mysql_fetch_array() error

kal78

Hello,

If anyone can help, I have the following error on my php code:
PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result.This is part of the PHP where the problem is and the message i get on the HTML page is something wrong in artistinfo table:

artistid is varchar(20) primary key in artistinfo table
song is varchar(20) from the same table

$query="select MAX(artistid) as artistid from artistinfo";
$result=mysql_query($query);
if($row=mysql_fetch_array($result))
{
$artistid = ++$row["artistid"];

}
else
{
die("something wrong in artistinfo table!");
}

Thank you for any response

NogDog

That error normally indicates that MySQL was unable to parse/process your query. Try the following to get more debug info:

$query="select MAX(artistid) as artistid from artistinfo";
$result=mysql_query($query) or die("Query failed ($query): " . mysql_error());

PS: wrapping your code in [ php] . . . [/php ] tags (without the spaces) will make it much easier for us to read your code.

kal78

NogDog wrote:
That error normally indicates that MySQL was unable to parse/process your query. Try the following to get more debug info:
$query="select MAX(artistid) as artistid from artistinfo";
$result=mysql_query($query) or die("Query failed ($query): " . mysql_error());
PS: wrapping your code in [ php] . . . [/php ] tags (without the spaces) will make it much easier for us to read your code.

Thanks for the reply. When i add die("Query failed ($query): " . mysql_error()); , it shows the following message:

Query failed (select MAX(artistid) as artistid from artistinfo): Can't connect to local MySQL server through socket '/tmp/mysql.sock' (46)

NogDog

Do you do a mysql_connect() earlier in your code? If so, do you verify that the connection is successful?

kal78

NogDog wrote:
Do you do a mysql_connect() earlier in your code? If so, do you verify that the connection is successful?

This is my entire code. Thanks for any help

<?php

$host="mysql";
$database="Music";
$dbusername="xxx";
$dbpassword="xxx";

$x=mysql_connect($host,$dbusername,$dbpassword);
$x1=mysql_select_db($database);

$query="select song from Discography where song='".$song."'";
$result=mysql_query($query);
if($row=mysql_fetch_array($result))
{
die(" Sorry that song " .$song. " already exists ! ");
}

$query="select MAX(artistid) as artistid from artistinfo";
$result=mysql_query($query) or die("Query failed ($query): " . mysql_error());
if($row=mysql_fetch_array($result))
{
$artistrid = ++$row["artistid"];

}
else
{
die("something wrong - artistinfo table!");
}
$query="INSERT into artistinfo VALUES('".$artistid."','".$name."','".$genre."')";
$result=mysql_query($query);
if (!($result))
{
die(" Artist information could not be stored");
}

$query ="INSERT into Discography VALUES('".$name."','".$song."','".$artistid."')";
$result=mysql_query($query);

if (!($result))
{
die("Artist song could not be stored ");
}

$query ="INSERT into login VALUES('".$artistid."','".$song."','".md5($password)."')";
$result=mysql_query($query);

if (!($result))
{
die(" Artist authentification information could not be stored");
}

echo " <center>Artist Added Successfully </center> ";
echo " <hr> ";
echo " Artistid/Login ID : ".$artistid." ";
echo $name," ";
echo " song : ",$song," ";
echo " Password : ";
for ($i=0;$i<strlen($password);$i++) {echo "*";}
echo " \n";
echo " Genre : ",$genre," ";

NogDog

Similar to the query debugging, you should verify that the connection works. At a minimum, you could do something like:

$x=mysql_connect($host,$dbusername,$dbpassword) or die("Unable to connect to DB: " . mysql_error());
$x1=mysql_select_db($database) or die("Unable to select DB: " . mysql_error());

bradgrafelman

When posting PHP code, please use the board's [PHP][/PHP] bbcode tags - they make the code much, much easier to read (as well as aiding the diagnosis of syntax errors). You should edit your post now and add these bbcode tags to help others in reading/analyzing your code.