Php5/Mysql Save function dosen't work

[Mokerhamer]

Hello,

I'v been working on a website. On the admin page you can edit the text of the site but for some reason it dosen't save when you push save could anyone help me out?

<?php
$connect = mysql_connect ("localhost","root","secret") or die ("Kon geen verbinding met MySQL Server tot stand brengen. MySQL error: " . mysql_error()); 
mysql_select_db("nout",$connect) or die("Kon geen verbinding maken met database. MySQL error: " . mysql_error()); 
$query = "SELECT * FROM main"; 
$result = mysql_query($query) or die("Kon query niet uitvoeren: " . mysql_error()); 
  if(mysql_num_rows($result) == 0) 
  { 
  echo "Niks gevonden neem a.u.b contact op met munier"; 
  } 
  else 
  { 
while($_POST = mysql_fetch_array($result))
  {
 echo "<INPUT type='tekst' name='tekst' value='$_POST[tekst]'></TD>";
  echo "<br />";
  }
 } 
        mysql_select_db('nout', $connect) or die (mysql_error());
 
        if (isset($_POST['submit']))
        {
                mysql_query("UPDATE `main` SET `tekst` =  '" . $_POST['text'] . "' WHERE id = '". (int) $_POST['id']."'") or die ("Kon query niet uitvoeren: " . mysql_error()); 
                echo 'Geupdate';
        }
 
        $result = mysql_query('SELECT * FROM `main`') or die (mysql_error());
 
        if (mysql_num_rows($result) < 1)
                die ('Geen tekst gevonden neem a.u.b contact op met munier voor hulp');
 
        while ($row = mysql_fetch_array($result))
        {
                echo 'TextID ' . $row['id'] . ': ' . $row['tekst'];
        }
?>
 
<form method="POST" action="<? echo $_SERVER['REQUEST_URI'] ?>">
<font color=red>Id Niet veranderen!</font><td>&nbsp;</td><td>&nbsp;</td>Id: <input TYPE='TEXT' NAME='id' VALUE='1'/>
<TEXTAREA ROWS='20' COLS='40' NAME='text'></TEXTAREA>
<INPUT TYPE='submit' NAME='submit' onClick='return checkForm();' />
</FORM>

Database table main

-- Tabel structuur voor tabel `main`
--

CREATE TABLE IF NOT EXISTS `main` (
  `id` int(50) NOT NULL auto_increment,
  `tekst` varchar(5000) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Gegevens worden uitgevoerd voor tabel `main`
--

INSERT INTO `main` (`id`, `tekst`) VALUES
(1, 'helloooo');

thanks// Mokerhamer

 

[Kofels]

Ahh... can you post a code more cleanly? Btw just maybe $sql = "insert into main values('$tekst')"; is wrong... and this row arleady exist?

 

[Mokerhamer]

Oke Korn helped me but it's not perfect yet. The submit function dosen't work yet help me out..

<?php
$connect = mysql_connect ("localhost","root","secret") or die ("Kon geen verbinding met MySQL Server tot stand brengen. MySQL error: " . mysql_error()); 
mysql_select_db("nout",$connect) or die("Kon geen verbinding maken met database. MySQL error: " . mysql_error()); 
$query = "SELECT * FROM main"; 
$result = mysql_query($query) or die("Kon query niet uitvoeren: " . mysql_error()); 
  if(mysql_num_rows($result) == 0) 
  { 
  echo "Niks gevonden neem a.u.b contact op met munier"; 
  } 
  else 
  { 
while($_POST = mysql_fetch_array($result))
  {
 echo "<INPUT type='tekst' name='tekst' value='$_POST[tekst]'></TD>";
  echo "<br />";
  }
 } 
        mysql_select_db('nout', $connect) or die (mysql_error());
 
        if (isset($_POST['submit']))
        {
                mysql_query("UPDATE `main` SET `tekst` =  '" . $_POST['text'] . "' WHERE id = '". (int) $_POST['id']."'") or die ("Kon query niet uitvoeren: " . mysql_error()); 
                echo 'Geupdate';
        }
 
        $result = mysql_query('SELECT * FROM `main`') or die (mysql_error());
 
        if (mysql_num_rows($result) < 1)
                die ('Geen tekst gevonden neem a.u.b contact op met munier voor hulp');
 
        while ($row = mysql_fetch_array($result))
        {
                echo 'TextID ' . $row['id'] . ': ' . $row['tekst'];
        }
?>
 
<form method="POST" action="<? echo $_SERVER['REQUEST_URI'] ?>">
<font color=red>Id Niet veranderen!</font><td>&nbsp;</td><td>&nbsp;</td>Id: <input TYPE='TEXT' NAME='id' VALUE='1'/>
<TEXTAREA ROWS='20' COLS='40' NAME='text'></TEXTAREA>
<INPUT TYPE='submit' NAME='submit' onClick='return checkForm();' />
</FORM>

Here is an export from my datababase table main

-- Tabel structuur voor tabel `main`
--

CREATE TABLE IF NOT EXISTS `main` (
  `id` int(50) NOT NULL auto_increment,
  `tekst` varchar(5000) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Gegevens worden uitgevoerd voor tabel `main`
--

INSERT INTO `main` (`id`, `tekst`) VALUES
(1, 'helloooo');

 

[Kofels]

Looks good

 

[Mokerhamer]

Looks good

The point it wont overwrite the new tekst with the old one..

 

[trou]

It's logicall problem.
Cause this part:

while($_POST = mysql_fetch_array($result))
  {
 echo "<INPUT type='tekst' name='tekst' value='$_POST[tekst]'></TD>";
  echo "<br />";
  }

overrides $_POST submited in form.

So it would be better to change variable names there and in query.

 

[Mokerhamer]

4220Niller helped me and now it works but before i could ask the last question he logged off..

i have a problem with this

echo "<INPUT type='tekst' name='tekst' value='".$sql["tekst"]."'>";

i want the text shower way bigger because it shows only 1 line but how?

 

[Kofels]

Use <textarea> in place of <input type="text" />

 

[Mokerhamer]

i used

echo "<textarea INPUT type='tekst' ROWS='20' COLS='40' name='tekst' value='".$sql["tekst"]."'></textarea>";

But it wont show me the text from the database only an empty typing place

 

[Kofels]

lol?

echo "<textarea ROWS='20' COLS='40' name='tekst'>".$sql["tekst"]."</textarea>";

BTW You use a bad HTML syntax ;s it will work but W3 won't pass it.

 

[Mokerhamer]

lol?

echo "<textarea ROWS='20' COLS='40' name='tekst'>".$sql["tekst"]."</textarea>";

BTW You use a bad HTML syntax ;s it will work but W3 won't pass it.

It worked thanks lol, dont worry i just started learning and i learn very fast i wont be n00bz0r anymore