passing id

[shinobi_master2]

hi
i have a problem..that maybe someone can help,

i try this program that let me say contact_menu.php as page 1,
and show_contact.php as page 2 i try to create another page that let me say insert.php as page 3

how can i carry the variable id from page 2 to page 3 let say from the table

desc mohon;

+--------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | | PRI | NULL | auto_increment |
| nama | varchar(60) | YES | | NULL | |
| unit | varchar(40) | YES | | NULL | |
| bahagian | varchar(40) | YES | | NULL | |
| peralatan | varchar(40) | YES | | NULL | |
| jawatan | varchar(40) | YES | | NULL | |
| tarikh | date | YES | | NULL | |
| kuantiti | varchar(20) | YES | | NULL | |
| catatan | varchar(20) | YES | | NULL | |
| status | varchar(20) | YES | | NULL | |
| pengurus | varchar(60) | | | | |
| tarikhterima | date | YES | | NULL | |
+--------------+------------------+------+-----+---------+----------------+
12 rows in set (0.00 sec)

i just need to insert data for status,pengurus and tarikhterima based on id..but it's not working...

Maybe community have idea....

Thank

PS : some file that you can try

I appriaciate your help

REGISTER_GLOBAL already ON

[paulnaj]

Hi shinobi,

On the contact_menu.php page, you've got a line ...

<a href=\"show_contact.php?id=$contact_id\">

... the link to show_contact.php, passing the id as a $_GET variable.

On page show_contact.php, you have a form that submits to the page proses2.php but are not passing the $_GET['id'] on. 😉

So, just after the line ...

<form method="post" action="proses2.php">

... put in a hidden 'id' field ...

<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>">

Remember, though that you now have to use $_POST['id'] in proses2.php because of the form's method="post".

Paul 🙂

PS. There may be other problems ... I notice that you have three select boxes with the same name (<select name="tarikh">). This could be another problem.

[shinobi_master2]

Hi Paulnaj....

Thank
I appriacte your help..but i still stuck here : proses2.php

//check for required query string variables
if (!$_POST[id]) {
header("Location:http://192.168.1.200/contact_menu.php");
exit;
} else {
//if form variables are present,start a session
echo "not passing";
}

//set up table and database names
$db_name ="YADIM";
$table_name ="mohon";

//connect to server and select database
$connection = @mysql_connect("localhost","root","") or die(mysql_error());
$db = @mysql_select_db($db_name,$connection) or die(mysql_error());

    //if valid,get information
    $sql ="INSERT INTO $table_name (pengurus,tarikhterima,status) VALUES ($pengurus,$tarikh,$status) WHERE id ='$_POST[id]'";
    $result = @mysql_query($sql,$connection) or die(mysql_error());

echo "$_POST[id]";

?>

I get an error

not passingYou have an error in your SQL syntax near 'WHERE id ='29'' at line 1

and actually 29 is the last id not id number 6 that i want to modified...

Have any idea...

Still learning

[paulnaj]

Hi shinobi,

If I were you, I would stop trying to write 'shortcut' code ... like embedding variables in double-quoted strings and leaving quotes off of array keys.

Although both are acceptable and often seen in tutorials, it is much easier to read, debug, etc writing things out in full.

Here's my version of your page ...

<?php
//check for required query string variables 
if(!isset($_POST['id'])){ 
	// Uncomment this once debugging finished
	// header("Location:http://192.168.1.200/contact_menu.php"); 

// Comment this out when debugging finished
die('$_POST[\'id\'] = has not been sent.');
}

//set up table and database names 
$db_name = "YADIM"; 
$table_name = "mohon"; 

//connect to server and select database 
$connection = @mysql_connect("localhost", "root", "") or die(mysql_error()); 
$db = @mysql_select_db($db_name, $connection) or die(mysql_error()); 

$sql = "
	INSERT INTO ".$table_name." (pengurus, tarikhterima, status) 
	VALUES ('".$pengurus."', '".$tarikh."', '".$status."') 
	WHERE id ='".$_POST['id']."'
"; 

// Print the sql statement to the browser ... this is what you want to check.
// Comment this out when debugging finished
echo '<pre>'.$sql.'</pre>'

// Uncomment this once debugging finished
// $result = @mysql_query($sql, $connection) or die(mysql_error()); 
?>

It might work just fine, but there still could be an issue with the INSERT query ... see how it goes.

Paul.

[shinobi_master2]

hi paul yearhh u are right the script is running but the data can't be sent to the table...

from proses2.php

it show this message, i try to change this script to update

UPDATE mohon SET pengurus = 'sdfd', status = 'LULUS', tarikhterima = '2004-11-2' where id = '6'

why ?? my brain is thinking 2...

anyway thank a lot Paul

[paulnaj]

Hi shinobi,

OK, nearly there ...

1) I think it's probably that you don't 'quote' numeric entries in SQL, so the id value in the WHERE clause should not be quoted ...

$sql = "
INSERT INTO ".$table_name." (pengurus, tarikhterima, status)
VALUES ('".$pengurus."', '".$tarikh."', '".$status."')
WHERE id =".$_POST['id']."
";

... which, for your particular example would produce ...

UPDATE mohon SET pengurus = 'sdfd', status = 'LULUS', tarikhterima = '2004-11-2' where id = 6

2) I notice that the date value you are attempting to INSERT may not be valid. The standard pattern is YYYY-MM-DD, so you'll need to fix the date before attempting to insert it.

In general, you should be error-checking ALL values to be inserted into a database. Never rely on the user (even if it's you) to put the data into the correct format. There is a 'law' of programming:

If there is a correct way of doing things and an incorrect way, the user will ALWAYS do the incorrect one!!!!

Paul 🙂