Problem with PHP tutorial

**theprof** · Sep 16, 2000 02:48

Hello, I'm encountering an error with newjoke.php at http://www.adolescentadulthood.com/phptest/newjoke.php

It gives me an error, and i dont know how to correct it. Can someone please take a look and help?

**Kevin Yank** · Sep 16, 2000 08:58

theprof,

Any chance you can make the code you're using available as a .TXT file on your sever? Blank out your database username/password of course, though. I can't diagnose the problem without seeing the code.

**theprof** · Sep 16, 2000 09:02

Sure. Its the same newjoke.php provided in your tutorial.

Code:

<!-- newjoke.php -->
<HTML>
<HEAD>
<TITLE> Add New Joke </TITLE>
</HEAD>
<BODY>
<?php

if ($submit): // A new joke has been entered
              // using the form.

  if ($aid == "") {
    echo("<P>You must choose an author " .
         "for this joke. Click 'Back' " .
         "and try again.</P>");
    exit();
  }

require ("dbcnx.php");

  $sql = "INSERT INTO Jokes SET " .
         "JokeText='$joketext', " .
         "AID='$aid'";
  if (mysql_query($sql)) {
    echo("<P>New joke added</P>");
  } else {
    echo("<P>Error adding new joke: " .
         mysql_error() . "</P>");
  }

  $jid = mysql_insert_id();

  $cats = mysql_query("SELECT ID, Name FROM Categories");

  while ($cat = mysql_fetch_array($cats)) {
    $cid = $cat["ID"];
    $cname = $cat["Name"];
    $var = "cat$cid"; // The name of the variable
    if ($$var) { // The checkbox is checked
      $sql = "INSERT IGNORE INTO JokeLookup " .
             "SET JID=$jid, CID=$cid";
      $ok = mysql_query($sql);
      if ($ok) {
        echo("<P>Joke added to category: $cname</P>");
      } else {
        echo("<P>Error inserting joke into category $cname:" .
             mysql_error() . "</P>");
      }
    } // end of if ($$var)
  } // end of while loop
?>

<P><A HREF="<?php echo($PHP_SELF); ?>">Add another Joke</A></P>
<P><A HREF="jokes.php">Return to Joke Search</A></P>

<?php
  else: // Allow the user to enter a new category

  $dbcnx = @mysql_connect("localhost", "root", "mypasswd");
  mysql_select_db("jokes");

  $authors = mysql_query("SELECT ID, Name FROM Authors");
  $cats    = mysql_query("SELECT ID, Name FROM Categories");
?>

<FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST>
<P>Enter the new joke:<BR>
<TEXTAREA NAME="joketext" ROWS=15 COLS=45 WRAP>
</TEXTAREA>
<P>Author:
<SELECT NAME="aid" SIZE=1>
  <OPTION SELECTED VALUE="">Select One
  <OPTION VALUE="">---------
<?php
  while ($author = mysql_fetch_array($authors)) {
    $aid = $author["ID"];
    $aname = $author["Name"];
    echo("<OPTION VALUE='$aid'>$aname\n"); 
  }
?>
</SELECT></P>
<P>Place in categories:<BR>
<?php
  while ($cat = mysql_fetch_array($cats)) {
    $cid = $cat["ID"];
    $cname = $cat["Name"];
    echo("<INPUT TYPE=CHECKBOX NAME='cat$cid'>$cname<BR>\n"); 
  }
?>
</P>
<P><INPUT TYPE=SUBMIT NAME="submit" VALUE="SUBMIT"></P>
</FORM>
<?php endif; ?>
</BODY>
</HTML>

**Kevin Yank** · Sep 16, 2000 09:33

As I mentioned in my article, this sample code omits all error checking in the interest of brevity. As a result, it does not tell you when one of your MySQL queries fails.

You should modify your code so that, when an error does occur, it is deteced and displayed. For example, consider the following line:

Code:

$cats = mysql_query("SELECT ID, Name FROM Categories");

Instead of just assuming this will work, check it:

Code:

$cats = mysql_query("SELECT ID, Name FROM Categories");
if (!$cats) {
  echo("MySQL Error: " . mysql_error());
  exit;
}

Now your Web page should display the actual error message that is being produced by your database. If you need any further help fixing the problem, feel free to ask!