Getting results in array

[mostafatalebi]

Hello

I need to get all of my result in an array, I mean each as an array's element.

 

So if I say $result[63] it right goes to 63th row and fetch the result.

 

I'm hostile to the column id, but if there don't be any other solution, I should go to it.

[Jessica]

Uhm. You are "hostile" to using the functionality designed to make life easier on you because....?

 

Are you saying you want all the results in an array, and then access the array key? Or you want to give a function the ID and get that row?

[mostafatalebi]

With Hostile I mean the id column of sql table do not update accordingly. OK if the id is the only solution I employ it.

My table has many many rows, and I need only the last six rows. I want to get all the rows and then by using array keys fetch them: $result[$size-1]; $result[$size-2] and ...;

 

If I could store only last six rows directly from the SQL statement, then life would be much easier.

[Jessica]

SELECT cols FROM tbl ORDER BY id_col DESC LIMIT 6

[mostafatalebi]

Thank you Jessica. Again you helped me

[Jessica]

See, it's a lot easier when you actually ask the question.

[mostafatalebi]

Sorry I again encountered a problem

 

The below part of script doesn't work with "prepare", while it easily runs with "query";

DEFINE("SELECT", "SELECT english, persian FROM ");

$statement = SELECT . $table . " ORDER BY id DESC LIMIT 6";

if($data = $database->query($statement))

{

$b = $data->fetch_array(MYSQLI_NUM);

echo $b[0] . " " . $b[1];

$data->close();

}

[Jessica]

Why are you creating your query using a constant, a variable table name, and a regular string?

 

Your code doesn't have anything about prepare() in it. 

[mostafatalebi]

It is due to security reason. I often use it. recently I read a book at which this method was also used. When I use prepare in place of query (and of course some more change such as execute() added) it errs that the method mysqli_stmt::fetch_array is undefined.

[Jessica]

POST THE CODE for crying out loud.

 

 

http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php

http://www.php.net/manual/en/class.mysqli-stmt.php

 

There is no fetch_array on a prepared statement. The error message is pretty clear. 

[Zane]

I am taking a shot in the dark here, but maybe this is what you are wanting?

 

$data = array();
while($r = mysql_fetch_array($query)) {
  $k = $r['id'];
  $data[$k] = $r;
}