No prompted error and data not inserting into child table - PHP, SQL

Question

I have a problem regarding "no prompted error and data cannot insert into child table".

I have 2 tables (users and useradvert); - The users is the PARENT - The useradvert is the CHILD

users (parent table-ID is primary key with auto increment)
- ID
- name
- username
- telno
- password
- date (timestamp)
useradvert (child table-ID is index with NO auto increment)
- ID
- name2
- color2
- hobby2

I have no problems creating a relation table. The 2 tables are now related.

Then I have a login page (login.php) - runs fine no problem..

And I have a user page (useracc-test.php)-> this is a page after a user log in successfully via login.php. They will be able to view their personal data and also enter their name again (any different nick name they like), color and hobby.This page display user's personal data and secondly, there is also a form where users can enter data like I said previously(name2,color and hobby).

I have no problem displaying the user's personal data in the user page (useracc.test.php). This data is retrieved from parent table "user". The problem I'm having is, I cannot insert data into the child table (useradvert). No error prompted.

<?php    
//useracc-test.php    
/**
 * Start the session.
 */
session_start();
ini_set('display_errors', 1);
error_reporting(E_ALL);
// require 'lib/password.php';
require 'connect-test.php';   

$userName= isset($_POST['username']) ? $_POST['username'] : '';    

$query = "SELECT id, name, username, telno FROM users WHERE username = ?";
$stmt = $conn->prepare($query);
$stmt->bind_param('s', $userName);
$stmt->execute();
$res = $stmt->get_result();
 ?>    
<html>
<head>
<style type="text/css">
#apDiv2 {
    position: absolute;
    left: 51px;
    top: 238px;
    width: 237px;
    height: 93px;
    z-index: 1;
}
#apDiv1 {
    position: absolute;
    left: 134px;
    top: 123px;
    width: 234px;
    height: 104px;
    z-index: 2;
}
#apDiv3 {
    position: absolute;
    left: 58px;
    top: 146px;
    width: 219px;
    height: 61px;
    z-index: 2;
}
#apDiv4 {
    position: absolute;
    left: 302px;
    top: 102px;
    width: 365px;
    height: 123px;
    z-index: 3;
}
</style>
<link href="SpryAssets/SpryTabbedPanels.css" rel="stylesheet" type="text/css">
<script src="SpryAssets/SpryTabbedPanels.js" type="text/javascript"></script>
</head>
<body>
Your Personal details:</p>
      <p><?php while($row = $res->fetch_array()): ?>
<p><?php echo $row['id']; ?></p>
<p><?php echo $row['name']; ?></p>
<p><?php echo $row['username']; ?></p>
<p><?php echo $row['telno']; ?>





  <?php     

  // $userid = $_POST['id'];
  $stmt=$conn->prepare("INSERT INTO useradvert (id,name2,color2,hobby2) VALUES (?,?,?,?)");
  $stmt->bind_param("isss", $id, $name2, $color2, $hobby2);
  $stmt->execute();
  if (!$stmt)
  { printf("Errormessage: %s\n", $mysqli->error);}
  else {

  echo "New records created successfully";}

$stmt->close();
$conn->close();

    ?>      

<form name="form2" method="post" action="useracc-test.php">
        <p>INSERT YOUR INTEREST:</p>
        <p>     
        </p>
          ID:
      <input name="id" type="hidden" id="id" value="<?php echo $row['id']; ?>">


  <p>Name :
          <input type="text" name="name2" id="name2">
        </p>
        <p>
          <label for="warna2"></label>
          Color :
          <input type="text" name="color2" id="color2">
        </p>
        <p>
          <label for="hobi2"></label>
          Hobby:
          <input type="text" name="hobby2" id="hobby2">
        </p>
        <p>
          <input type="submit" name="submit" id="submit" value="submit">
       </p>
        <p>&nbsp;</p>
      </form>       

               <?php endwhile; ?>

               </body>
               </html> 
--
-- Table structure for table `useradvert`
--

CREATE TABLE IF NOT EXISTS `useradvert` (
  `id` int(10) unsigned NOT NULL,
  `name2` varchar(60) COLLATE utf8_unicode_ci NOT NULL,
  `color2` varchar(60) COLLATE utf8_unicode_ci NOT NULL,
  `hobby2` varchar(60) COLLATE utf8_unicode_ci NOT NULL,
  KEY `id` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

-- --------------------------------------------------------

--
-- Table structure for table `users`
--

CREATE TABLE IF NOT EXISTS `users` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(60) COLLATE utf8_unicode_ci NOT NULL,
  `telno` varchar(11) COLLATE utf8_unicode_ci NOT NULL,
  `username` varchar(60) COLLATE utf8_unicode_ci NOT NULL,
  `password` varchar(60) COLLATE utf8_unicode_ci NOT NULL,
  `date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`),
  UNIQUE KEY `username` (`username`),
  KEY `id` (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=96 ;

--
-- Dumping data for table `users`
--

INSERT INTO `users` (`id`, `name`, `telno`, `username`, `password`, `date`) VALUES
(95, 'Test Name', '09999999999', 'test@test.com', '$2y$12$fqdmAQk5c8qk8Eh2TWy2n.AdNO.lFjqmi2ruSzk8tsVXcK71OcPae', '2015-12-24 05:00:13');

