[RESOLVED] Manipulate numeric values inside string

Inpresif

Hi,

We're trying to edit a dbase with values in it, but we have little luck getting it done.

This is what we have:

"This sentence has not 5.50 words in it, not 10 either"

We're increasing numbers by 20%, so the output must be:
"This sentence has not 6.60 words in it, not 12 either"

Can this be done?

Thanks.

ps. Wow, last login here was YEARS ago ;-)

johanafm

If you really want each and every number to be incremented, no matter where in a string it appears

$s = preg_replace('#\d+#e', "1.2 * '$0'", $s);

Inpresif

johanafm;10961827 wrote:
If you really want each and every number to be incremented, no matter where in a string it appears
$s = preg_replace('#\d+#e', "1.2 * '$0'", $s);

Yeah, by setting up (and tiresome cleaning) of the dbase/csv file we took that into account, numbers that should not be changed we just wrote full out, like Ten instead of 10.

Thanks for the code, we'll give it a shot.

NogDog

A slight modification to cope with decimals:

<?php
$text = "This sentence has not 5.50 words in it, not 10 either";
$modified = preg_replace('#\d+(\.\d+)?#e', '1.2 * $0', $text);
echo $modified;

Inpresif

NogDog;10961837 wrote:
A slight modification to cope with decimals:
<?php
$text = "This sentence has not 5.50 words in it, not 10 either";
$modified = preg_replace('#\d+(\.\d+)?#e', '1.2 * $0', $text);
echo $modified;

Going to test in an hour when the kid is in bed, will this work for both . and , ?

$modified = preg_replace('#\d+(.|,\d+)?#e', '1.2 * $0', $text);

NogDog

I would make it:

'#\d+([.,]\d+)?#e'

However, it will probably give undesired results for something like "12,345.67". If that is a valid possibility, then more fiddling with the regexp would be required. (Frankly, I'd rather look at a redesign of the process if possible. 🙂 )

Inpresif

NogDog;10961850 wrote:
I would make it:
'#\d+([.,]\d+)?#e'
However, it will probably give undesired results for something like "12,345.67". If that is a valid possibility, then more fiddling with the regexp would be required. (Frankly, I'd rather look at a redesign of the process if possible. 🙂 )

Nah, xx,xxx.xx is out of the question, it involves % en $ discounts;-)

Inpresif

The plot thickens...

$text = "This sentence has not 5.50 no 1, 2, 3 or 4,50 words in it, not 10 either";
$modified = preg_replace('#\d+([.,]\d+)?#e', '1.2 * $0', $text);
echo $modified;

will give an error:

Parse error: syntax error, unexpected ',' in C:\tst.php(3) : regexp code on line 1

Fatal error: preg_replace() [<a href='function.preg-replace'>function.preg-replace</a>]: Failed evaluating code: 1.2 * 4,50 in C:\tst.php on line 3

Any idea?

bradgrafelman

Instead of using '$0' in the eval code, send the '$0' value to a function as a string (e.g. surround it with quotes) that switches the comma to a decimal.

EDIT: E.g. instead of '1.2 $0' you'd have '1.2 myFunc("$0")' instead.

Inpresif

bradgrafelman;10961855 wrote:
Instead of using '$0' in the eval code, send the '$0' value to a function as a string (e.g. surround it with quotes) that switches the comma to a decimal.

EDIT: E.g. instead of '1.2 $0' you'd have '1.2 myFunc("$0")' instead.

Great, i needed a function somewhere anyway, to always get two decimals, like 1.23

However, tried this and did not get lucky:

$text = "This sentence has not 5.50 no 1, 2, 3 or 4,50 words in it, not 10 either";
$modified = preg_replace('#\d+([.,]\d+)?#e', '0.8 * round("$0",2)', $text);

Result:

This sentence has not 4.4 no 0.8, 1.6, 2.4 or 3.2 words in it, not 8 either

4,50 became 4, which became 3.2. Should be 3.6, actually my goal is this:

This sentence has not 4.40 no 0.80, 1.60, 2.40 or 3.20 words in it, not 8 either

I'll keep searching;-)

Inpresif

Ok, this comes close, as long as i do not use comma's as separator:

$modified = preg_replace('#\d+([.,]\d+)?#e', 'number_format("$0" * 0.8,2, ",", ".")', $text);

This prints out:
This sentence has not 4,40 no 0,80, 1,60, 2,40 or 3,60 words in it, not 8,00 either

for this sentence:
$text = "This sentence has not 5.50 no 1, 2, 3 or 4.50 words in it, not 10 either";

Which is almost perfect for us Dutchies. I always want things perfect, so the only thing that bugs me is the round number 8. I'd like that to be just 8, not 8,00 ;-)

NogDog

Ah, the problem is multiplying a number with a comma in it. I didn't check the calculations, but this parses:

<?php
$text = "This sentence has not 5.50 no 1, 2, 3 or 4,50 words in it, not 10 either";
$modified = preg_replace('#(\d+)([.,](\d+))?#e', '1.2 * "$1.$3"', $text);
echo $modified;

Output:

This sentence has not 6.6 no 1.2, 2.4, 3.6 or 5.4 words in it, not 12 either

Inpresif

Nice, thanks.