I guess I should have walked away then come back to it.  I was just trying
to get this code to work with little effort as possible.  I guess, I should
have done it right in the first place.  What I didn't realize is that I
mistakenly made

 $sql= "SELECT * FROM cd_list WHERE $searchstring LIKE '%searchstring%'

I should have made $searchstring to $searchtype that resolved the initial
problem with the code.

Thank you Nathan.  I guess from now on I wont be trying to shortcut coding

Also thank you all for you assistance in this matter.

Thank You


Daniel Negrón
Lotus Notes Administrator / Developer
KB Electronics, Inc.


                    <etherwolf@sop       To:     "Daniel Negron/KBE" 
<[EMAIL PROTECTED]>                   
                    ris.net>             cc:                                           
                                         Subject:     Re: Invalid Argument ???  Not 
sure how to debug this          
                    04:54 PM                                                           

The simplest thing to do to narrow down the source of the problem is change
line 9 to read:

$result = mysql_query($sql,$db) or exit(mysql_error()."<BR>".$sql);

This will halt the script before it even gets to line 12 and give you the
error MySQL had as well as
the query that was sent to the database. This is a very good practice for
the rest of your php code.

Your ...LIKE %searchstring%... bit with no $ will give you all rows

I take it that is not your intention... You probably need to change that

...LIKE %".$searchstring."%...

Also, why are both your column name and desired value the same? Should you
be selecting WHERE
columnname LIKE %$searchstring%?

If you are still unsuccessful in resolving this, I'll help out however I


# Nathan

----- Original Message -----
From: "Daniel Negron/KBE" <[EMAIL PROTECTED]>
Sent: Sunday, March 17, 2002 1:38 AM
Subject: Invalid Argument ??? Not sure how to debug this

Hi all and Good Morning.

I have been struggling with this code:  I was wondering if someone could
lend me a hand.

I am working with this code and can't get the error to disappear.

After running I get

| Warning: Supplied argument is not a valid MySQL result resource in C:\My
Documents\My   |
| Webs\php\TMPibtpit3zq0.php on line 12

Line 12 happens to be the WHILE statement.  I am not sure if it is the
while statement or the results its looking for.

TIA  So far, i havent been able to sleep atleast until I get this.

if (isset($searchstring))

     $db = mysql_connect("localhost","webuser","");
     $sql= "SELECT * FROM cd_list WHERE $searchstring LIKE '%searchstring%'
ORDER BY artist ASC";
     $result = mysql_query($sql,$db);
     echo "<TABLE BORDER=2>";
     echo "<tr><td><b>Artist</b><td><b>Title</b><td><b>Options</b></tr>";
     while ($myrow = mysql_fetch_array($result))
               echo "<tr><td>".$myrow["artist"]."<td>".$myrow["title"];
               echo "<td><a href=\"edit.php?id=".$myrow["cd_id"]."
          echo "</TABLE>";
          <form method="POST" action="<? $PHP_SELF ?>">
          <table border="2" cellspacing="2">
          <tr><td>Insert Your Search String Here.</td>
          <td>Search Type</td></tr>
          <td><input type="text" name="searchstring" size="28"></td>
          <td><select size="1" name="searchtype">
               <option selected value="artist">Artist</option>
               <option value="title">Title</option>
               <option value="category">Category</option>
          <p><input type = "submit" value="Submit" name="B1">
          <input type="reset" value="Reset"></p>



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