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Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 in (file path)

Posted on 2008-06-25
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I just installed Firefox 3, and it automatically entered root as the username in my website's login form and some password (I don't know what it is, since I don't have a password set for my development server).

Just to see what happens, I submit the username and password.  And I receive this error...
Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 in (path to my login file)

In IE and firefox 2 it did not automatically populate the form with that username and password.  But, in IE, when I use root as the user name and I copy the password over, I recieve the same error message.

I've attached my corresponding log in file
loginHTML.txt
loginPHP.txt
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Question by:jabrthel
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3 Comments
 

Author Comment

by:jabrthel
ID: 21871443
Well, I just cleared all my private data in firefox 3, so that solved one problem... I guess it was pulling stuff from odd places.  If I place root and that same password, it still gives me a problem.  However, as long as I don't place that password in, it seems to be fine...
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Author Comment

by:jabrthel
ID: 21871460
Well, I just figured out that the password being enter was localhost... so I guess my question would be, why is that particular password giving the above error, when all others are fine?
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Accepted Solution

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Steve Bink earned 2000 total points
ID: 21920334
I would guess that you are receiving the error not because of the password, but because of the loginID.  Your code to query the database (84-88 in loginPHP.txt):

      mysql_connect("localhost","root","");
      mysql_select_DB("database") or die("unable to select database");
      $query = mysql_query("select * from member where loginID like '$loginID'");
      $passwordSQL = mysql_result($query, 0, "password");
      mysql_close();

If $loginID does not exist in table member, then $query will return with an empty result set.  Since there would be no records, jumping to the first record would fail, presenting the error you are seeing.  You can resolve this by explicitly checking for a single-record result set, taking action if anything looks strange.  You should check that mysql_num_rows($query)==1.  If it does not (no records, or more than one record), then something went wrong.

You can test this theory by using the same user name with any password, as well as trying a real user name with the same password.
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