[Resolved] mysql_result giving error

ryland22

I can't figure out why I'm getting an error. I pass two variable via a querystring and retrieve them using GET, run a query on my database using one of the variables, and I keep getting the following error when I try to write the data to the page using mysql_result:

mysql_result(): Unable to jump to row 1 on MySQL result index 2

Below is my code:

<?php
$thread =$_GET['thread'];
$sub =$_GET['sub'];
?>

<?php
//retrieve thread info

$host = "";
$uname="root";
$pass="";
$database="discuss";
$tablename="threads";
$connection = mysql_connect($host, $uname, $pass);
$result =mysql_select_db("discuss");
$query ="SeLECT * from threads where thread_num = $thread";
$result = mysql_query($query);
?> 

<?php
echo mysql_result($result, $sub, 4);
echo "<br>";
echo mysql_result($result, $sub, 3);
echo "<p></p>";
?>

dotbob

this sounds like an "out of range" error.

i.e. the query returned no results or $sub doesn't contain a valid row number.

Try retrieving and echoing the number of rows returned(

echo mysql_num_rows($result)

). This way you will know if your $sub var is out of the range of returned rows.

nemonoman

I think I understand your problem better now;

You are using mysql_result 2 times:

Each time you use mysql_result, it advances the row pointer to the next returned row of the query result.

In this instance, I would guess that you have a unique id, ie only one row from threads will be returned.

The first time you use mysql_result, it looks to the first (and only) returned row.

The second time you call it, it looks to the second (non-existent) row, and blows up.

What you really mean, I think, is to get 2 values from this query, right?

Why not just make it explicit:

$query ="SeLECT FirstValueIWant, SecondValueIWant from threads where thread_num = $thread";
$result = mysql_query($query);

$row=mysql_fetch_array($result);
//likely this query just gets you 1 specific row, right? if multiple rows returned, you need a while loop.

echo $row['FirstValueIWant'];
echo $row['SecondValueIWant '];

dotbob

mysql_result() advances the row???

Since when?

//this statement returns the fifth field on row num specified by $sub var
echo mysql_result($result, $sub, 4); 
echo "<br>"; 
//this statement returns the fourth field on row num specified by $sub var
echo mysql_result($result, $sub, 3);

Therefore, if the $sub var contains a value < 0 or > the actual number of rows returned, it will return an error as the row it is trying to reference doesn't exist.

nemonoman

[man]mysql_result[/man]

from the user notes

daendersNOSPAM at yahoo dot com
10-Jul-2002 04:25
If you came to this function like I did, looking for a function that grabs info from the result but does not increment it to the next result row, you are out of luck.

This function appears to move the result to the next row just like mysql_fetch_assoc() and the others.

dotbob

From the mysql man

mysql_result
(PHP 3, PHP 4 )

mysql_result -- Get result data
Description
mixed mysql_result ( resource result, int row [, mixed field])

mysql_result() returns the contents of one cell from a MySQL result set. The field argument can be the field's offset, or the field's name, or the field's table dot field name (tablename.fieldname). If the column name has been aliased ('select foo as bar from...'), use the alias instead of the column name.

When working on large result sets, you should consider using one of the functions that fetch an entire row (specified below). As these functions return the contents of multiple cells in one function call, they're MUCH quicker than mysql_result(). Also, note that specifying a numeric offset for the field argument is much quicker than specifying a fieldname or tablename.fieldname argument.

Calls to mysql_result() should not be mixed with calls to other functions that deal with the result set.

nemonoman

Thanks for quoting some of the text of the manual page that I posted. I don't understand the context, however. You asked why your code broke, I told you. When you appeared surprised by my answer, I documented my response. That the PHP manual often sucks is common knowledge, and one of the reasons I make it a point to read the user comments and examples included on line.

On the other hand, you could keep your code as is, and wait for PHP to change the nature of the mysql_result function to reflect your idea of how it ought to work.

dotbob

I think we're on crossed wires here.

Firstly, I didn't ask why my code broke, I was replying to the thread authors request for help.

Secondly, I appreciate the fact that users comments can "sometimes" be helpful. This, however, doesn't change the fact that the mysql_result() function DOES NOT advance rows automatically.

I agree with your opinion of the PHP manual...

...mostly.

However, the vast majority of common function documentation within the manual is pretty good.

Have you read it?

On the other hand, you could keep your code as is, and wait for PHP to change the nature of the mysql_result function to reflect your idea of how it ought to work.

Have you ever even used mysql_result()???? Based on your previous answers on this thread, I am beginning to doubt it.

nemonoman

DotBob, I apologize.

I tried and gave up using this function in my early PHP days. Now I don't doubt that my interpretation of the results was due to my own stupidity then.

You are right and I am wrong.

Thanks for setting me straight.

dotbob

Not a problem,

Cheers.

mtimdog

It seems like your actually printing out rows 4 and 3 instead of columns 4 and 3 of the first result row (if I'm reading the man right, which I hardly do).

Anyway, try using mysql_fetch_array like previously suggested and then echo out the column using column name, or the numeric id(4 and 3).

HTH

dotbob

nope.

it's definately to do with the fact that he is trying to reference a row which doesn't actually exist within the result set.

$result - this is the resource identifier
$sub - this is the row offset
3 - this is the column offset (in this case the fourth column)

mysql_result($result, $sub, 3);

ryland22

Hey guys, thanks for all the input. The error was that the row didn't exist, I have since corrected it.