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這是什么錯誤?

這是什么錯誤?

錯過了年華 2017-06-21 19:20:00
class?MysqlConn { public?$dbtype; public?$host; public?$dbname; public?$user; ????public?$pwd; function?__construct($dbtype,$dbname,$host,$user,$pwd) { $this->dbtype=$dbtype; $this->host=$host; $this->dbname=$dbname; $this->user=$user; $this->pwd=$pwd; if($this->dbtype=="mysql"?||?$this->dbtype=="mssql") { ???? ????$dsn="$this->dbtype:host=$this->host;dbname=$this->dbname"; ????} else { ????$dsn="$this->dbtype:dbname=$this->dbname"; ????} try? { $conn?=?new?PDO($dsn,?$this->user,?$this->pwd);? //初始化一個PDO對象,就是創建了數據庫連接對象$pdo $conn->query("set?names?utf8"); return?$conn; ????}? catch(PDOException?$e) { ???? ????die("Error!:?"?.?$e->getMessage()?.?"<br/>"); ????}???? }? } $conn=new?MysqlConn("mysql","dzxx_db","localhost","admin","123");//數據庫連接類實例化,執行連接操作,返回連接標識
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3 回答

?
天使之翼死神之鐮

TA貢獻81條經驗 獲得超102個贊

$dsn的值有問題,正確的應該分別是?$dsn=$this->dbtype.":host=".$this->host.";dbname=".$this->dbname;和$dsn=$this->dbtype.":dbname=".$this->dbname;

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1 反對 回復 2017-06-22
點擊展開后面1
?
錯過了年華

TA貢獻56條經驗 獲得超22個贊

class?ConnDB{
	
	var?$dbtype;
	var?$host;
????var?$user;
????var?$pwd;
????var?$dbname;
????
	//構造方法
????function?ConnDB($dbtype,$host,$user,$pwd,$dbname){
		$this->dbtype=$dbtype;
?????????	$this->host=$host;
		$this->user=$user;
		$this->pwd=$pwd;
		$this->dbname=$dbname;
	}

????//實現數據庫的連接并返回連接對象
????function?GetConnId(){
?????	
????	if($this->dbtype=="mysql"?||?$this->dbtype=="mssql"){
????		$dsn="$this->dbtype:host=$this->host;dbname=$this->dbname";
		}else{
			$dsn="$this->dbtype:dbname=$this->dbname";
		}????
		try?{
????		$conn?=?new?PDO($dsn,?$this->user,?$this->pwd);?	//初始化一個PDO對象,就是創建了數據庫連接對象$pdo
			$conn->query("set?names?utf8");
????		return?$conn;
		}?catch?(PDOException?$e)?{
????		die?("Error!:?"?.?$e->getMessage()?.?"<br/>");
		}

?????	
????}	
}

上面這個是原本的類,可沒見需要像你那樣寫啊



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反對 回復 2017-06-23
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