Data Input

[Quijote]

Hello
with this script I need to show the user (utenti) list and after I need to
input in the table RICHIESTA the field COGNOME e NOME.
This is the code but don't go. Someone can tell me where is the mistake???
Many Thanks!!
<select name="user" id="user">
<?
$q= mysql_query("select * from utenti order by user");
while ($res=mysql_fetch_array($q))
{
echo "<OPTION VALUE=\"$res[user]\">$res[user] --
$res[reparto]</OPTION>";
}
mysql_free_result($q);

$rs = mysql_query("SELECT * FROM utenti WHERE user = ".$user);

$cognome = mysql_result($rs, 0, "cognome");
$nome = mysql_result($rs, 0, "nome");
mysql_query("INSERT INTO richiesta (cognome,nome) VALUES
($cognome,$nome)");
?>
</select>

[Pedro Graca]

Quijote wrote:

with this script I need to show the user (utenti) list and after I need to
input in the table RICHIESTA the field COGNOME e NOME. <select name="user" id="user">
<?
$q= mysql_query("select * from utenti order by user");

(snip)
Do some error-checking in your scripts!

For a SQL select action I do:

<?php
$conn = mysql_connect($host, $user, $pass)
or die('Impossible to connect to '.$host.' now: '.mysql_error());
$sql = 'select * from utenti order by user';
$q = mysql_query($sql)
or die('Error '.mysql_error().' in query '.$sql);
while ($res = mysql_fetch_array($q)) {
/* whatever */
}
// ...
?>

This way, if there's an error connecting or in the select string I get
feedback about it.

In your second query to the "utenti" table you are using a variable
named $user. Where did it come from?
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[Quijote]

"Pedro Graca" <he****@hotpop.com> ha scritto nel messaggio
news:c0*************@ID-203069.news.uni-berlin.de...

In your second query to the "utenti" table you are using a variable
named $user. Where did it come from?
--

Thanks Pedro, in the code there is a mistake the right line is

$rs = mysql_query("SELECT * FROM utenti WHERE user=".$_POST[user]);

[Quijote]

"Pedro Graca" <he****@hotpop.com> ha scritto nel messaggio
news:c0*************@ID-203069.news.uni-berlin.de...

Do some error-checking in your scripts!

For a SQL select action I do:

<?php
$conn = mysql_connect($host, $user, $pass)
or die('Impossible to connect to '.$host.' now: '.mysql_error());
$sql = 'select * from utenti order by user';
$q = mysql_query($sql)
or die('Error '.mysql_error().' in query '.$sql);
while ($res = mysql_fetch_array($q)) {
/* whatever */
}
// ...
?>

Thanks, the problem for me is not to list the users (utenti) this is
possible but the problem is to update the fields COGNOME and NOME that are
in table UTENTI and I need to fill in table RICHIESTA connecting the posted
user with the table UTENTI ... I hope it's clear :-). Many many thanks again
....

[Pedro Graca]

Quijote wrote:

"Pedro Graca" <he****@hotpop.com> ha scritto nel messaggio
news:c0*************@ID-203069.news.uni-berlin.de...

Quijote wrote:

$rs = mysql_query("SELECT * FROM utenti WHERE user = ".$user);

$cognome = mysql_result($rs, 0, "cognome");
$nome = mysql_result($rs, 0, "nome");
mysql_query("INSERT INTO richiesta (cognome,nome) VALUES ($cognome,$nome)");
Do some error-checking in your scripts!
Do some error-checking in your scripts!

$rs = mysql_query("select * from utenti where user=".$_POST["user"]);
Are you sure? Why not doing some error-checking? :)
I guess $_POST["user"] is a string, and as such, should be enclosed in
single quotes.
mysql_query("INSERT INTO richiesta (cognome,nome) VALUES ($cognome,$nome)");

Are you sure? Why not doing some error-checking? :-))
Don't $cognome and $nome need to be enclosed in single quotes?
Thanks, the problem for me is not to list the users (utenti) this is
possible but the problem is to update the fields COGNOME and NOME that are
in table UTENTI and I need to fill in table RICHIESTA connecting the posted
user with the table UTENTI ... I hope it's clear :-). Many many thanks again

It helps to do error-checking.

Ii does not hurt your script noticeably (if at all) and it is a great
tool to aid in debugging when something goes wrong.
Happy Coding :-)
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