Better Way??

[yendor]

This works fine, but it seems like there is should be a better way.

foreach($code as $i=>$val)
{
$query= "SELECT id FROM codes WHERE code='$val'";
$result = mysql_query($query);
$code_id[$i] = mysql_fetch_row($result);
}

It gives me the correct code id for the appropriate code, but I have to reference each code id like this.

echo $code_id[0][0];
echo $code_id[1][0];
echo $code_id[2][0];
etc....

Instead of like this

echo $code_id[0];
echo $code_id[1];
echo $code_id[2];

Many thanks,

Rodney

[brandonschnell]

$row = mysql_fetch_row($result);
$code_id[$i] = $row["id"];

[[deleted]]

"
foreach($code as $i=>$val)
{
$query= "SELECT id FROM codes WHERE code='$val'";
$result = mysql_query($query);
$code_id[$i] = mysql_fetch_row($result);
}
"

question: what's in $i exactly?
You are using it in an EACH statement, which suggests that it is an array,
but you are also using it in brackets, which suggests it is an intereger...

[yendor]

The for each loop pulls corresponding code_id's from a database and puts them in an array like this. $i is just the index for the arrays. It is an integer.

$code[0] = G corresponds to $code_id[0] = 14
$code[1] = PG corresponds to $code_id[1] = 18
$code[2] = R corresponds to $code_id[2] = 54
etc...

After it creates the $code_id array however I have to index it like this to get the correct values.

print("{$code_id[0][0]}<br>");<-- will print 14
print("{$code_id[1][0]}<br>");<-- will print 18
print("{$code_id[2][0]}<br>");<-- will print 54
etc..

It works, I just didn't want the extra[0]. Brandons solution will work, only I was hoping I didn't miss something so simple. DUH on my part!

[yendor]

actually it will have to be.

$code_id[$i] = $row[0];

[[deleted]]

That'll work, but you're back to square ont if you ever decide to use two fields in there...

[yendor]

True,True, but it won't ever change in this program.

Thanks Guys!

[[deleted]]

"True,True, but it won't ever change in this program."

... where have I heard that before.... hmm... :-)