Mysql database to html page

Question

I have a database from which I would like to display two columns in a table.

I have the following code however, this does not work and I get

"; } mysqli_close($con); as my answer.[solved after removing echo "<br />"]

Code follows:

<html>
<style>
div.transbox
  {
  width:850px;
  height:500px;
  margin:30px 50px;
  background-color:#ffffff;
  border:1px solid black;
  opacity:0.9;
  padding: 20px;
  filter:alpha(opacity=60); /* For IE8 and earlier */
  }



  ul#navigation
  {
  link-style: none;
   padding: 5px 20px 5px 20px
  margin: 0;
  }
  ul#navigation li
  {
  display: inline;
  }
  h1#heading
  {
    color:#ffffff;
        background-color:#000000;
        text-decoration:none;
font-family:georgia, times, serif;
font-size:2em;
padding:10px;
border-bottom:3px solid #ff6600;
  }
  ul#navigation a {
        color:#ffffff;
        background-color:#000000;
        text-decoration:none;
font-family:georgia, times, serif;
font-size:1.126em;
padding:10px;
border-bottom:3px solid #ff6600;}

  div.line
  {
  display: inline;
  }

 ul#navigation a:hover {
        color:#000000;
        background-color:#ff6600;
        padding:10px;
        border-bottom:3px solid #000000;}
  </style>

<head><h1 id="heading">Food Review </h1>
<ul id="navigation">
<li><a href="form4.html">Review</a></li>
<li><a href="Ratings.html">Ratings</a><li>
</ul>
</head>

<body background = img.jpg>
<div class="transbox">
<div class="work">
<?php
define('DB_NAME','form');
define('DB_USER','root');
define('DB_PASSWORD','toor');
define('DB_HOST','localhost');
$link = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD);
if(!$link)
{
    die('could not connect : ' . mysql_error());
}
$db_selected = mysql_select_db(DB_NAME,$link);
if(!$db_selected)
{
    die('Can\'t use ' .DB_NAME . ': ' .mysql_error());
}
$result = mysqli_query($link,"SELECT fname,Ratings from demo1");

while($row = mysqli_fetch_array($result))
{
echo $row['fname']." ".$row['Ratings'];
}
mysqli_close($link);
?>
</div>
</div>
</body>
</html>
}

I did the following changes but still no results are displayed.

The html you have within the <head> tag is invalid. That code belongs in the body. — ChrisLively
– ChrisLively, Commented Oct 8, 2013 at 22:54
Wow that is a seriously malformed HTML document. Why do you have actual display content within the <head> element? — Mike Brant
– Mike Brant, Commented Oct 8, 2013 at 22:57
Also mysqli_close($con); seems faulty, i don't see $con being defined anywhere, shouldn't you use $link ? — schwarz
– schwarz, Commented Oct 8, 2013 at 22:59
I would recommend you to use a framework to better organize your work and your learning guide them. I use CodeIgniter for example. — alditis
– alditis, Commented Oct 8, 2013 at 23:04
not sure this help you but you selected database via mysql_select but other database commands are mysqli — Pooya Estakhri
– Pooya Estakhri, Commented Oct 8, 2013 at 23:15

user2092317 · Accepted Answer · 2013-10-08 23:30:16Z

2

$link = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD);

Since you initialized the connection with $link you should close $link

 mysql_close($link);

and change the following line

echo "<br />";

with

echo "<br>";

and you are mixing mysql with mysqli please check and use only one which suits you

edited Oct 8, 2013 at 23:30

answered Oct 8, 2013 at 22:55

user2092317

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6 Comments

user2626335 Over a year ago

Ok i did this and still i get no result

user2092317 Over a year ago

try to remove the line echo "<br />"; and check

user2626335 Over a year ago

i removed it now nothing is displayed

user2092317 Over a year ago

$num_rows = mysqli_num_rows($result); echo "$num_rows Rows\n"; //try to check if records exists

user2626335 Over a year ago

i do not get a reply but i stored the values in the table

|

kitingChris · Accepted Answer · 2013-10-08 23:20:18Z

You are mixing mysql functions and mysqli. Have a look here: http://php.net/manual/en/mysqli.query.php how to use mysqli properly

<?php

define('DB_NAME','form');
define('DB_USER','root');
define('DB_PASSWORD','toor');
define('DB_HOST','localhost');

$mysqli = new mysqli(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);

/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

$result = $mysqli->query("SELECT fname,Ratings from demo1");

while($row = $result->fetch_array())
{
$rows[] = $row;
echo $row['fname']." ".$row['Ratings']."<br />";
}

$mysqli->close();
?>

There is also a procedural style but I highly recommend the object oriented ;)

EDIT: Also a good idea is to enable error reporting for development. Add this to the beginning of your php file:

<?php
ini_set('display_errors',1); 
error_reporting(E_ALL);
?>

geomagas · Accepted Answer · 2013-10-08 23:21:50Z

1

I'm pretty sure some data from the demo1 table contains characters that mess with html and breaks the tag nesting.

Try replacing this:

echo $row['fname']." ".$row['Ratings'];

with

echo htmlentities($row['fname']." ".$row['Ratings']);

EDIT: Also, replace mysql_connect with mysqli_connect and mysql_select_db with mysqli_select_db

edited Oct 8, 2013 at 23:21

answered Oct 8, 2013 at 23:15

geomagas

3,2801 gold badge19 silver badges28 bronze badges

2 Comments

user2626335 Over a year ago

i still didnt get anything

user2626335 Over a year ago

It did work but thanks a lot for the effort sir i really appreciate it

Marc B · Accepted Answer · 2013-10-08 22:55:30Z

0

Your PHP install is broken and not functioning, so your PHP code is being sent to the browser, which interprets the <?php tag as a broken/unknown HTML tag, and the entire contents of the page are ignored, until you reach

echo "<br />";

where the /> serves as the closing bracket for this "malformed" <?php tag.

That's why you only see the mysqli_close business in your browser, because everything else is hidden as a broken tag.

answered Oct 8, 2013 at 22:55

Marc B

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