[[SOLVED] Javascript Maths - Using a full stop]

any ideas anyone?

[[SOLVED] Javascript Maths - Using a full stop]

Hi all

 

I have the following javascript maths being actioned from the input of a html form

 

if (document.myform.fieldr.value && document.myform.fieldw.value && document.myform.fieldjj.value && document.myform.fieldkk.value)
{
fieldrval = document.myform.fieldr.value
fieldwval = document.myform.fieldw.value
fieldjjval = document.myform.fieldjj.value
fieldkkval = document.myform.fieldkk.value
calc1 = fieldwval - fieldrval
calc2 = calc1 / fieldwval
calc3 = fieldjjval - fieldkkval
calc4 = calc3 / fieldjjval


document.myform.fieldy.value=calc2.calc4
}

 

At the moment this wont work because of the full stop between

 

calc2.calc4

 

on the last line of code

 

How can I output the value with a full stop in between calc2 and calc4?

 

Hope that makes sense

 

Thanks

 

Dave

[[SOLVED] Average out of 10]

Managed to get it right, in the end it was

 

printf('% (%0.1f<br>', $x, $pc/10);

 

Thanks for all the help

[[SOLVED] Average out of 10]

oh thanks mate.

 

what would be the best way to just display the value, so

 

20 (80.0)

2 (8.0)

8 (32.0)

4 (16.0)

5 (20.0)

 

would be come

 

80.0

8.0

32.0

16.0

20.0

 

and then deviding them by 10

 

is there a prefered way of doing that?

 

[[SOLVED] Average out of 10]

How can I remove the % sign from the code above outputs?

[[SOLVED] Average out of 10]

cool, guess I could just devide it by 10 to get a average out of 10

 

or is there a more logical way of doing that

[[SOLVED] Average out of 10]

Plus i'm not sure that's producing the right results, discomatt comments with outputting the following as an example looked right

 

200 (10)

150 (7.5)

90 (4.5)

50 (2.5)

10 (0.5)

 

I mean, that's the kinda result I want to product

 

any good ideas anyone?

[[SOLVED] Average out of 10]

Cool, well that produced something, but do you know how to get it to output as 1 didget after the decimal place, so

 

0.04

 

becomes

 

0.4

[[SOLVED] Average out of 10]

any ideas? still at it with no luck yet

[[SOLVED] Average out of 10]

Hi all

 

I tried altering it to

 

$stuff  "SELECT levelcode FROM levels";
$levels = array();
while ($stuff = $stuff)
{
  $levels[] = $stuff['levelcode'];
}

$total = array_sum($levels);
foreach ($levels AS $value)
{
  echo $value.' = '.round($value / $total, 2).'<br />';
}

 

not sure if I did that right, but it doesnt seem to produce any result at all.

 

Can anyone help?

[[SOLVED] Average out of 10]

discomatt, what query should I use with that?

 

akitchin, your query wants me to do a GROUP and if I do that, it doesnt seem to add up multiple values assigned to that userid

 

any ideas?

[[SOLVED] Average out of 10]

any ideas?

[[SOLVED] Average out of 10]

 

Well I guess im just trying to get a score out of 10 depending on their value compared to other in the table

[[SOLVED] Average out of 10]

Yes, I think that method would work, how would it be possible to do with the top value being 10%

[[SOLVED] Average out of 10]

Hi all

 

Wondering if someone can help.

 

I have the following query

 

$sql = "SELECT * FROM `levels`";
$show = @mysql_query($sql,$connection) or die(mysql_error());
while ($row = mysql_fetch_array($show))
{
print $row['levelcode'].'<br>';
}

 

which returns a list of level codes, the output at the moment looks like

 

72

234

4

7655

43

 

What I want to do is display next to each code returned, an average score out of 10 compared to the other codes.

 

I don't 100% know to write the query or even what the result should be, but it would look something like

 

72 - (2.6)

234 (7.4)

4 (0.4)

7655 (9.9)

43 (1.4)

 

Of course those numbers are incorrect, hence why I am trying to work out the query to use.

 

I thought maybe it would be something like

 

 

select count(*) as cnt, SUM(levelcode) as sum, (SUM(levelcode)/count(*)) as avg FROM levels;

 

but that doesnt stick to an average of 10, plus it gives an average for the whole database and not each row individually, plus it gives 4 digits result after the fullstop, such as 27.9375, whereas I need to stick to 1 digit such as 5.4

 

Could someone please help me work this one out, it's slowly making me dumber

 

Thanks

 

Dave

[Help with a QUERY]

Just stuck trying to get just the rows of the table "members" that have a particular code, but then auto skipping the ones that dont have that code.

 

it cant work with

 

$return = "SELECT id, name, code FROM members WHERE id = '".$id."' AND code = '' ";

 

because as soon as a row is returned without the correct code you are asking it for, it simply returns an empty row.

[Help with a QUERY]

If the code doesnt exist on the row it's returning I guess it almost needs to skip that row and move onto the next, but the mods i've tried have not worked yet.

[Help with a QUERY]

Yep, that seems to do the job until you reach a row which doesnt have that code im asking for.

 

Because you're grabbing the next ID in the sequence, if that ID it returns doesnt equal what im asking then it just displays nothing.

 

If that makes sense?

 

Any ideas?

[Help with a QUERY]

Thanks Minase

 

The problem is that the code is held in the table "members" and not "scores", therefore that QUERY wouldn't be able to find the field code.

 

Do you have anything else up your sleve you maybe able to advise on?

 

Thanks

[Help with a QUERY]

Minase

 

Your code worked spot on.

 

I'm wondering if anyone can help me further this code, basically I want to add another field to the "members" table called "code", so the table would look like

 

ID     NAME     CODE

1      David      A

2      Bob        B

3      Sarah     A

 

and then I want to alter the QUERY so that it only selects members with A in the code field and skips any member with B in their CODE field.

 

Can anyone help?

 

Masssive thanks to everyone who has helped so far

 

Dave

[Help with a QUERY]

Thanks again Minase

[Help with a QUERY]

afraid it still returns a row with a EMPTY value for some reason, not sure why

 

my little mod seems to have solved it

 

thanks for all your help

[Help with a QUERY]

It's exactly like your code.

 

I added

 

if(isset($_GET['start']))
{
$id = 1;
}

 

just under

 

$id = $r['member_id'] + 1;

 

which seems to work

[Help with a QUERY]

Thanks very much

 

It just seems when you reach the last row of the members table, it returns an empty ID

 

I think that might be because we get the last ID of the members table and add 1 and that ID wouldnt exist

[Help with a QUERY]

It doesn't seem to be for me, I have the following

 

$select = mysql_query("SELECT * FROM scores ORDER BY ID DESC LIMIT 1");
$r = mysql_fetch_array($select);
$id = $r['member_id'] + 1;
if(!mysql_query("SELECT id, name FROM members WHERE id = '".$id."' "))
{
$return = "SELECT id, name FROM members ORDER BY id ASC LIMIT 1";
}
else
{
$return = "SELECT id, name FROM members WHERE id = '".$id."' ";
}

$show = @mysql_query($return,$connection) or die(mysql_error());
while($r = mysql_fetch_array($show))
{
print $r['name']
}

 

But when it gets to "member" ID number 3 (which is the last one) it seems to not display anything, which makes me think it isn't looping.

 

Any ideas?