It's best if you can supply the full code. Code may not need to be attached,
you can just put in a text file and send the link so that we all can read.
Error in line 42 DOES NOT mean the error is in line 42, sometimes the error
can be in the couple of lines before line 42.
Best wishes,
Indri
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> From: <[EMAIL PROTECTED]>
> Reply-To: <[EMAIL PROTECTED]>
> Date: Mon, 9 Apr 2001 18:15:52 -0500
> To: <[EMAIL PROTECTED]>
> Subject: [PHP-DB] Help please
>
> Hi all,
>
> I have a website that I am trying to develop a "add news" page for.
> Basically people who have the authority to add news for the site can type in
> their username, password, the title of the article, and the article itself
> and submit it to the site (which then verifies the username and password and
> if successfull adds it to the DB so the news will appear the next time
> somone browses to the website.
>
> This file is an included file into the main page...the $db variable is
> valid, and works with all the other pages so far... I am really stumped and
> getting frustrated.
>
> I apologise in advance if I am including too much code in my example....the
> place with the error is a couple lines from the bottom of the code sample.
>
>
> My problem is that I am getting an error when i try to submit the form, code
> sample is below.
>
> The error is:
>
> Warning: Supplied argument is not a valid MySQL-Link resource in
> /usr/www/kyid/public_html/addnews.php on line 42
>
>
>
> <?php
> $size=20;$cols=46;
> echo "<div align=\"center\">";
> if(!$submit){
> echo "<form method=\"post\" action=\"addnews.php\">\n";
> echo "<table><tr><td> </td></td></table>\n";
> echo "<table width=\"450\" border=\"0\" cellspacing=\"1\"
> cellpadding=\"2\" align=\"center\">\n";
> echo "<tr valign=\"top\">\n";
> echo "<td bgcolor=\"#6A7292\" class=\"text\">* UserName<br>\n";
> echo "<input type=\"text\" name=\"uname\" class=\"input\"
> maxlength=\"20\" size=\"".$size."\" value=\"\">\n";
> echo "<br>\n";
> echo "</td>\n";
> echo "<td bgcolor=\"#6A7292\" class=\"text\">* Password<br>\n";
> echo "<input type=\"password\" name=\"password\" class=\"input\"
> size=\"".$size."\" value=\"\">\n";
> echo "</td>\n";
> echo "</tr>\n";
> echo "<tr valign=\"top\">\n";
> echo "<td bgcolor=\"#565d77\" class=\"text\">* Headline<br>\n";
> echo "<input type=\"text\" name=\"headline\" class=\"input\"
> maxlength=\"40\" size=\"".$size."\" value=\"\">\n";
> echo "<br>\n";
> echo "</td>\n";
> echo "</tr>\n";
> echo "<tr valign=\"top\">\n";
> echo "<td colspan=\"2\" bgcolor=\"#565d77\" class=\"text\">*
> Story<br>\n";
> echo "<textarea name=\"newstext\" cols=\"".$cols."\"
> wrap=\"VIRTUAL\" rows=\"10\" class=\"input\"></textarea>\n";
> echo "</td>\n";
> echo "</tr>\n";
> echo "<tr valign=\"top\">\n";
> echo "<td bgcolor=\"#7B819A\" align=\"center\"
> colspan=\"2\">\n";
> echo "<input type=\"submit\" name=\"submit\" value=\"Add
> Article\" class=\"submit\"><input type=\"reset\" name=\"reset\"
> value=\"Clear\" class=\"submit\">\n\n";
> echo "</td>\n";
> echo "</tr>\n";
> echo "</table></form>\n";
> }else{
> /* More stuff would go in here, but this isn't working. */
> /******** THE LINE BELOW THIS IS THE ERROR LINE ********/
> $validate = mysql_query("SELECT username, user_password FROM users
> WHERE username=".$uname,$db);
> }
> echo "</div>";
>
> Thanks in advance for any help, or please let me know if this should be on
> php-general instead...
>
> Again, thanks
> Keith Young.
>
>
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