Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Fatal error: Call to undefined function mysqli_result()

Can someone please tell me why this doesnt work, when I tried to switch my old sql to sqli:

$query = "SELECT * FROM `product_category`";
$result = mysql_query($query, $connect) or die("could not perform query: " . mysql_error());
$num_rows = mysql_num_rows($result);

for ($i=0; $i < $num_rows; $i++)
{
    $ID = mysql_result($result,$i,"ID");
    $name = mysql_result($result,$i,"name");
    $description = mysql_result($result,$i,"description");

to:

$query = ("SELECT * FROM `product_category`");
$result = mysqli_query($connect, $query) or die("could not perform query");
$num_rows = mysqli_num_rows($result);

for ($i=0; $i < $num_rows; $i++)
{
    $ID = mysqli_result($result, "ID");
    $name = mysqli_result($result,$i,"name");
    $description = mysqli_result($result,$i,"description");`

it keeps giving me an error of: "Fatal error: Call to undefined function mysqli_result()"

like image 731
Cesarg2199 Avatar asked Jul 17 '13 18:07

Cesarg2199


2 Answers

Don't use this kind of code. It's highly inefficient. Use mysqli_fetch_assoc() instead:

while($row = mysqli_fetch_assoc($result)) {
   $id = $row['ID'];
   $name = $row['name']; 
   etc..
}

One SINGLE database operation, rather than the 3+ you're trying to do.

like image 98
Marc B Avatar answered Oct 24 '22 07:10

Marc B


if (!function_exists('mysqli_result')) {
  function mysqli_result($res, $row, $field=0) {
    $res->data_seek($row);
    $datarow = $res->fetch_array();
    return $datarow[$field];
  }
}

You can create this function.

like image 6
Roman Panevnyk Avatar answered Oct 24 '22 06:10

Roman Panevnyk