes($author);
    $title = addslashes($title);
    $price = doubleval($price);
  }

  $dbhost = 'localhost';
  $dbuser = 'XXXXXX_XXXXXX;
  $dbpass = 'XXXXXXXXX';
  
  //Connect to Database
  @ $db = new mysqli($dbhost, $dbuser, $dbpass);

  //Select Database
  $dbname = 'XXXXXX_bookorama';

  if (mysqli_connect_errno()) 
  {
     echo 'Error: Could not connect to database.  Please try again later.';
     exit;
  }

  $query = "insert into books values 
            ('".$isbn."', '".$author."', '".$title."', '".$price."')"; 
  $result = $db->query($query);
  if ($result)
      echo  $db->affected_rows.' book inserted into database.'; 

  $db->close();
?>
</body>
</html>

Any help would be greatly appreciated. Thanks in advance!

nloding · February 21, 2007

You're missing a closing ' on $dbuser ... That's the only thing I see.

BTW, "PHP and MySQL Web Development" is a great book ... try to add to it's examples, it really helped me.

mcclellanfsu · February 21, 2007

Thanks for the quick response!

The ' in $dbuser was actually there...I just accidentally deleted it when I was putting the XXXX in there.

I agree on the book. That was my plan, to add more but I thought I should get the original working first!

Now I am getting another error (but at least it's not a blank screen this time)....

Fatal error: Call to a member function on a non-object in /booktest/results.php on line 44

Here is the code (i'll mark line 44)....

<body>
<h1>Book-O-Rama Search Results</h1>
<?php
  // create short variable names
  $searchtype=$_POST['searchtype'];
  $searchterm=$_POST['searchterm'];

  $searchterm= trim($searchterm);

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

 $db=mysql_connect ("localhost", "XXXXXXXX_XXXXXXX", "XXXXX") or die ('I cannot connect to the database because: ' . mysql_error());
      mysql_select_db ("XXXXXXX_bookorama");


  $query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
  $result = $db->query($query); // THIS IS LINE 44

  $num_results = $result->num_rows;

  echo '<p>Number of books found: '.$num_results.'</p>';

  for ($i=0; $i <$num_results; $i++)
  {
     $row = $result->fetch_assoc();
     echo '<p><strong>'.($i+1).'. Title: ';
     echo htmlspecialchars(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>';
  }
  
  $result->free();
  $db->close();

?>
</body>

Any help will be greatly appreciated!

nloding · February 22, 2007

Fatal error: Call to a member function on a non-object in /booktest/results.php on line 44

You're using two forms of a mysql object -- you create $db using regular mysql, but then try to call it using mysqli objects. Google the difference ... mysqli is slower, but more functionality but requires PHP5.

<?php
//$db=mysql_connect ("localhost", "XXXXXXXX_XXXXXXX", "XXXXX") or die ('I cannot connect to the database because: ' . mysql_error());
//mysql_select_db ("XXXXXXX_bookorama");
//here's the mysqli way:

$db = new mysqli("localhost", "user", "pass", "XXX_bookorama");

//now $db->query works
//either that or replace $db->query with mysql_query ...

  $query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
  $result = $db->query($query); // THIS IS LINE 44
?>

mcclellanfsu · February 22, 2007

Thank everyone very much for their quick responses! You are helping me out greatly!

I have one last problem now.

The error I am receiving is:

Fatal error: Call to a member function on a non-object in /booktest/results.php on line 65

My code is:

<body>
<h1>Book-O-Rama Search Results</h1>
<?php
  // create short variable names
  $searchtype=$_POST['searchtype'];
  $searchterm=$_POST['searchterm'];

  $searchterm= trim($searchterm);

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

 $db=mysql_connect ("localhost", "XXXXX_booktst", "XXXX") or die ('I cannot connect to the database because: ' . mysql_error());
      mysql_select_db ("XXXXXXX_bookorama");


  $query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
  $result = mysql_query($query); //Change made here -- used to be $db->query

  $num_results = $result->num_rows;

  echo '<p>Number of books found: '.$num_results.'</p>';

  for ($i=0; $i <$num_results; $i++)
  {
     $row = $result->fetch_assoc();
     echo '<p><strong>'.($i+1).'. Title: ';
     echo htmlspecialchars(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>';
  }
  
  $result->free(); //This is Line 65
  $db->close();

?>
</body>

Once again any help will be greatly appreciated. Thanks!

mcclellanfsu · February 22, 2007

Ok, I changed my code up a little bit, but now I am getting a new error....

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /booktest/insert_book.php on line 39

This is line 39:

 mysql_free_result($result);

The rest of my code:

<body>
<h1>Book-O-Rama Book Entry Results</h1>
<?php
  // create short variable names
  $isbn=$_POST['isbn'];
  $author=$_POST['author'];
  $title=$_POST['title'];
  $price=$_POST['price'];

  if (!$isbn || !$author || !$title || !$price)
  {
     echo 'You have not entered all the required details.<br />'
          .'Please go back and try again.';
     exit;
  }
  if (!get_magic_quotes_gpc())
  {
    $isbn = addslashes($isbn);
    $author = addslashes($author);
    $title = addslashes($title);
    $price = doubleval($price);
  }

//Connect to Database

 $db=mysql_connect ("localhost", "mcclella_booktst", "testing") or die ('I cannot connect to the database because: ' . mysql_error());
      mysql_select_db ("mcclella_bookorama");


  $query = "insert into books values ('".$isbn."', '".$author."', '".$title."', '".$price."')"; 
  $result = mysql_query($query);
  if ($result)
      echo  $db->affected_rows.' The book has been sucessfully inserted into the database.'; 

mysql_free_result($result); //This is line 39
mysql_close(); 

?>

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Problem with results not showing up.

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