当我尝试将旧的sql切换到sqli时,有人可以告诉我为什么这不起作用:
$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");
到:
$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,$i,"description");`
它一直给我一个错误:“致命错误:调用未定义的函数mysqli_result()”
答案 0 :(得分:16)
不要使用这种代码。这是非常低效的。请改用mysqli_fetch_assoc()
:
while($row = mysqli_fetch_assoc($result)) {
$id = $row['ID'];
$name = $row['name'];
etc..
}
一个SINGLE数据库操作,而不是你想要做的3+。
答案 1 :(得分:0)
function mysqli_result($res, $row, $field=0) {
$res->data_seek($row);
$datarow = $res->fetch_array();
return $datarow[$field];
}
您可以创建此功能