本文介绍了请给我这个错误的解决方案的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
<?php
class database {
public $sql = '';
public $query = '';
public $con;
function __construct(){
self::db();
}
static function db()
{
if($_SERVER['HTTP_HOST'] == 'localhost'){
$host = 'localhost';
$username = 'root';
$pass = '';
$db = 'alltoitskin';
}else{
$host = 'alltoitskin.db.7649825.hostedresource.com';
$username = 'alltoitskin';
$pass = 'Skinlogica123';
$db ='alltoitskin';
}
$con=mysqli_connect($host, $username ,$pass)or die("mysql not connected");
mysqli_select_db($con,$db) or die('Invalid database selected');
}
function result_array($query){
$this->query = $query;
self::db();
$result = array();
$Q=mysqli_query($con,$query);
if($Q && mysqli_num_rows($Q) > 0){
while($tmp = mysqli_fetch_assoc($Q)){
$result[] = $tmp;
}
}
return $result;
}
function result_row($query){
$this->query = $query;
self::db();
$result = array();
//$Q=mysql_query($query);
$Q=mysqli_query($con,$query);
if($Q && mysqli_num_rows($Q) > 0){
$tmp = mysqli_fetch_assoc($Q);
return $tmp;
}
return $result;
}
function insert($data = array(), $table = ''){
$t = sizeof($data);
if(empty($data) || $table == ''){
return false;
}
$this->sql = 'INSERT INTO '.$table.' SET ' ;
$i = 1;
foreach($data as $k => $v){
$this->sql .= $k." = '".$v."' ";
if($i < $t ){
$this->sql .= ' , ';
}
$i++;
}
//echo $this->sql;
$Q = mysql_query($this->sql);
if($Q){
return mysql_insert_id();
}
return false;
}
function num_rows($query){
$this->query = $query;
self::db();
$result = array();
$Q=mysql_query($query);
if($Q){
return mysql_num_rows($Q);
}
return 0;
}
function update($query){
$this->query = $query;
$Q = mysql_query($query);
if($Q && mysql_affected_rows() > 0 ){
return true;
}
return false;
}
function last_query(){
return $this->query;
}
}
?>
我尝试过:
What I have tried:
Notice: Undefined variable: con in C:\xampp\htdocs\grip-engineering\classes\database.php on line 62
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\grip-engineering\classes\database.php on line 62
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