PHP MySQL Error Fetch Array

seanmayhew

The following code produces this result from the Kevin Yank Book:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\www\code\chapter7\joke.php on line 15

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THE CODE:

<html>
<head>
<title> Joke </title>
</head>
<body>
<?php

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

// Get the joke text from the database
$joke = mysql_query("SELECT JokeText FROM Jokes ".
"WHERE ID=$id");
$joke = mysql_fetch_array($joke);
$joketext = $joke["JokeText"];

// Filter out HTML code
$joketext = htmlspecialchars($joketext);

// If no page specified, default to the
// first page ($page = 0)
if (!isset($page)) $page = 0;

// Split the text into an array of pages
$textarray=split("[PAGEBREAK]",$joketext);

// Select the page we want
$joketext=$textarray[$page];

// Bold and italics
$joketext = eregi_replace("[b]","",$joketext);
$joketext = eregi_replace("[eb]","",$joketext);
$joketext = eregi_replace("[i]","",$joketext);
$joketext = eregi_replace("[ei]","",$joketext);

// Paragraphs and line breaks
$joketext = ereg_replace("\r","",$joketext);
$joketext = ereg_replace("\n\n","",$joketext);
$joketext = ereg_replace("\n"," ",$joketext);

// Hyperlinks
$joketext = ereg_replace(
"[L]([-./a-zA-Z0-9!&%#?,'=:~]+)[EL]",
"<a href=\"\1\">\1</a>", $joketext);
$joketext = ereg_replace(
"[L=([-./a-zA-Z0-9!&%#?,'=:~]+)]".
"([-_./a-zA-Z0-9 !&%#?,'=:~]+)[EL]",
"<a href=\"\1\">\2</a>", $joketext);

if ($page != 0) {
$prevpage = $page - 1;
echo("<a href=\"$PHP_SELF?id=$id&page=$prevpage\">".
"Previous Page</a>");
}

echo( "$joketext" );

if ($page < count($textarray) - 1) {
$nextpage = $page + 1;
echo("<a href=\"$PHP_SELF?id=$id&page=$nextpage\">".
"Next Page</a>");
}

?>
<a href="index.php">Back to the front page</a>
</body>
</html>

Jayson

I believe that is what it does when your result is empty, isn't it?

I've had this problem too, and what I did was to change it so it is like:

if($result) {
    $row = mysql_fetch_array($result);
    // do something
}
else {
    // empty set!
    // do something else
}

That way if there is no result, it doesn't try to fetch a row, thus you avoid the error.

if($result) {
    while($row = mysql_fetch_array($result) {
        // do something
    }
}

The later option obviously if you have multiple rows to read. There may be a better solution, but this is what I have done and it works alright for me. If someone knows a better way, please comment.

barand

Have you got a value in $id ?

if not, your query is

"select joketext from jokes where id ="

which will give an error