Solved

number checker

Posted on 2008-09-29
8
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Last Modified: 2012-05-05
hello there,
I would like to create something so that if I give a number lets say (150)
the code checks from 0 until 150 and once it finds it then say "number found!"
0
Comment
Question by:XK8ER
  • 5
  • 3
8 Comments
 
LVL 17

Expert Comment

by:nanharbison
ID: 22595327
this should work

<?PHP
$checkthisnumber = $_GET['thisnumber'];
 
$i=1;
while ($i < 151) 
{
	
 if ($i == $checkthisnumber) 
 	{
		echo "number found";
		break;
	}
$i++;
}
 
?>

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LVL 17

Accepted Solution

by:
nanharbison earned 500 total points
ID: 22595338
or if you want to show the number

<?PHP
$checkthisnumber = $_GET['thisnumber'];
 
$i=1;
while ($i < 151) 
{
	
 if ($i == $checkthisnumber) 
 	{
		echo "number found, it is: ". $checkthisnumber;
		break;
	}
$i++;
}
 
?>

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0
 
LVL 1

Author Comment

by:XK8ER
ID: 22595352
yes, the last one is good.. how can I make it show a number not found?
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LVL 1

Author Comment

by:XK8ER
ID: 22595378
lets say I type number 6 it should be like

1 not found
2 not found
3 not found
4 not found
5 not found
6 number found
0
 
LVL 1

Author Comment

by:XK8ER
ID: 22595414
ok I think I got it!
$checkthisnumber = '50';
 
$i=1;
while ($i < 151) 
{
        
 if ($i == $checkthisnumber) 
        {
                echo $checkthisnumber . ' number found';
                break;
        }
      			else
        				echo $i . ' not found<br>';
$i++;
}

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0
 
LVL 17

Expert Comment

by:nanharbison
ID: 22595423
like this

<?PHP
$checkthisnumber = $_GET['thisnumber'];
 
$i=1;
while ($i < 151) 
{
	
 if ($i == $checkthisnumber) 
 	{
		echo "number found, it is: ". $checkthisnumber;
	} else {
		echo $checkthisnumber." was not found ";
	}
$i++;
}
 
?>

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0
 
LVL 17

Expert Comment

by:nanharbison
ID: 22595451
oops, that was wrong

<?PHP
$checkthisnumber = $_GET['thisnumber'];
$checkthisnumber = 100;
$i=1;
while ($i < 151) 
{
	
 if ($i == $checkthisnumber) 
 	{
		echo "number found, it is: ". $checkthisnumber;
	} else {
		echo $i." was not found<br /> ";
	}
$i++;
}
 
?>

Open in new window

0
 
LVL 17

Expert Comment

by:nanharbison
ID: 22595477
use the break statement if you want the loop to stop when the number is found.
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