我正在尝试将我当前的代码转换为 MySQLi 扩展......但我对这段代码有疑问......
$sql = "SELECT COUNT(*) FROM users WHERE username='$username'";
$result = mysql_query($sql);
if (mysql_result($result, 0) > 0) {
$errors[] = 'The username is already taken.';
}
“mysql_result”的 mysqli 函数是什么?我无法让它工作。
我现在的MySqLi代码就是连接和
$sql = "SELECT COUNT(*) FROM users WHERE username='$username'";
$result = $mysqli->query($sql);
if (mysql_result($result, 0) > 0) {
$errors[] = 'The username is already taken.';
}
但我需要更改mysql_result,但在MySQLi 中找不到相反的函数..
最佳答案
我通常使用 MySQLi 的 object-y 接口(interface):
<?php
$db = new mysqli($hostname, $username, $password, $database);
$result = $db->query('SQL HERE');
// OOP style
$row = $result->fetch_assoc();
$row = $result->fetch_array();
while($row = $result->fetch_row())
{
// do something with $row
}
// procedural style, if you prefer
$row = mysqli_fetch_assoc($result);
$row = mysqli_fetch_array($result);
while($row = mysqli_fetch_row($result))
{
// do something with $row
}
?>
完整的结果方法列表在这里:http://www.php.net/manual/en/class.mysqli-result.php
关于php - Mysql 到 mysqli php,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7551469/