此错误的解决方案是什么:
警告:mysqli_free_result()期望参数1为mysqli_result,
中给出的null
$result = HSDB::getInstance()->selectlast_classification();
#while ($row = mysqli_fetch_array($result)); {
while ($row_cnt = $result->num); {
$fkclassification = $row["pkid"];
}
mysqli_free_result($result);
答案 0 :(得分:1)
错误解释。您必须为函数mysqli_free_result()
答案 1 :(得分:0)
<强>结果强>
仅限过程样式:mysqli_query(),mysqli_store_result()或mysqli_use_result()返回的结果集标识符。
当mysqli_query的返回值因错误而为假时,可能会发生此错误。您可以通过检查所有mysqli_*()
次调用的返回值来避免这种情况。例如
$result = mysqli_query($mysqli, 'select * from mytable) or die(mysqli_error($mysqli));
或
$result = mysqli_query($mysqli, 'select * from mytable);
if ($result === false) {
// error handling
}
答案 2 :(得分:-2)
如果你有
$myConnection = mysqli_connect($db_host, $db_username, $db_pass, $db_name)
or die ("could not connect to mysql");
$sql = "SELECT blabla from blalala where bla=1";
$query = mysqli_query($myConnection,$sql) or die (mysqli_error($myConnection));
while ($row=mysqli_fetch_array($query)){
//do something
}
比你能做的更多:
mysqli_free_result($query);