Notice: Uninitialized string offset: 1

my function is :
function sele_orde($tb, $id, $ord)
{
$row[] = '';
$sql = mysql_query("select * from $tb order by $id $ord");
while($re = mysql_fetch_array($sql))
{
$row[] = $re;
}
return $row;
}

and call here :

foreach(sele_orde('user_tb', 'id', 'desc') as $val)
{

echo $val[1]."::::".$val[2]."::::".$val[3];
echo "<br>";
}
but here show the error and after showing result:
Notice: Uninitialized string offset: 1 in D:\xampp\htdocs\ls\developer.php on line 103

Notice: Uninitialized string offset: 2 in D:\xampp\htdocs\ls\developer.php on line 103

Notice: Uninitialized string offset: 3 in D:\xampp\htdocs\ls\developer.php on line 103

1::::firstname::::lastname

Plz help me...........

 

first print this, lets see what you have there

print_r ( sele_orde('user_tb', 'id', 'desc') )

 

is what you wrote the same as doing this?

$row[0]=$re[0];
$row[1]=$re[1];
$row[2]=$re[2];

and if there were more than 1 match, it would only return the last result?

 

$row[] = '';
your first row has an empty string and your array will looks like this
array('',array(0=>1,2=>'firstname',3=>'lastname'))


replace it with $row = array(); instead of $row[] = '';

 

Gee... no one noticed the error... ok, here is the solution.

First understand the message:

Assuming that $string = "Hello";
echo $string[0] will return "H".

Now, your code works but there's just an issue: you are declaring the array in a wrong way. When you want to declare an array you cannot do this:

$array[] = "";

Code (markup):

Because what you are doing is to assign to the $array a subsequential value of "empty", hence you get the error that at offset 1 there are no chars to show.
Instead, what you have to do is to declare the empty array like this

$array = array();

Code (markup):

Why? Because like this the foreach will have just the exact numbers of rows to show instead of a empty row at the beginning and then the others.