douxing1850 2017-05-02 05:44
浏览 23

mysql数据库连接错误

whenever I give sign up data and press sign up button data isn't inserted into the table and it says "failed". I m not getting why is that so. HTML code is as follow.

<form action="connection.php" method="post">
<label for="fname" style="color: white; margin-top: 0%;">Full Name</label>
<input type="text" id="FulName" name="FullName" placeholder="Full Name">
<br>
<label for="email" style="color: white;">Email</label>
<input type="text1" id="Email" name="UserEmail" placeholder="User Email">
<br>
<label for="fname" style="color: white;">User Name</label>
<input type="text2" id="UserName" name="UserName" placeholder="User Name">
<br>
<label for="password" style="color: white;">Password</label>
<input type="Password" id="Pass" name="Pass" placeholder="User Password">
<br>                
<input type="submit" value="Sign Up">
</form> 

php Code is as follow.

define('DB_HOST', 'localhost');
define('DB_NAME', 'nfakonline');
define('DB_USER','root');
define('DB_PASSWORD','');
$con=mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect    MySQL: " . mysqli_error());
$db=mysqli_select_db(DB_NAME,$con) or die("Failed !" . mysqli_error());

Error: Failed

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4条回答 默认 最新

  • dqtl46964 2017-05-02 05:46
    关注

    In mysqli you have to pass connection object as first argument

    $db=mysqli_select_db($con,DB_NAME) or die("Failed !" . mysqli_error($con));
    
    评论

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