php連接數(shù)據(jù)庫的類，但是選擇數(shù)據(jù)庫的時(shí)候總是出錯(cuò)，怎么回事呢？

構(gòu)造函數(shù)錯(cuò)咯

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function?__construct($host,$user,$pass,$database){

$this?-?host=$host;

$this?-?user=$user;

$this?-?pass=$pass;

$this?-?database=$database;

//?echo?$db;

$conn?=?mysql_connect($host,$user,$pass);

$db?=?mysql_select_db($this?-?database,$conn);

if($db){

echo?"數(shù)據(jù)庫成功";

}else{

echo?"數(shù)據(jù)庫失敗";

}

}

//給你改咯哈這個(gè)類

class?register{

private?$host;????????//The?host?address

private?$user;????????//The?user

private?$pass;????//The?password

private?$database;??????????//The?database

private?$conn;

//Connect?with?the?database

function?__construct($host,$user,$pass,$database){

$this?-?host=$host;

$this?-?user=$user;

$this?-?pass=$pass;

$this?-?database=$database;??????????

$conn?=?mysql_connect($host,$user,$pass)or?die("連接失敗!");

mysql_select_db($this?-?database,$conn)?or?die("選擇數(shù)據(jù)庫失敗!");

$this-conn=$conn;

}

//Add?a?user

function?addUser($data){

$dataObj????=?????$data;//類里邊不要把變量寫得太死

$userInfo???=?????json_decode($dataObj,true);??//change?the?data?from?the?type?of?json?to?array.

$username???=?????$userInfo[0];

$password???=?????$userInfo[1];

$repassword?=?????$userInfo[2];

//The?sql?to?add?the?new?user

@$sql_addUser?=?EOF

insert?into?users?values("","$username","$password","$repassword");

EOF;

//?var_dump($sql_addUser);break;

$result?=?mysql_query($sql_addUser,$this-conn);//指定連接，少些麻煩

//?var_dump($result);break;

if($result){

echo??1;

}else{

echo?0;

}

}

}

php數(shù)據(jù)庫選擇數(shù)據(jù)庫時(shí)出錯(cuò)了

錯(cuò)誤描敘清楚

$sql="select * from xh_article";

$res=mysql_query($sql);

echo $res;

----

echo 肯定不行了 var_dump

echo 只能輸出 字符串 數(shù)字 之類的 不能輸出數(shù)組 對(duì)象

PHP選擇數(shù)據(jù)庫失敗

親！你把127.0.0.1換成 localhost， 然后在看一下行否！如果換不行的話可能是你的Apache沒有開啟！ 你可以到這個(gè)網(wǎng)站上看一下PHP數(shù)據(jù)庫的連接！ 希望能夠幫到你！