[RESOLVED] Help with mysqli_fetch_field problem

[php_steve]

I've spent ages trying to figure this out but no joy, please help...

my test code reads
$mysqli = mysqli_connect("localhost","user","password","database");

$sql = "SELECT url FROM useful_urls WHERE urlName = 'redirect'";

if($res = mysqli_query($mysqli,$sql)){

while ($query_result = mysqli_fetch_field($res)) {
printf ("URL: %s\n", $query_result->url);                     //line 21

	}
mysqli_free_result($res);
}

I get the following error message (and the url in question is not printed to the browser)

[Fri Mar 11 01:28:30 2011] [error] [client 127.0.0.1] PHP Notice: Undefined property: stdClass::$url in /var/www/delete.php on line 21, referer: http://localhost/index.php

If I change the code to use mysqli_fetch_array - it works perfectly!!

i.e. while ($query_result = mysqli_fetch_array($res, MYSQLI_ASSOC)) {
printf ("URL: %s\n", $query_result['url']);

thanks.

[bradgrafelman]

Did you read the manual page for [man]mysqli_fetch_field/man? It doesn't do what you think it does.

[php_steve]

thanks for reply,

sorry, I did read it but obviously I didn't understand it! (I only started writing PHP this week)

I've now got the code working fine using the three incarnations of mysqli_fetch_array (ASSOC,NUM,BOTH) and mysqli_fetch_row and mysqli_fetch_obj but is there a more simple why of retrieving a single field value that doesn't require the WHILE loop?

thanks

[johanafm]

manual page should be [man]mysqli-result.fetch-field[/man]

That depends on what you mean by "requires a while loop".

$row = $result->fetch_assoc();

# even though you could iterate over all columns
foreach ($row as $field => $value)
{
	if ($field == 'someField')
	{
		echo $value
	}
}
# You do not have to, since you can access each specific field directly
echo $row['someField'];

If you are aware of this and rather talk about specifying a certain row (and column), rather than just a certain column in any row, then you should probably use a sql query that retrieves one row, for example

SELECT some, fields FROM table WHERE pirmary_key = 23

If that is not an option, but you do know in which row(s) your desired data is found, you could also use either data_seek or fetch_all

$rows = $result->fetch_all(MYSQLI_ASSOC);

# Row 16 in the array built from the result set (row 1 has index 0)
echo $rows[15]['someField'];

# row 16 in the result set
$result->data_seek(15);
$row16 = $result->fetch_assoc();

[php_steve]

thanks johanafm, that's answered my question.

cheers

[halles]

php_steve;10976154 wrote:

I've spent ages trying to figure this out but no joy, please help...

my test code reads
$mysqli = mysqli_connect("localhost","user","password","database");

$sql = "SELECT url FROM useful_urls WHERE urlName = 'redirect'";

if($res = mysqli_query($mysqli,$sql)){

while ($query_result = mysqli_fetch_field($res)) {
printf ("URL: %s\n", $query_result->url);                     //line 21

	}
mysqli_free_result($res);
}

I get the following error message (and the url in question is not printed to the browser)

[Fri Mar 11 01:28:30 2011] [error] [client 127.0.0.1] PHP Notice: Undefined property: stdClass::$url in /var/www/delete.php on line 21, referer: http://localhost/index.php

If I change the code to use mysqli_fetch_array - it works perfectly!!

i.e. while ($query_result = mysqli_fetch_array($res, MYSQLI_ASSOC)) {
printf ("URL: %s\n", $query_result['url']);

thanks.

HI.
it is an syntax error in line: while ($query_result = mysqli_fetch_field/COLOR)
the right syntax is :

while ($query_result = mysqli_fetch_fields/B)
field with an 's' at the end..