CODEWITHSUNDEEP
phpmysql
michalis
michalis

Reputation: 19

PHP Can't enter data into database table

I am trying to enter the data that I get from the two variables stuname and book in the table's username and book columns !! I only want to enter data into those two columns since the id column is auto increment and the date is auto updated with time stamp!!! Each time I run my code I enter my data into the two text fields and when I press submit I get this message!!

Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\assignment.php on line 35

Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\assignment.php on line 36

Here is my Code:

<?php

$servername = "localhost";
$Username = "root";
$Password = "admin";
$Dbname = "nfc";
$conn = mysqli_connect($servername, $Username, $Password, $Dbname);
if (mysqli_connect_errno())
{
 echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
echo "Connected successfully";
if(isset($_POST["stuname"])&&($_POST["book"]))
{
$stuname =  $_POST["stuname"];
$book =$_POST["bookname"]; 
$sql = "INSERT INTO library (id, username, book, date)
VALUES ('', '$stuname', '$book','')";

       mysqli_select_db($conn, 'nfc') or die(mysqli_error($con));
       $retval = mysqli_query( $sql, $conn );
       if(! $retval )
        {
            die('Could not enter data: ' . mysql_error());
        }


        else 
        {
            echo "Success";
        }

        echo " to stuname ". $stuname;
        echo " to book ". $book;
}
?>
<form id="form1" name="form1" method="post" action="#">
  <p>
    <label for="1">student name</label>
    <input type="text" name="stuname" id="1"  />
  </p>
  <p>
    <label for="12">book name</label>
    <input type="text" name="bookname" id="12" />
  </p>
  <input name="submit" type="submit" value="Submit" />

</form>

Upvotes: 0

Views: 990

Answers (2)

Devon Bessemer
Devon Bessemer

Reputation: 35337

The first problem was solved by @Ghost in the comments.

Now on to the rest of the problems:

1. Your database design is faulty

This should have failed immediately because you are inserting an empty value for id. id should be a primary key and therefore should be unique. An auto-increment doesn't work if you insert an empty value.

2. Your insert statement is faulty

You should exclude an auto-increment column in the INSERT statement and should not use an empty value for date. If date is a timestamp, you should either use NULL if the time is supposed to be empty or use NOW() to use the current timestamp.

3. You shouldn't be using insert on this page according to your comments.

You should be using UPDATE or REPLACE instead of INSERT if you are trying to update the existing row but you should be using the primary key to signify which row you are replacing. Right now, it looks like you don't have a primary key, so refer to my 1st point.

4. Security concerns: Your query is subject to SQL injections.

You use user input ($_POST) directly in a query. Any malicious user can take advantage of this and extract, delete, or manipulate data in your database. You should be using prepared statements, or at the very least escape functions.

Upvotes: 1

sten
sten

Reputation: 61

In the mysqli_query you should put the conn first and then the query itself

$retval = mysqli_query( $conn, $sql );

Upvotes: 1

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