--
-- Constraints for dumped tables
--

--
-- Constraints for table `useradvert`
--
ALTER TABLE `useradvert`
  ADD CONSTRAINT `useradvert_ibfk_1` FOREIGN KEY (`id`) REFERENCES `users` (`id`);

Answer 1

In your connection you're using mysqli , in your query you're using mysql_

$query = sprintf("select name, username, telno FROM users WHERE username='%s'", mysqli_real_escape_string($conn, $userName));

(Change to mysqli_real_escape_string )

Answer 2

The first problem, as lucap stated, is that you try to insert data in a table with no auto increment ID, without providing that ID. if you look at the mysqli error you should with print($conn->error); you should see an error message about that.

But there is some other errors in your script that make it fail. Let's take this step by step.

$query = sprintf("select name, username, telno FROM users WHERE username='%s'", mysql_real_escape_string($userName));
$result = $conn->query($query);

You should test the password in this query, as right now it consider as logged an user who just send as post data the username, which is a severe security issue. Beside that, there is two error: you don't get the ID of the user in the query, and mysql_real_escape_string is deprecated. Considering you're using mysqli, I suggest to proceed in this way to sanitize input data:

$sql = sprintf("SELECT ID, name, username, telno FROM users WHERE username = '%s' AND password = '%s'", mysqli_real_escape_string($conn, $_POST['username']), mysqli_real_escape_string($conn, $_POST['password']));
$result = $conn->query($sql);
$result = $result->fetch_row();

Of course you should hash the password with the appropriate encryption you're using, because of course you don't store the password raw :)

So this:

<?php while($row=$result->fetch_assoc()): ?>
    <?= $row['name'] ?>
</p>
<p>
    <?= $row['username'] ?>
</p>
<p>
<?= $row['telno'] ?>
<?php endwhile; ?>

Will become:

<p><?php echo $row['name']; ?></p>
<p><?php echo $row['username']; ?></p>
<p><?php echo $row['telno']; ?>

So let's get on what you stumbled upon:

$sql = "INSERT INTO useradvert(name2,color2,hobby2)VALUES('$name2', '$color2','$hobby2')";
if ($conn->query($sql) === TRUE) {
    if (isset($_POST['submit'])){
        $name2=$_POST['name2'];
        $color2=$_POST['color2'];
        $hobby2=$_POST['hobby2'];
        echo "You have succesfylly inserted your data. Insert another data again?";
    } 
    else 
    {
        die ("Error: " . $sql . "<br>" . $conn->error );
    }

    $conn->close();
}

This can't work for the following reasons:

As stated above you don't provide the ID.

$name2 , $color2 and $hobby2 just contain true because of this:

 $name2=isset($_POST['name2']); $color2=isset($_POST['color2']); $hobby2=isset($_POST['hobby2']);

I guess what you wanted to do was in fact:

 $name2 = isset($_POST['name2']) ? $_POST['name2'] : ''; $color2 = isset($_POST['color2']) ? $_POST['color2'] : ''; $hobby2 = isset($_POST['hobby2']) ? $_POST['hobby2'] : '';

And beside those two points, this query isn't secure as it insert raw data in the query. So the final code would be:

$sql = sprintf("INSERT INTO useradvert (ID, name2, color2, hobby2) VALUES (%d, '%s', '%s', '%s')", (int)$result['ID'], mysqli_real_escape_string($conn, $name2), mysqli_real_escape_string($conn, $color2), mysqli_real_escape_string($conn, $hobby2));
if($conn->query($sql) === true) {
    // do your stuff
}

I think that's all, but if you still encounter issues after fixing all this, I suggest to check on $conn->error to see any error from mysqli, and do var_dump of your data to see if everything is ok. Be sure too that you turned on php errors with the following:

ini_set('display_errors', 1);
error_reporting(E_ALL);

Answer 3

我自己解决了这个问题......我忘记了 if(isset($_POST['submit']))..LoL.我做了一些小改动只是为了添加上面的isset..现在一切正常..

Answer 4

you must specify the id in your insert statement. add id in select statement

$query = sprintf("select id,name, username, telno FROM users WHERE username='%s'", mysql_real_escape_string($userName));

and then:

$id=$row['id'];
$sql = "INSERT INTO useradvert(id,name2,color2,hobby2)VALUES('$id','$name2', '$color2','$hobby2')";

No prompted error and data not inserting into child table - PHP, SQL

Question

4 answers

solution1
0 2015-12-22 16:10:59

solution2
0 2015-12-22 16:19:12

solution3
0 2015-12-24 12:27:25

solution4
-1 2015-12-22 15:32:00

No prompted error and data not inserting into child table - PHP, SQL

Question

4 answers

solution1 0 2015-12-22 16:10:59

solution2 0 2015-12-22 16:19:12

solution3 0 2015-12-24 12:27:25

solution4 -1 2015-12-22 15:32:00

solution1
0 2015-12-22 16:10:59

solution2
0 2015-12-22 16:19:12

solution3
0 2015-12-24 12:27:25

solution4
-1 2015-12-22 15:32:00