If statement

bluscale4

I was going through this tutorial and it states to put in a variable in an if statement that has not been initialized. From my understanding, it seems the the desired affect being sought is that the if statement would return flase if the variable is not declared but instead there is an error message.

Here is the php file being mentioned, the link to the tutorial follows underneath. Thanks ahead of time to who ever clears this up for me.

<?php

$db = mysql_connect("localhost", "root");

mysql_select_db("mydb",$db);

$PHP_SELF='display_2nd_set.php';
// display individual record

if ($id) 😕
{
$result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);

$myrow = mysql_fetch_array($result);

printf("First name: %s\n<br>", $myrow["first"]);

printf("Last name: %s\n<br>", $myrow["last"]);

printf("Address: %s\n<br>", $myrow["address"]);

printf("Position: %s\n<br>", $myrow["position"]);

}

else
{
//show employee list

$result = mysql_query("SELECT * FROM employees",$db);

if ($myrow = mysql_fetch_array($result)) 
{
// display list if there are records to display

	do 
	{
	        printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["first"], $myrow["last"]);
	} 
	while ($myrow = mysql_fetch_array($result));

} 

else 
{
// no records to display

	echo "Sorry, no records were found!";	
}

}

</body>
</html>

http://webmonkey.wired.com/webmonkey/99/21/index3a_page3.html?tw=programming

devinemke

depending on your error reporting settings you may see and warning about undeclared variable. perhaps the tutorial implies that the variable would have been set earlier. anyway, use [man]isset[/man] to check:

if (isset($id) && $id)
{
	// do something if $id is set and evaluates to boolean TRUE
}

onion2k

Putting 'if ($id) {' is the approximately the same as putting 'if ($id != "") {'. The code is fine. What error message are you getting?

bluscale4

Notice: Undefined variable: id in f:\program files\easyphp1-7\www\display_2nd_set.php on line 13

bluscale4

isset gets rid of the error message but it doesnt seem to become true when it should. (it doesnt print out table information)

laserlight

My guess is that $id is coming from say, the query string, and so you should use $_GET['id'] instead.

bluscale4

ok that actually makes a lot of sense, but i still get errors. this is the intial error i get before anything is passed to get:

Notice: Undefined index: id in f:\program files\easyphp1-7\www\display_2nd_set.php on line 13
Bob Smith
John Roberts
Brad Johnson

after:

Notice: Undefined variable: id in f:\program files\easyphp1-7\www\display_2nd_set.php on line 15

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in f:\program files\easyphp1-7\www\display_2nd_set.php on line 17
First name:
Last name:
Address:
Position:

laserlight

first check with isset($GET['id']), then use $GET['id'] or even just assign it to $id.

You might want to post your updated code in forum bbcode tags.

bluscale4

<html>
<body>

<?php

$db = mysql_connect("localhost", "root");

mysql_select_db("mydb",$db);

$PHP_SELF='display_2nd_set.php';
// display individual record

if (isset($_GET['id']))
{
	$result = mysql_query("SELECT * FROM employees WHERE id=$id",$db);

$myrow = mysql_fetch_array($result);

printf("First name: %s\n<br>", $myrow["first"]);

printf("Last name: %s\n<br>", $myrow["last"]);

printf("Address: %s\n<br>", $myrow["address"]);

printf("Position: %s\n<br>", $myrow["position"]);

} 

else 
{
//show employee list

$result = mysql_query("SELECT * FROM employees",$db);

if ($myrow = mysql_fetch_array($result)) 
{
// display list if there are records to display

	do 
	{
	        printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["first"], $myrow["last"]);
	} 
	while ($myrow = mysql_fetch_array($result));

} 

else 
{
// no records to display

	echo "Sorry, no records were found!";	
}
}

?>

</body>
</html>

laserlight

yep, you should use
$id = $GET['id'];
after the isset, or use
SELECT * FROM employees WHERE id={$GET['id']}

oh, but realistically you should validate $_GET['id'] to ensure that there is no SQL injection.

bluscale4

ok thanks, im sure it'll work now, im getting around to validation soon. at the moment i have no idea what youre talking about but i should soon. thanks for the help.