Generated form, how to get POST content and insert it in mysql

[Peuplarchie]

Good day all,

            I'm working on a thing which lets the member create, edit, delet it's own table and data.

 

I have a form in a table which list the table data, this form should add a row.

Here is it's code :

 

$tid = $_GET['id'];
$tname = $_GET['tname'];

$con = mysql_connect($server, $login, $password);
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db($db, $con);

$query = "SELECT * FROM $tname";
$resultv = mysql_query($query); 
if (($resultv)||(mysql_errno == 0))
{
  echo "<table width='100%' border=\"1\">\n";
  echo "<tr>";
  echo "<td align='center'><b>Options</b></td>";  
  if (mysql_num_rows($resultv)>0)
  {
          //loop thru the field names to print the correct headers
          $i = 0;
          while ($i < mysql_num_fields($resultv))
          {
       echo "<th>". mysql_field_name($resultv, $i) . "  <a href=\"tableau.php?act=delcol&opt=tableau&tname=".$tname."&cname=".mysql_field_name($resultv, $i)."\" title=\"DELETE COLUMN : ". mysql_field_name($resultv, $i) . "\"><img src=\"Images/stock_delete-column.png\" align=\"bottom\"></a></th>\n";
       
       $i++;
    }
    echo "</tr>";
   
echo "<tr><form  name=\"formadd\" method=\"post\" action=\"tableau.php?add=row\">\n";
          //loop thru the field names to print the correct headers
          $i = 0;
  echo "<td align='center'><input type=\"submit\" value=\"Ajouter\"></td>";
          while ($i < mysql_num_fields($resultv))
          {
          echo "<td align='center'><input type=\"text\" name=\"". mysql_field_name($resultv, $i) . "\"</td>\n";
          $i++;
          }
echo "</form></tr>\n";  
   
    //display the data
    while ($rowv = mysql_fetch_array($resultv,MYSQL_ASSOC))
    {
      echo "<tr><td><br></td>\n";
      foreach ($rowv as $data)
      {
       if ($data == "") {
        echo "<td align='center'><br></td>\n";
       }else{	
        echo "<td align='center'>". $data . "</td>\n";
       }
      }
    }
  }else{
    echo "<tr><td colspan='" . ($i+1) . "'>No Results found!</td></tr>\n";
  }

echo "</table>";  
}else{
  echo "Error in running query :". mysql_error();
} 

 

 

 

My issue is to receive the data and insert it in the database.

I have know clue how to put the POSTs into variable so I can Insert it in the query..

 

Can somebody help me ?

 

Take care !

[Christian F.]

The trick is to do all of the PHP processing before you output any content to the browser. Doing that it'll be easy to determine whether or not you'll need to (re-)display the form, and by the time you know that you should already have retrieved and validated any user input.

At that point it's as simple as using echo htmlspecialchars (), to escape the output (security measure) and add it to the relevant places in the form.

 

PS: The $_POST superglobal is a variable already, an array to be exact. Read more about it in the PHP manual.

[Peuplarchie]

Getting closer...

 

$sql1= 'SHOW TABLES FROM ' .$db .' LIKE '.$tname;
$res1= mysql_query($sql1) OR die('Query 1 error:<br />' .mysql_error());

echo '<table border=1 cellpadding=1 cellspacing=0 width="90%">';
while ( $row= mysql_fetch_row($res1) )
{
    echo '<tr>';
    echo '<td valign="top" align="center">' .$tname. '</td>';
    echo '<td>';
    $sql2= 'SHOW CREATE TABLE ' .$tname;
    $res2= mysql_query($sql2) OR die('Query 2 error:<br />' .mysql_error());
    while ( $table_def= mysql_fetch_row($res2) )
    {
        for ($i=1; $i<count($table_def); $i++)
        {
            echo '<pre>' .$table_def[$i]. '</pre>';
        }
    }
    echo '</tr>';
}
echo '</table>'; 

[Christian F.]

OK, congratulations..?

(I don't quite see what you're getting at with your last post.)

[Peuplarchie]

hahaha...

 

I'm adding my development so if another have the same issue he can see.

If....

 

This way also it gives more info on my issue, i guess.

 

And it shows that I'm working on it...

 

[Peuplarchie]

But I'm far from what I'm trying to do.

 

[Nina555]

I am having similar issue with this code. I want to add "title". This script does not have write title option and I would like to add it to script, but it already has the add video option. That's why I tried to do something like this and I just can't get the script to insert title in the database, if you could help me with this one I would be very thankful. Here is the srcipt

<?
session_start();
include("../func/db_info".".php");
include("logincheck".".php");
logincheck();

$surl = $_SERVER['HTTP_HOST'] ;

if (strip_tags($_POST['add'])){
$vidid=mysql_real_escape_string(strip_tags($_POST['vidid']));

$url = "$vidid";
parse_str( parse_url( $url, PHP_URL_QUERY ), $my_array_of_vars);
$t1=mysql_real_escape_string(strip_tags($_POST['tit1']));
$title="$t1";
parse_str( $title, $my_vars );
mysql_query("INSERT INTO `videos` (`id`, `videoid`, `views`, `title`) VALUES ('', '".$my_array_of_vars['v']."', '0', '".$my_vars['v']."');") or die (mysql_error());
$message="<img src=\"../img/tick-icon.jpg\">";
}
if (strip_tags($_POST['rem'])){
$vidid2=mysql_real_escape_string(strip_tags($_POST['vidid2']));

$url = "$vidid2";
parse_str( parse_url( $url, PHP_URL_QUERY ), $my_array_of_vars );

mysql_query("DELETE FROM videos WHERE id=".$my_array_of_vars['v']."");

$message2="<img src=\"../img/tick-icon.jpg\">";
}
?><head>
<html xmlns:fb="http://ogp.me/ns/fb#">
<title>Admin Panel</title>
<META name="robots" content="NOINDEX,NOFOLLOW" />
<link type="text/css" rel="stylesheet" href="../css/admin.css" />
<link href="../img/favicon.ico" rel="icon" type="image/x-icon" />
<body>
<a href="index.php"><img src="../img/logo.png" border="0"></a>
<div class="banner"><a href="index.php">Delete</a> - <a href="add.php">Add & Remove</a> - <a href="logout.php">Logout</a></div>
<div align="center"><form id="form1" method="post" action="">
Youtube URL: <input type="text" name="vidid" id="vidid">
Title: <input type="text" name="tit1" id="tit1">
<input name='add' type='submit' id='add' value='Add Video' /> <? echo"$message";?><br><br>

Website URL: <input type="text" name="vidid2" id="vidid2">
<input name='rem' type='submit' id='rem' value='Remove Video' /> <? echo"$message2";?><br><br>
Example: <strong>http://<? echo"$surl"; ?>/watch.php?v=25</strong>
</div>
</form>
</body>

[PFMaBiSmAd]

STOP adding your post onto the end of existing threads. Your previous post was split into it's own thread - http://forums.phpfreaks.com/topic/270197-cant-insert-title-into-database-split-from-existing-thread/

 

No matter how similar your problem is to an existing problem, it's a different problem.

[Nina555]

Sorry