CODEWITHSUNDEEP
phphtmlmysqlsql
user223062
user223062

Reputation: 65

Display a specific value from database in a html

Im pretty new on making webpages. But i´m doing a homepage with forms to Insert to my database. Thats no problem, my problem is that I want to show a specific column from the last row. And the code that I've got so far is this:

<html>
<body>

<form action="insert.php" method="post">
 Publiceringsdag (OBS! En dag tidigare an foregaende):<br>
 <?php
 $con=mysqli_connect("localhost","rss","Habb0","kalender");
 if (mysqli_connect_errno())
 {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
 }
 $lastPub = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1")
 or die(mysql_error());
 echo $lastPub
 ?>
 <br>
 <input type="text" name="pub"><br>
 <input type="submit">
 </form>

 </body>
 </html>

Upvotes: 1

Views: 205

Answers (3)

zavg
zavg

Reputation: 11061

Actually, it is not a very good idea to use the deprecated mysql_ functions. Look at PDO or Mysqli instead.

Meanwhile, in your current implementation you just need to fetch your data after the query execution:

$con = mysql_connect("localhost", "rss", "Habb0", "kalender");

if (mysql_connect_errno())    
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

$lastPub = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1")
  or die(mysql_error());

if($row = mysql_fetch_assoc($lastPub)))
    $result = $lastPub['pub'];

Now the result should be in your $result variable.

EDIT: I just noticed that in your code you use mysqli_connect, mysqli_connect_errno and mysql_query, mysql_error at the same time. But they belongs to different PHP extensions.

Upvotes: 3

sulmanpucit
sulmanpucit

Reputation: 456

Try this.

    <html>
<body>

<form action="insert.php" method="post">
 Publiceringsdag (OBS! En dag tidigare an foregaende):<br>
 <?php
 $con=mysql_connect("localhost","rss","Habb0") or die("Failed to connect to MySQL: " . mysql_error());
 $db=mysql_select_db("kalender",$con) or die("Failed to connect to MySQL: " . mysql_error());
 $result = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1");
 $data = mysql_fetch_array($result);

 echo $data['pub'];
 ?>
 <br>
 <input type="text" name="pub"><br>
 <input type="submit">
 </form>

 </body>
 </html>

Upvotes: 0

Igoooor
Igoooor

Reputation: 407

You must fetch the result first:

$lastPub = mysql_query("SELECT DISTINCT pub FROM event ORDER BY `id` DESC LIMIT 1")
 or die(mysql_error());
$result = mysql_fetch_array($lastPub);
echo $result['pub'];

Upvotes: 0

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