错误：调用未定义的函数mysqli_result()

Question

有人能告诉我，当我尝试将旧的sql切换到SQLI时，为什么这不起作用吗：

$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");

致：

$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");`

它一直给我一个错误：“致命错误：调用未定义的函数mysqli_result()”

Answer 1

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

您可以创建此函数。

Answer 2

不要使用这种代码。效率很低。改用mysqli_fetch_assoc():

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

一个单一的数据库操作，而不是你尝试做的3+。