我正在尝试制作一个html表单,填充后将插入到我的sql数据库中。 但是我得到这个错误,没有选择数据库[重复]

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I get the error

no database selected

I am using wamp. I am very new to php and sql. here is the php html. the inputs comes from a html file first when the submit button is pressed.

<?php

$link = mysqli_connect("localhost", "erthiph", "");

if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
    $db_select = mysql_select_db( 'inewsfeed' ,$connection);
    if (!$db_select) {
        die("Database selection failed:: " . mysql_error());
        }
}

$Qid = mysqli_real_escape_string($link, $_POST['Qid']);
$Mclass = mysqli_real_escape_string($link, $_POST['Mclass']);
$Sclass = mysqli_real_escape_string($link, $_POST['Sclass']);
$Question = mysqli_real_escape_string($link, $_POST['Question']);
$Answer = mysqli_real_escape_string($link, $_POST['Answer']);
$Doc = mysqli_real_escape_string($link, $_POST['Doc']);
$Time = mysqli_real_escape_string($link, $_POST['Time']);

$sql = "INSERT INTO feed (Qid, Mclass, Sclass, Question, Answer, Doc, Time) VALUES ('$Qid', '$Mclass', '$Sclass', '$Question', '$Answer', '$Doc', '$Time')";
if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

mysqli_close($link);
?>
</div>

What @saty said is correct.You can try like this also,

mysqli_connect("localhost","username","password","dbname");

Because you are mixing mysql add mysqli

Instead of this

mysql_select_db( 'inewsfeed' ,$connection);

use

mysqli_select_db($link,"inewsfeed");

Read http://php.net/manual/en/mysqli.select-db.php