Function won't return values

[Anon]

I have the following function below, and when I call the function it won't return any values, however if I put an echo inside the function (right before the return call) then it will show the values, but I cannot get those values outside the function with the function call. Anyone know why this is? Function and call is below. Thanks!

//Automatically finds words that are in the dictionary and replaces them with links in the form input
function DictEncode($forminput){
//Take form input and convert it into an array of words
$formwords = preg_replace("/[^{[:alnum:]|[:space:]]/","",$forminput);}
$wordarray = explode(" ",$formwords);

//Connect to database and find words in the dictionary that match words in the above array
$db=mysql_connect("localhost","name","pass");
mysql_select_db("database",$db);
$query = "SELECT * FROM dict WHERE word IN ('".implode("', '", $wordarray)."');";
$myresult = mysql_query($query);

//Go through the results of the word matching and change the word into a link and resave it as the new form input
while ($row = mysql_fetch_array($myresult)) {

	$origword = " ".$row["word"]." ";
	$theword = "/".$row["word"]."/"; 
	$worddef = $row["definition"];

	$caption = "Definition: ".$origword."";
	$replace = "<a href='javascript:void(0);' onMouseOver='overlib(\"".$worddef."\", CAPTION,\"".$caption."\")' onMouseOut='nd()'>".$origword."</a>";

	$forminput = preg_replace($theword, $replace, $forminput);

}

echo $forminput;
return $forminput;

}
//End of function that automatically checks and adds links for Dictionary definitions that are in articles

//Call the Dictionary Encode function
$NewsMsg = DictEncode($NewsMsg);

[Anon]

Actually, I just did a test, and it will return the value of the function, it just won't write it to the database.

[junebug]

well, i don't see where you are writing to a database in that chuck of script.

[Anon]

Sorry, it's just after the call to the function....

$offset_time=offset_time();

			$query="insert into $article_file (ArtCat,ArtDate,ArtTitle,ArtLead,ArtMsg) values ('$ArtCat','$offset_time','$NewsTitle','$NewsLead','$NewsMsg')";
				$insert=mysql_query($query,$db);
				echo "<br>";
				echo "Done!";

[Anon]

This is the value of $NewsMsg that it doesn't like to write to the database.

<a href='javascript:void(0);' onMouseOver='overlib("Application Service Providers are third-party firms that manage and deliver software-based services to customers across a wide area network. ASPs allow companies to out-source some or all aspects of their technology requirements.", CAPTION,"Definition: asp ")' onMouseOut='nd()'> asp </a>

[Anon]

Are you escaping the ' characters? You either need to use the addslashes() function on the data you are sending through to the DB, or have Magic_Quotes on.

Also, try testing for errors after running the query:

mysql_query($query) or
die (mysql_error());

will print the error the DB is returning when the query does not work.

[Anon]

Yes I am escaping the characters and magic_quotes is on. If I were to echo the variable, it works perfectly, but it still won't add it to the database.

[Anon]

This is the sql error I get, but it says it is with that same line as above.

You have an error in your SQL syntax near '<a href="javascript:void(0);" onMouseOver="overlib('Application Service Provider' at line 1