You can't start variables with a number.
---John Holmes...
----- Original Message -----
From: "Hutchins, Richard" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, September 26, 2002 4:18 PM
Subject: [PHP-DB] Return Array from Function
> Can somebody tell me what I might be doing wrong in the code below? I'm
> trying to use a function to perform an often-used db query and return the
> resulting resource to the calling script. I keep getting this error
though:
>
> Parse error: parse error, unexpected T_LNUMBER, expecting T_VARIABLE or
'$'
> in c:\spidey\public\addleveltwo.php on line 51
>
> I've looked and looked, but I can't find what could be causing a parse
> error. Most likely, I've just been looking at it for too long. Any fresh
> eyes available?
>
> <?php
> include("../protected/getContent.php");
>
> getLevelOnes($docnum); //user-defined function from included file.
> Definition appears after this snippet.
>
> $num=mysql_num_rows($levelones);
>
> if($num<1)
> {
> echo("There is no level one (Section) content for this
> document in the database.");
> exit();
> }
>
> while($row=mysql_fetch_array($levelones))
> {
> $1contentID=$row["1contentID"];
> $parentID=$row["parentID"];
> $1child=$row["childID"];
> $1content=$row["1content"];
> $1ordinal=$row["ordinal"];
>
> echo("<tr><td>".$1content."</td><td><a
> href=\"contentMgr.php?docnum=".$docnum."\"
> target=\"_right\">Add</a></td></tr>");
> }
> ?>
>
> <?php
> function getLevelOnes($docnum)
> {
> $sql="SELECT 1contentID, parentID, childID, 1content, 1ordinal FROM
> levelone WHERE 1contentID='$docnum'";
> $levelones=mysql_query($sql)
> or die(mysql_error());
> return $levelones;
> }
> ?>
>
>
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