trouble creating array of field names

[sickel2]

I am simply trying to get a list of fields from a table. I am getting an: "Undefined variable: current_array" error. Here's my code:

function get_fields($user)
{
$conn = db_connect();
$sql = "show fields from current";
$result = $conn->query($sql);
$totalFields = mysqli_num_fields($result);
for ( $i = 0; $i < $totalFields; ++$i )
{
$current_array[$i] = mysqli_fetch_array($result);
}
return $curent_array;
}

Any ideas? And is there a simpler way to create an array out of the field names for a table?

[catweasel]

use mysqli_fetch_fields

[danebrood]

$sql= 'DESCRIBE current';
$results = mysql_query($sql) or die('Query error: ' . mysql_error());

$field_array = array();
while($row = mysql_fetch_array($results)) {
    $field_array[] = $row['Field'];
} 

[sickel2]

I've tried both techniques to no avail. Let's see if I executed correctly:

Catweasel, I tried to use mysqli_fetch_fields with this code:

$sql = "select * from current";
$result = $conn->query($sql);
$current = mysqli_fetch_fields($results) ;
return $current;

I get nothing

Danebrood, I tried your code word for word and I got a "Query error: No database selected" error. I added the databasename to the table name (so describe sickel.current) and I got "Query error: Access denied for user ''@'localhost' to database 'sickel' ". This should not happen since I used the db_connect function that has my password and user name and which never fails.

Any ideas?

[keviin6]

Yry to print the variable which has the username and password, I belive that you are using the global varible which holds username and password inside the function. If I am right you need to declare the variables as global inside the function too.

[sickel2]

I wasn't sure what you meant by make the variables global within the function. What I just did was substitute "$conn = mysqli_connect('localhost', 'myusername', 'mypassword', 'mydatabase);" in place of "$conn = db_connect();" I still got the "Query error: Access denied for user ''@'localhost' to database 'mydatabase' " Was that what you were suggesting?

[danebrood]

For the example I gave, you'd want something like the following to occur prior:

mysql_connect('localhost', 'root', '') or die('Connect error: ' . mysql_error());
mysql_select_db('sickel') or die('Database selection error: ' . mysql_error()); 

Without the mysql_select_db() function, you would get the error you received.

[sickel2]

Can you tell me how to include "mysql_select_db('sickel') or die('Database selection error: ' . mysql_error()); " ?

Would it be a variable like $conn = mysql_connect('localhost', 'root', '') or die('Connect error: ' . mysql_error()); ?

[danebrood]

when you call that function, it just sets the current mysql connection as using that database; no other steps are required.

If you use my two examples together, using your own connection variables (i.e. your own host, username, password, and database), it should all work out as is.


Note that my examples use the mysql functions, rather than the mysqli.

[sickel2]

Still getting the "Database selection error: Access denied". Here's my code:

function get_fields($valid_user)
{
$conn = mysqli_connect('localhost', 'myusername', 'mypassword', 'mydatabase');
mysql_select_db('mymovief_RobSickel') or die('Database selection error: ' . mysql_error());
$sql= "describe mydatabase.current";
$result = $conn->query($sql);
$field_array = array();
while($row = mysql_fetch_array($results))
{
$field_array[] = $row['Field'];
}
return $field_array;
var_dump($field_array);
}

Any ideas?

[catweasel]

Any ideas?

try this way -

$mysqli = new mysqli("localhost", "myusername", "mypassword", "mydatabase");

if (mysqli_connect_errno()) {
    printf("Connect failed.\n");
    exit();
}

function get_fields($user) {
    global $mysqli;
    $query = "SELECT * from current";
    if ($result = $mysqli->query($query)) {
        $finfo = $result->fetch_fields();
        foreach ($finfo as $val) {
            printf("Name:     %s\n", $val->name);
            printf("Table:    %s\n", $val->table);
            printf("max. Len: %d\n", $val->max_length);
            printf("Flags:    %d\n", $val->flags);
            printf("Type:     %d\n\n", $val->type);
        }    
    $result->close();
    }
    return $finfo;
} 

[sickel2]

Thanks for the code but I'ver got the "Call to a member function query() on a non-object" error.

[danebrood]

your mixing mysqli (note the i) and mysql. And referring to two databases ("mydatabase" and "mymovief_RobSickel"). My example is strictly using mysql (no i on the end). Using only the mysql functions:

mysql_connect(HOST, USERNAME, PASSWORD) or die('Connect error: ' . mysql_error());
mysql_select_db(DATABASENAME) or die('Database selection error: ' . mysql_error()); 

$sql= "describe TABLENAME";
$result = mysql_query($sql);
$field_array = array();
while($row = mysql_fetch_array($results))
{
  $field_array[] = $row['Field'];
}
return $field_array;
var_dump($field_array); 

Change HOST, USERNAME, PASSWORD, DATABASENAME, and TABLENAME to the correct strings.

Using mysqli is probably similiar using the object oriented syntax, but you don't want to mix it with the mysql functions. They work differently:

MySQL
mysqli