round () Syntax Help

[aljohnstone]

This works

echo $data3['value'];

 

But this doesn't

echo round($data3['value'], 2);

 

Can someone please advise me as to what I am doing wrong?

[kenrbnsn]

What do you mean by "this doesn't"?

 

Ken

[aljohnstone]

echo $data3['value'];

returns: 1,384.1000

 

echo round($data3['value'], 2);

returns: 1

 

I need it to round off the last two decimal places: 1,384.10

[Pikachu2000]

You need to remove the comma. It can't be type-juggled into a numeric data type with the comma in it.

[aljohnstone]

The number which is dynamically generated has changed to 1,384.1500

 

If I take out the comma it returns 1,384.2500

[Pikachu2000]

Although I'm not sure what you're getting at in your last post, round simply isn't going to work on it at all if there's a comma in it.

[aljohnstone]

My understanding is that in round() the comma seperates the parameters. The first parameter is the number that is being rounded which in this case is the value of the variable $data3['value'] and the second parameter is to how many places after the decimal which is in this case 2.

 

If I remove the comma it does not limit the number of decimal places it just messes up the number that is returned.

[Pikachu2000]

No, no, no . . .

 

1,384.2500

^ That comma

[cgsmith105]

What Pikachu is saying is in the PHP manual which he linked to.

Note: PHP doesn't handle strings like "12,300.2" correctly by default. See converting from strings.

 

[lalnfl]

You could just do this, it would save you all the hassle:

 

$value = $data3['value'];

$value = round($value, 2);

echo "$value";

[Pikachu2000]

How do you figure that would help? The value in that array element has a comma in it, so round() still won't work on it.

[aljohnstone]

Oh OK. Sorry for being so thick.

 

Can I strip out the comma and then apply round()?

[aljohnstone]

I got it!!!

 

$data4 = str_replace(",", "", $data3['value']);

echo round($data4, 2);

 

Thanks for your help!