Is there any problem using loop with mys

[Anon]

The following is part of the scripts in my wepages.
Is there any problem using loop with mysql+php?
It can't do looping, the query statment only execute once, and then the error statment "Couldn't add client." is printed.
Is there any wrong?
<?
$connection = mysql_connect("localhost","database","password")
or die ("Couldn't connect to server.");

$db = mysql_select_db("database", $connection)
or die ("Couldn't select database.");

$fund_no=count($fund_id);
for($index=0; $index<$fund_no;$index++)
{
$sql= "insert into select_fund(user_id,client_id,fund_id) values('$user_id','$client_id','$fund_id[$index]')";
$result = mysql_query($sql,$connection) or die("Couldn't add client.");
}
........
?>

notes:the $fund_id was selected by user, and the $fund_id can be selected more than one and store in array.

[[deleted]]

I've said it before and I'll say it again untill somebody listens:

Use error checking!

The simple "or die('couldnt bla)" is pointless.

Step one:
make your script print the mysql_error() instead of just "or die".
Step Two:
If you dont do step one, at least make your script PRINT the query before you execute it so you can see what you are trying to do.

[Anon]

Thanks for reply me,vincent!

i'm a new in php, i try to use mysql_erorr instead of "or die"

$sql= "insert into select_fund(user_id,client_id,fund_id) values('$user_id','$client_id','$fund_id[$index]')";
//line37
$result = mysql_query($sql,$connection) or mysql_error(&$sql);

and a warning statement is printed:

Warning: Supplied argument is not a valid MySQL-Link resource in //add_client.php on line 38

but i still can't understand what's the problem, could you help me more?

[Anon]

O..sorry,i know it is mysql_error($connection) instead of mysql_error(&$sql) now.
but there are no error msg of mysql_error($connection), and i also try to add a echo"" statement within the loop.
$fund_no=count($fund_id);
for($index=0; $index<$fund_no;$index++)
{
$sql= "insert into select_fund(user_id,client_id,fund_id) values('$user_id','$client_id','$fund_id[$index]')";
$result = mysql_query($sql,$connection) or mysql_error($connection);
echo"ok";
}

if the $fund_no is 2, the echo statement is printed two times, but actually the query just can execute in the first time, and i also check that there are only one row is insert in the table.

[[deleted]]

Abby, this is how I would do this. Have a look and see if you understand what is going on.
Be carefull when you copy/paste this code, some lines are broken in two by this forum.

$fund_no=count($fund_id);
for($index=0; $index<$fund_no;$index++)
{
$sql= "insert into select_fund(user_id,client_id,fund_id)
values('$user_id','$client_id','$fund_id[$index]')";
if ($result = mysql_query($sql,$connection))
{
echo "Query worked, index=$index<BR>";
}
else
{
echo 'Query error:'.mysql_errno($connection).mysql_error($connection).'<br>';
echo 'The query was: '.$sql.'<BR>';
echo "Index was $index<BR>";
}
}

[Anon]

Thanks a lot,vincent!
I know what's the problem now.