UPDATE на таблица- MySQL

sle

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Не иска да ги Update-не ?

<form action="index.php" method="post" name="edit">
<input type="submit" name="edit" />
</form>

<?PHP

$query = mysql_query("SELECT * FROM posledni");
while($rows = mysql_fetch_array($query)){
$a=$rows['asd'];
$b=$rows['asdasd'];
$c=$rows['asdasdasd'];
$d=$rows['asdasdasdasd'];
$e=$rows['asdasdasdasdasd'];

echo '<tr bgcolor="#D7EBFF">
<td align="center"><input type="text" value="'.$rows['asd'].'" />','</td>
<td align="center"><input type="text" value="'.$rows['asdasd'].'" />','</td>
<td align="center"><input type="text" value="'.$rows['asdasdasd'].'" size="5" /></td>
<td align="center"><input type="text" value="'.$rows['asdasdasdasd'].'" size="5" /></td>
<td align="center"><input type="text" value="'.$rows['asdasdasdasdasd'].'" size="6" /></td>
</tr>';
}


if($_POST['edit'])
{
$nq = "UPDATE posledni SET match='$a', tips='$b', Handicaps='$c', Score='$d', Verdict='$e' WHERE id='$id'";
$nr = mysql_query($nq);
}
?>
 
PHP:
<form action="index.php" method="post" name="edit">
<?PHP
$query = mysql_query("SELECT * FROM posledni");
while($rows = mysql_fetch_array($query)){ 
echo '<tr bgcolor="#D7EBFF">
<td align="center"><input type="text" name="asd" value="'.$rows['asd'].'" />','</td>
<td align="center"><input type="text" name="asdasd" value="'.$rows['asdasd'].'" />','</td>
<td align="center"><input type="text" name="asdasdasd" value="'.$rows['asdasdasd'].'" size="5" /></td>
<td align="center"><input type="text" name="asdasdasdasd" value="'.$rows['asdasdasdasd'].'" size="5" /></td>
<td align="center"><input type="text" name="asdasdasdasdasd" value="'.$rows['asdasdasdasdasd'].'" size="6" /></td>
</tr>
<input type="submit" name="edit" />
</form>';
}


if($_POST['edit'])
{
$a=$_POST['asd'];
$b=$_POST['asdasd'];
$c=$_POST['asdasdasd'];
$d=$_POST['asdasdasdasd'];
$e=$_POST['asdasdasdasdasd'];
$nq = "UPDATE `posledni` SET `match`='$a', `tips`='$b', `Handicaps`='$c', `Score`='$d', `Verdict`='$e' WHERE `id`='$id'";
mysql_query($nq);
}
?>
 
Код:
<form action="index.php" method="post" name="edit">
<input type="submit" name="edit" />

<?PHP

$query = mysql_query("SELECT * FROM posledni WHERE id='$id'");
$rows = mysql_fetch_array($query);
echo '  <tr bgcolor="#D7EBFF">
        <td align="center"><input type="text" value="'.$rows['asd'].'" />','</td>
        <td align="center"><input type="text" value="'.$rows['asdasd'].'" />','</td>
        <td align="center"><input type="text" value="'.$rows['asdasdasd'].'" size="5" /></td>
        <td align="center"><input type="text" value="'.$rows['asdasdasdasd'].'" size="5" /></td>
        <td align="center"><input type="text" value="'.$rows['asdasdasdasdasd'].'" size="6" /></td>
    </tr>
</form>';


if($_POST['edit'])
{
    $a = $rows['asd'];
    $b = $rows['asdasd'];
    $c = $rows['asdasdasd'];
    $d = $rows['asdasdasdasd'];
$nq = "UPDATE posledni SET match='$a', tips='$b', Handicaps='$c', Score='$d', Verdict='$e' WHERE id='$id'";
mysql_query($nq);
}
?>
 
А защо може да не иска да добави ??

Код:
<?php
echo '
<form action="index.php"  method="post" name="edit">
<input type="submit" value="Добави!"  name="edit" />';

echo '
<tr bgcolor="#D7EBFF">
<td  align="center"><input type="text" name="asd" value="" />','</td>
<td  align="center"><input type="text" name="asdasd" value="" />','</td>
<td  align="center"><input type="text" name="asdasdasd" value="" size="5" /></td>
<td  align="center"><input type="text" name="asdasdasdasd" value="" size="5" /></td>
<td align="center"><input type="text" name="asdasdasdasdasd" value="" size="6" /></td>
</tr>
</form>';

if($_POST['edit'])
{
$a=$_POST['asd'];
$b=$_POST['asdasd'];
$c=$_POST['asdasdasd'];
$d=$_POST['asdasdasdasd'];
$e=$_POST['asdasdasdasdasd'];

$query=mysql_query("INSERT INTO 'posledni' ('match', 'tips', 'hyhy', 'Score', 'Verdict') VALUES('$a', '$b', '$c', '$d', '$e')");
}

}
?>
 

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