If clause output problem

Question

vanpersie 0 Light Poster

13 Years Ago

Hi
I have a code that should:-
if the number entered more than 60
the following should be printed
"Congrats!!!<p>"
"Grade1"
if the number entered between 60 and 40 the following should be printed
"Grade2"
else the script should print
"fail"
but the code print fail all times
can someone help me please.

<html>
<body>
<?php
$A=$_GET['n1text'];
if($A>60)
 {
 echo "Congrats!!!<p>";
echo "Grade1";
}
 else 
{
if($A <60 and $A >= 40)
{
 echo "Grade2";
 } 
 else  
{
echo "\n Fail";
}
} 
 
?>
</body>
</html>

php

4 Contributors
20 Replies
370 Views
2 Days Discussion Span
Latest Post 13 Years Ago Latest Post by ko ko

almostbob 866 Retired: passive income ROCKS

13 Years Ago

try

<?php $A=floatval($_GET['n1text']); // if it is not possible for $a to have a decimal component or if the decimal part is not significant intval() can be used
if($A>=60) { echo "Congrats!!!<br>Grade1"; }
elseif($A <60 and $A >= 40) { echo "Grade2";  }
else { echo "Fail"; } ?>

60 was excluded in the op, > 60 or < 60, no =60

$a is likely being treated as text, by convention all the request arrays are parsed as text, so the original value may be interpreted as "60"

Edited 13 Years Ago by almostbob because: n/a

almostbob 866 Retired: passive income ROCKS

13 Years Ago

show the code used to call the file,
is ?n1text=60 in the link
is $_get the right array, $_post $_request ?

Zagga

13 Years Ago

Hi vanpersie,

Are you sure the correct value for $_GET is being passed through? After you assign $A try echoing it to see if it is what you expect.

Zagga

ko ko 97 Practically a Master Poster

13 Years Ago

<input type="text" value="n1text">

Replace above with below one:

<input type="text" name="n1text">

almostbob commented: good eyes zero +13

Zagga commented: Well spotted +5

ko ko 97 Practically a Master Poster

13 Years Ago

<html>
<body>
<?php
 $A = floatval($_GET['n1text']);
 
if($A>=60)
 {
 
 echo "Congrats!!!<p>Grade1";
 }
elseif ($A <60 && $A >= 40)
 {
 
 echo "Grade2";
 } 
else  
{
 
echo "\n Fail";
} 
?>
</body>
</html>

Try with above. It works on my PC.

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vanpersie 0 Light Poster · Answer 1 · 2011-05-28T05:03:56+00:00

<html>
<body>
<?php
$A=floatval($_GET['n1text']);
if($A>=60)
 {
 echo "Congrats!!!<p>Grade1";
 }
elseif ($A <60 and $A >= 40)
 {
 echo "Grade2";
 } 
else  
{
echo "\n Fail";
} 
?>
</body>
</html>

Same Result!.

vanpersie 0 Light Poster · Answer 2 · 2011-05-28T06:06:00+00:00

show the code used to call the file

<html>
<body>
<b>Grades</b>
<form method=get action="Grade.php">
enter your percentage:
<input type="text" value="n1text">
<br>
<input type="submit" name="GRADE" Value="GRADE">
<br>
</body>
</html>

vanpersie 0 Light Poster · Answer 3 · 2011-05-28T07:01:22+00:00

Hi vanpersie, After you assign $A try echoing it to see if it is what you expect.

Zagga

each time the output is 0 Fail

<html>
<body>
<?php
$A;
$A=floatval($_GET['n1text']);
if($A>=60)
 {
echo $A;
 echo "Congrats!!!<p>Grade1";
 }
elseif ($A <60 and $A >= 40)
 {
echo $A;
 echo "Grade2";
 } 
else  
{
echo $A;
echo "\n Fail";
} 
?>
</body>
</html>

Zagga · Answer 4 · 2011-05-28T07:07:00+00:00

Hi again,

I meant after line 5 add a line like this

echo "The value of the variable is: " . $A;

What does it echo?

vanpersie 0 Light Poster · Answer 5 · 2011-05-28T07:25:18+00:00

Hi again,

What does it echo?

The value of the variable is: 0 Fail

Zagga · Answer 6 · 2011-05-28T07:37:45+00:00

The value of the variable is: 0 Fail

Your code is assuming the variable is a number and checking for numerical values. '0 Fail' is a string not an integer.

You need to ensure $A is a numerical value for your current code to work.

almostbob 866 Retired: passive income ROCKS · Answer 7 · 2011-05-28T08:01:17+00:00

floatval() is to convert the text variable to a numeric <form method=get action="Grade.php"> should be <form method='get' action="Grade.php"> dependant on html doctype used, it makes a big difference

vanpersie 0 Light Poster · Answer 8 · 2011-05-30T01:15:11+00:00

Hello Again
I have done the changes that are posted in the two later replies
but with sorry is the same
when I get rid from the floatval function
The output is:-
The value of the variable is: Fail
Hope you help
thank you

ko ko 97 Practically a Master Poster · Answer 9 · 2011-05-30T04:01:26+00:00

Echo the value first, and ensure that the parameter takes the correct data. Put this line on the first line of 'Grade.php'.

echo $_GET['n1text'];

If it returns the correct value, then try with floatval() and check again. Hope this help.

ko ko 97 Practically a Master Poster · Answer 10 · 2011-05-30T04:02:33+00:00

Echo the value first, and ensure that the parameter takes the correct data. Put this line on the first line of 'Grade.php'.

echo $_GET['n1text'];

If it returns the correct value, then try with 'floatval()' and check again. Hope this help.

vanpersie 0 Light Poster · Answer 11 · 2011-05-30T04:59:18+00:00

Any Number is outputted with fail

50 Fail

<html>
<body>
<?php

echo $_GET['n1text'];
 
 
if($A>=60)
 {
 
 echo "Congrats!!!<p>Grade1";
 }
elseif ($A <60 and $A >= 40)
 {
 
 echo "Grade2";
 } 
else  
{
 
echo "\n Fail";
} 
?>
</body>
</html>

vanpersie 0 Light Poster · Answer 12 · 2011-05-30T05:16:50+00:00

Oh
Idon't try the floatval function
After using it the code work perfectly
thank you

<html>
<body>
<?php

 
$A=  floatval($_GET['n1text']);
 
 
if($A>=60)
 {
 
 echo "Congrats!!!<p>Grade1";
 }
elseif ($A <60 and $A >= 40)
 {
 
 echo "Grade2";
 } 
else  
{
 
echo "\n Fail";
} 
?>
</body>
</html>

ko ko 97 Practically a Master Poster · Answer 13 · 2011-05-30T05:23:12+00:00

Mark this thread as Solve then, and it will help us to save our time.

vanpersie 0 Light Poster · Answer 14 · 2011-05-30T05:36:33+00:00

vanpersie 0 Light Poster

13 Years Ago

ok

Edited 13 Years Ago by vanpersie because: n/a

ko ko 97 Practically a Master Poster · Answer 15 · 2011-05-30T05:51:31+00:00

Click 'Mark this Thread as Solve' link below of these answers.