[SOLVED] "Undefined offset" notices driving me crazy

[johnc71]

I've tried everything and can't seem to get rid of the annoying "undefined offset" notices.  The codedoes what it suppose to do but I would like ot know what I am doing wrong here.  I searched for similar posts but can't seem to find the answer for my code. 

 

Thanks in advance for your help!

 

$sql1="SELECT * FROM mytable where active='yes'";
$result1=mysql_query($sql1);
while($rows1=mysql_fetch_array($result1)){
	$count='4';
	$id[]=$rows1['id'];
	for($i=0;$i<$count;$i++){
		$sql2="UPDATE mytable SET active='no' WHERE id='$id[$i]'";
		$result2=mysql_query($sql2);
		}
	}

[phpSensei]

you're referring to an array key that doesn't exist

[teng84]

$sql1="SELECT * FROM mytable where active='yes'";
$result1=mysql_query($sql1);
while($rows1=mysql_fetch_array($result1)){
	$count='4';
	$id[]=$rows1['id'];
}
if(isset($id) || count($id)>0){
	$sql2="UPDATE mytable SET active='no' WHERE `id` in (".implode(',',$id).")";
	$result2=mysql_query($sql2);

}

 

try that.. and what is the error line? and the exact message if possible

[johnc71]

Your code will set all "active" values to be "no". 

 

I edited my original post with more clarifications and exact error I am receiving.  Thanks for your help!

 

Here is the exact code:

 

<?php
$host="myhost"; 
$username="myusername";
$password="mypassword";
$db_name="mydatabase";

mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$sql1="SELECT * FROM mytable where active='yes'";
$result1=mysql_query($sql1);
while($rows1=mysql_fetch_array($result1)){
	$count='4';
	$id[]=$rows1['id'];
	for($i=0;$i<$count;$i++){
		$sql2="UPDATE mytable SET active='no' WHERE id='$id[$i]'";
		$result2=mysql_query($sql2);
		}
	}

 

Here is the result:

Notice: Undefined offset: 1 in /mycode.php on line 15

Notice: Undefined offset: 2 in /mycode.php on line 15

Notice: Undefined offset: 3 in /mycode.php on line 15

Notice: Undefined offset: 2 in /mycode.php on line 15

Notice: Undefined offset: 3 in /mycode.php on line 15

Notice: Undefined offset: 3 in /mycode.php on line 15

 

[wildteen88]

Could you explain fully what you're trying to do? I find the following code difficult to understand:

		$count='4';
	$id[]=$rows1['id'];
	for($i=0;$i<$count;$i++){
		$sql2="UPDATE mytable SET active='no' WHERE id='$id[$i]'";
		$result2=mysql_query($sql2);

 

Specifically the $id[]=$rows1['id'] line.

 

$rows1['id'] will return a non array value (I presume a number), however you're adding the value of $row['id'] to the $id[] array. Then next your telling PHP to run the code within the for loop 4 times, which iterates through the the $id array. << This happens everytime PHP loops through the while loop.

 

I am not surprised why you're getting the notices. As most of the time keys 2 and 3 will not be set.

[johnc71]

Thanks for your reply. What I am trying to do is, connect to db, update first 4 records where active is "yes" to "no".

 

I am planning to sell pin numbers so lets say I have DB with 100 records. Let's say customer buys 4 pins. I would like to go to DB, retrive first 4 available pins ("Active" = yes) and then set the "Active" field to "No" so that same pins will not be resold.

 

Hope this makes sense.

 

[sasa]

try

$sql1="SELECT * FROM mytable where active='yes'";
$result1=mysql_query($sql1);
while($rows1=mysql_fetch_array($result1)){
	$id[]=$rows1['id'];
}
$count=4;
for($i=0;$i<$count;$i++){
	$sql2="UPDATE mytable SET active='no' WHERE id='$id[$i]'";
	$result2=mysql_query($sql2);
}

[wildteen88]

It would be better if you added LIMIT 4 to the end of your query. This way rather than MySQL having to return all rows which are active, it'll just return 4 results.

 

I'd then use this code:

$sql1 = "SELECT * FROM mytable where active='yes' LIMIT 4";
$result1 = mysql_query($sql1);

while($rows1 = mysql_fetch_array($result1))
{
    $id[] = $rows1['id'];
}

if(is_array($id))
{
    $sql2 = "UPDATE mytable SET active='no' WHERE id IN(" . implode(',', $id) . ')';
    mysql_query($sql2);
}

[johnc71]

Thanks everyone!  It works fine now.