Warning: 0 is not a Mysql result index

vicent

Hi everybody .

I am trying to evaluate a mysql result because when i use that result in a script (ex: $row=mysql_fetch_array($result) ) it shows me the alert "Warning: 0 is not a mysql result index..." .
I tried with empty($result) , !$result, !isset($result) but didn't work.

Can anyone help me?

thanks a lot.

spacemax

You can catch this exception with a if()

if ($row = mysql_fetch_array($result)) {
do {
?><?=$row["text"]?><?
} while($row = mysql_fetch_array($result));
} else {}

trooper

From he snippet of informaio you have provided it look s like you are returning no records. Are you expecting some records to be returned ? If so check you sql statement carefully

if you are expecting no records to be returned then i believe that you should have an if statement to catch this.

vicent

Thank you very much.
It works perfectly.

Anon

I was also getting this warning, and have implemented your solution, but keep getting parse errors when I use the do {?>}.

if($result = mysql_query($sql)) {
do { ?><?=$result["text"]?><? }
while $result = mysql_query($sql));
}
is where I am up to .. I actually found that your message had two versions, one in context, and one out of thread ... I have tried both, but cant resolve the parse error situation with them. Could it be a problem with the ?><? ?

Anon

Try to drop any '?><?' .
You must start your php script with '<?' and use '?>' to finish it.

Anon

I understand the usage of <? ?> ... I was simply copying what had appeared in the following post ...

Author: Dominik Nufer (195.65.87.136) Profile
Date: 2001-03-07 04:14:14

You can catch this exception with a if()

if ($row = mysql_fetch_array($result)) {
do {
?><?=$row["text"]?><?
} while($row = mysql_fetch_array($result));
} else {}

.. it didnt make sense to me, but then I'm relatively new to this, and thought I would submit myself to someone who was a bit more experienced ...

Unfortunately I became VERY confused when this same post (listed out of thread context) reads as follows ...

Author: Dominik Nufer
Date: 2001-03-07 04:14:14
Subject: RE: Warning: 0 is not a Mysql result index
You can catch this exception with a if()

if ($row = mysql_fetch_array($result)) {
do {
?> } while($row = mysql_fetch_array($result));
} else {}

Now, I have tried both these solutions, trying to tailor them to my situation, but am not having any luck.

My code reads as follows:

8# verify user from admin_login.htm
9
10$sql = "SELECT * FROM admin_login WHERE (username=$admin_user && password=$admin_pass)";
11
12# Execute the query and put results in $result
13
14 if($result = mysql_query($sql)) {
15 do { ?> <? =$result("text") ?> <? }
16 while $result = mysql_query($sql));
}

I get a parse error on line 15.

Before this, I also have a require file, which I have checked and double checked for errors ... but the ?><? on line 15 worries me... can someone clarify this please.

Anon

You can try:

8# verify user from admin_login.htm
9
10$sql = "SELECT * FROM admin_login WHERE (username=$admin_user && password=$admin_pass)";
11
12# Execute the query and put results in $result
13
14 if($result = mysql_query($sql)) {
15 do {$row = mysql_fetch_array($result));
16 $your_field=$row["your_field"] }

#Now you can use the value of $your_field.
.
.
.

Please, check '(', ')', '{', '}', ';', '<?' and '?>'

Bye

cldeans

Thanks for your help ... managed to work it out with some help from miester ...

verify user from admin_login.htm

$sql = "SELECT * FROM admin_login WHERE (username='$admin_user' && password='$admin_pass')";

Execute the query and put results in $result

// $r becomes the result index - basically a pointer to the results of
// your query.
$r = mysql_query($sql);

// The @ supresses errors that might be given by a bad $r (which is then
// no results were returned by the query)
if(@mysql_num_rows($r))
{
while($row = mysql_fetch_array($r))
{
echo $row['table_field'];
}
}

Works perfectly!