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

推荐答案




这篇关于请给我这个错误的解决方案的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

09-05 16:21
查看更多