我想知道我的查询是否成功(好吧,我成功地执行了插入查询)但是在第25行得到了这个错误:“mysqli_num_rows()期望参数1是mysqli_result,布尔给出”我已经检查每一行,我甚至使用mysqli_affected_rows并失败。感谢任何帮助,这是我的代码:
<?php
include "connection.php";
$fn = $_POST['first_name'];
$ln = $_POST['last_name'];
$uname = $_POST['username'];
$pwd = $_POST['password'];
$stat = $_POST['status'];
if($fn=="" || $ln=="" || $uname=="" || $pwd=="" || $stat==""){
echo "ALL FIELD MUST BE FILLED!";
}
else {
$sql = "SELECT * from user WHERE username='$uname'";
$result = mysqli_query($conn, $sql);
$number = mysqli_num_rows($result);
if ($number != 0) {
echo "Username $uname is not available";
}
else {
$options = ['cost' => 12];
$pwd = password_hash($pwd, PASSWORD_BCRYPT, $options);
$query = "INSERT INTO user
(username,password,first_name,last_name,status) VALUES
('$uname','$pwd','$fn','$ln','$stat')";
$res = mysqli_query($conn, $query);
$nmb = mysqli_num_rows($res);
echo $nmb;
}
}
?>
答案 0 :(得分:1)
$nmb = mysqli_num_rows($res);
无效,因为$res
是INSERT查询的结果。 The documentation for mysqli_query
说:
失败时返回FALSE。对于成功的SELECT,SHOW,DESCRIBE或EXPLAIN查询,mysqli_query()将返回一个mysqli_result对象。对于其他成功的查询,mysqli_query()将返回TRUE。
因此,$res
始终为false
或true
,您将无法使用mysqli_num_rows()
或mysqli_affected_rows()
。问题是,无论如何,计算行数确实没有意义。您正在尝试插入一条记录,因此它要么是一条或零,所以真或假的结果应该足以让您检测查询是否成功。
答案 1 :(得分:-1)
改变这个:
$res = mysqli_query($conn, $query);
$nmb = mysqli_num_rows($res);
echo $nmb;
为此:
$res = mysqli_query($conn, $query);
if ($res) {
echo "Insert ok";
}
else
{
echo "Problem to insert";
}