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Read form mysql, insert in form, modify and update in mysql

Help, I have a big probem. I wanto to read data from an mysql table and then write this data into a form so that the user can update this data. The code could seem somethin like this, but it doesn't work

<?php

include 'complete.php';

    $user_name = "exa";
    $password = "exa";
    $database = "exa";
    $server = "localhost";

$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);

if ($db_found) {

$SQL = "SELECT * FROM tb_exa WERE username='$CCGuser' AND password='$CCGpassw'";
$result = mysql_query($SQL);

while ($db_field = mysql_fetch_assoc($result)) {
$nome=$db_field['name'];
$cognome=$db_field['gender'];
}

mysql_close($db_handle);

}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}


 
	

	<table style="width: 100%">
	<form action="process_u.php" method="post">
		<tr>
			<td>Name: </td>
			<td> <input type="text" name="name" value="$nsme"></td>
		</tr>
		<tr>
			<td>Gender: </td>
			<td> <input type="text" name = "gender" value="$gender"></td>
		</tr>

 <input type="submit" value="GO" name="Submit">&nbsp; <input type="reset" name="reset" value="Reset" /></td>
			<td>&nbsp;</td>
		</form></tr>
	</table>


My problem is: How to read form the sql table and write the result on the form.

*the variables $CCGuser and $CCGpssw are written in complete.php (see top of the code) that's included in here

naui95
Newbie Poster
2 posts since Nov 2011
Reputation Points: 10
Solved Threads: 0
 

There are a couple of mistakes in your code:
1. you declare $nome=$db_field['name']; on line 19 and then use value="$nsme" on line 39 (typo?)
2. you declare $cognome=$db_field['gender']; on line 20 and then use value="$gender" on line 43 (shouldn't it be value="$cognome")
3. your <'php opening tag is not followed by a ?> closing tag somewhere before html code
4. your html code is within PHP block (see 3. above) which causes errors
5. in html you must echo variables enclosed in php tags

See the example of how this code should look like:

<?php

include 'complete.php';

    $user_name = "exa";
    $password = "exa";
    $database = "exa";
    $server = "localhost";

$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);

if ($db_found) {

$SQL = "SELECT * FROM tb_exa WERE username='$CCGuser' AND password='$CCGpassw'";
$result = mysql_query($SQL);

while ($db_field = mysql_fetch_assoc($result)) {
$nome=$db_field['name'];
$cognome=$db_field['gender'];
}

mysql_close($db_handle);

}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}

?>
 
    

    <table style="width: 100%">
    <form action="process_u.php" method="post">
        <tr>
            <td>Name: </td>
            <td> <input type="text" name="name" value="<?php echo $nome;?>"></td>
        </tr>
        <tr>
            <td>Gender: </td>
            <td> <input type="text" name = "gender" value="<?php echo $cognome;?>"></td>
        </tr>

 <input type="submit" value="GO" name="Submit">&nbsp; <input type="reset" name="reset" value="Reset" /></td>
            <td>&nbsp;</td>
        </form></tr>
    </table>
broj1
Posting Whiz
359 posts since Jan 2011
Reputation Points: 29
Solved Threads: 43
 

Thx, actually I was asking something mouch simpler I foud the answer here http://www.daniweb.com/web-development/php/threads/16765 .

Still thx very much!! broj1

naui95
Newbie Poster
2 posts since Nov 2011
Reputation Points: 10
Solved Threads: 0
 

This article has been dead for over three months

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