Php 在成功的查询中获取fetch_object()错误
我得到了这个错误:Php 在成功的查询中获取fetch_object()错误,php,mysql,Php,Mysql,我得到了这个错误: Fatal error: Call to a member function fetch_object() on a non-object in C:\.....php on line 136 代码如下: if ($result = $mysqli->query("INSERT INTO partlistlist ( P_fam , P_code , P_name, P_var , P_ver , P_lnk , P_fol , P_Notes , P_status
Fatal error: Call to a member function fetch_object() on a non-object in C:\.....php on line 136
代码如下:
if ($result = $mysqli->query("INSERT INTO partlistlist ( P_fam , P_code , P_name, P_var , P_ver , P_lnk , P_fol , P_Notes , P_status , P_op ) VALUES ( \"".$mod_fam ."\", \"".$mod_code ."\", \"".$mod_name ."\", \"".$mod_var ."\", 1 , \"".$mod_lnk ."\", \"".$mod_fold ."\", \"".$mod_note ."\", \"".$mod_stat ."\", \"". $_SESSION['wh_pwd_usr']."\" ) "));
{
echo "<br>Articolo creato con successo";
$created_id = $mysqli->insert_id;
if (!$result = $mysqli->query("SELECT * FROM partlistlist WHERE P_id = $created_id LIMIT 1"))
echo "Error";
$row = $result->fetch_object();
$P_id = $created_id;
if (!$result_clonepart = $mysqli->query("SELECT * FROM partlist WHERE K_pid = $old_P_id"))
{
echo "Error";
}
else
{
if ($result_clonepart->num_rows > 0)
{
while ($row_clonepart = $result_clonepart->fetch_object())
{
if (!$result_clonepart = $mysqli->query("INSERT INTO partlist ( K_ref ,K_pid ,K_iid ,K_qty ,K_subpid, K_stat ) VALUES ( \"".$row_clonepart->K_ref ."\" ,\"". $P_id ."\" ,\"".$row_clonepart->K_iid ."\" ,\"".$row_clonepart->K_qty ."\" ,\"".$row_clonepart->K_subpid ."\" ,\"".$row_clonepart->K_stat ."\")"))
echo "Error";
}
}
}
}
如果num\u rows检查良好,$result\u clonepart怎么可能不是对象?在while循环中,您正在重新分配
$result\u clonepart
(您可能不想做的事情)。如果该查询失败,脚本将停止并出现致命错误
使用不同的变量名(如果您不想使用相同的变量名)或在失败时中断循环(如果您有意重新分配)
如果在执行后打印
$result\u clonepart
,您会得到什么<代码>模具(变量转储($result\u clonepart))代码>对象(mysqli_结果)#3(5){[“当前_字段”]=>int(0)[“字段计数”]=>int(8)[“长度”]=>NULL[“num_行”]=>int(1)[“类型”]=>int(0)}这看起来非常不安全。您确定您的用户参数正确吗?使用mysqli
时,您应该使用参数化查询,并将用户数据添加到查询中。不要使用字符串连接来实现这一点,因为您将创建严重的。实际上,这永远不会出现在internet上,它只是一个本地管理网站-顺便说一句,它们被转义并经过非常仔细的检查(据我所知(这并不多))@SimoneApprendista,只要您使用的是mysqli_real_escape_string
,trim
并根据正则表达式验证它们。在不使用参数化查询的情况下,应对sql查询提供良好的保护。
while ($row_clonepart = $result_clonepart->fetch_object())
while ($row_clonepart = $result_clonepart->fetch_object()) {
// VVVVVVVVVVV Here is the reassignment
if (!$result_clonepart = $mysqli->query("INSERT INTO partlist ( K_ref ,K_pid ,K_iid ,K_qty ,K_subpid, K_stat ) VALUES ( \"".$row_clonepart->K_ref ."\" ,\"". $P_id ."\" ,\"".$row_clonepart->K_iid ."\" ,\"".$row_clonepart->K_qty ."\" ,\"".$row_clonepart->K_subpid ."\" ,\"".$row_clonepart->K_stat ."\")"))
echo "Error";
//You could insert a break statement if you are reassigning the variable intentionally
}