Arrays are Smelly

saltious

Hi.

Im having trouble with arrays.

i run the following query.

select * from table where fieldName = $variable order by primaryKey;
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result);

$a = '0'; $b = count($row);
while ($a <= $b) {
$pr = $row[$a];
echo "$pr<br>";
$a++;
}

the first row seems to print out. however after that it just get a load of Notice: Undefined offset: 13 in D:\dir\adminbyadmin.php on line 39

i get 13 of them they start at 13 and go up to 26 - no matter what the results of the query should be.

anyone know whats gone wrong?

cheers.

NogDog

Read the "Return Values" section of the [man]mysql_fetch_array[/man] page of the manual, and then either add the MYSQL_NUM constant for the optional second parameter, or else just use [man]mysql_num_rows/man, instead.

saltious

Hi.

thanks for this. i am onto a winner now.

i now need help with a tougher challenge.

heres my code:

$query = "SELECT * FROM calls where assigned = '$admindude' order by callid";

$result = mysql_query($query) or die(mysql_error());

$row = mysql_fetch_array($result);

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

printf("ID: %s  Name: %s Logged By: %s", $row["callid"], $row["problem"], $row["userid"]);
echo "\n";
echo "<br>";
echo "<br>";

} - this works fine.

you see how im outputing a field userid from the query. well, rather than put out the user id i need the user name. which is held in a field in another table. essentially i need to select name from usertable where userid = $row['userid'].

however, i tried but failed.

anyone any ideas?

many thanks all.

efficacious

you probably failed your attempt to snatch the info from the other table because you didn't define the other table to use.. or you had some sort of syntax error..

this is very easy to do .. just simply write out a new query to the new table. However your userID coloumn has to be in the other table or you will not be able to retrieve it that way..

Basically you just need some sort of a reference to it in order to select just that one.

wat is your tables structures like?

saltious

Hi, thanks.

basically the above query will work for the one table and i think im confortable with that. its just i dont know how to write the second query. that takes the $row["userid"] and matches that to a userid in the userstable. i then need to pull field called forename and surname from the second table.

do you see what i mean. or have i not answered what you needed?

cheers.

leatherback

Hi,

in your case I would simply get all the details in one go:

"SELECT callid, problem, username FROM calls left join userstable on calls.userid = userstable.userid where assigned = '$admindude' order by callid";