我正在使用以下代碼。但是我收到了這個警告信息。我不知道爲什麼會發生這種情況。警告mysqli_query()至少需要2個參數
mysqli_query() 期望至少2個參數....
配置:
class org_type
{
private $conn = '';
function __construct()
{
global $con;
$this->conn = $con;
}
public function create(){
$this->create_org_type();
}
$sql = "CREATE TABLE IF NOT EXISTS `org_type`
(
`orgTypeId` int NOT NULL AUTO_INCREMENT,
PRIMARY KEY(orgTypeId),
`orgTypeName` text,
`status` varchar(150),
`create_date` INT(11),
`desp` TEXT DEFAULT ''
)";
mysqli_query($this->conn,$sql);
}*/
public function insert_orgType($orgName,$sts,$editVal){
$now = time();
mysqli_query($this->conn, "INSERT INTO `org_type`(orgTypeName,status,create_date,desp) VALUES ('".$orgName."','".$sts."',".$now.",'".$editVal."')") or die (mysqli_error($this->conn));
$insId = mysqli_insert_id($this->conn);
return $insId;
}
public function select_orgType($id=0){
$str = '';
$arrResult = array();
if($id != 0){
$str .= " AND orgTypeId = $id ";
}
$sql = "SELECT * FROM org_type WHERE 1 $str order by orgTypeId ASC";
$result = $this->conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$arrResult[] = $row;
}
return $arrResult;
}
}
}
你還沒有使用的連接'$這個 - > conn'在'insert_orgType' –
使用。如果我使用$ insId查詢= mysqli_insert_id($ this-> conn),得到相同的錯誤 – Ived
'mysqli_query($ this-> conn,QUERY)' –