Grabbing text from a site and adding to variables?

[sean123]

I'm creating a highscore grabber for an online game but I'm having a problem. I want it to grab the users stats and put each stat in a variable so kills would be in 1 and deaths would be in another. I've done the grabbing but not the putting each stat in a variable. This is what I get after I grab the stats:

 

username  Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632

 

I want it so it's like this:

 

$name = "username"; 

$level = "25";

$ratio = "1.54";

$rank = "3266 th";

$kill = "73761";

$death = "9977";

$exp = "674534632";

 

And removing anything before the colon. Is this possible?

[phpSensei]

You need to add some sort of comma or ";"...

 

I took off the Username Level and stuff and put the data in seperate columns

 

25;1.54;3266th;73761;9977;674534632';

 

 

Try

 

<?php
$stats = 25;1.54;3266th;73761;9977;674534632';
            //username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632 

$break = explode(';',$stats);
$name = "username";
$level = str_replace($break[0]);
$ratio = str_replace($break[1]);
$rank = str_replace($break[2]);
$kill = str_replace($break[3]);
$death = str_replace($break[4]);
$exp = str_replace($break[5]);
?>

[doni49]

Will it ALWAYS be formatted like this?  Have a look at the substring functions.  www.php.net/substr

 

Here's one portion:

$deaths = substr($var, stripos($var, "Deaths: "),stripos($var, "Experience:"));//<--should return Deaths: 9977
$deaths = substr($deaths, stripos($var, " "));  //  <--should return 9977

 

Here's a quick explanation of what's happening...

substr($var, 3, 6); //  <--will return the 3rd thru the 6th character (zero based--the first char is actually char zero)
stripos($var, "Deaths:  ");//<--will return the position within the variable where "Deaths:  " begins.

[phpSensei]

You need to add some sort of comma or ";"...

 

I took off the Username Level and stuff and put the data in seperate columns

 

25;1.54;3266th;73761;9977;674534632';

 

 

Try

 

<?php
$stats = 25;1.54;3266th;73761;9977;674534632';
            //username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632 

$break = explode(';',$stats);
$name = "username";
$level = str_replace($break[0]);
$ratio = str_replace($break[1]);
$rank = str_replace($break[2]);
$kill = str_replace($break[3]);
$death = str_replace($break[4]);
$exp = str_replace($break[5]);
?>

 

little error

 

<?php
$stats = '25;1.54;3266th;73761;9977;674534632';
            //username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632 

$break = explode(';',$stats);
$name = "username";
$level = str_replace($break[0]);
$ratio = str_replace($break[1]);
$rank = str_replace($break[2]);
$kill = str_replace($break[3]);
$death = str_replace($break[4]);
$exp = str_replace($break[5]);
?>

 

Donie, Its easier for him to format it like that anyways, he doesnt need to names like LEVEL, he can just add that himself later.

[doni49]

Donie, Its easier for him to format it like that anyways, he doesnt need to names like LEVEL, he can just add that himself later.

 

Agreed--that would be easier.  But since he's screen scraping, I'm guessing that he doesn't have access to the data to reformat it.  If he CAN reformat it, I'd go with your suggestion.  If he CAN'T, my suggestion would work well.

[phpSensei]

Donie, Its easier for him to format it like that anyways, he doesnt need to names like LEVEL, he can just add that himself later.

 

Agreed--that would be easier.  But since he's screen scraping, I'm guessing that he doesn't have access to the data to reformat it.  If he CAN reformat it, I'd go with your suggestion.  If he CAN'T, my suggestion would work well.

 

As I said you can just add it later, like right when you set the variables you can always "reformat"

 

$name = "username";
$level = "User Level:" . str_replace($break[0]);

[Ken2k7]

sean123,

This is a pain in the ass to write But I'll try.

 

<?php
function charAt($str, $pos) {
    return (substr($str, $pos, 1)) ? substr($str, $pos, 1) : -1;
}
function remBdrySp ($str) {
    $f = 0; $l = strlen($str)-1;
    while ($f < $l) {
        if (charAt(" ", $f) == -1 && charAt(" ", $l) == -1) break;
        if (charAt(" ", $f) != -1) $f ++;
        if (charAt(" ", $l) != -1) $l++;
    }
    return substr($str, $f, $l+1);
}
$daStr = "
username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632";
$name = substr($daStr, 0, strpos($daStr, " "));
$daStr = explode(": ", $daStr); $e = array();
for ($k=1; $k < count($daStr); k++) {
    $e[$k - 1] = substr($daStr[$k], 0, strrpos($daStr[$k], " "));
}
$level = $e[0];
$ratio = $e[1];
$rank = $e[2];
$kill = $e[3];
$death = $e[4];
$exp = $e[5];
?>

 

Ken

[sean123]

Thanks I'm using your code Ken2k7 but I'm having a problem. I added it to my code and when it grabs the users stats theres some spaces at the start before it says the username so your code doesn't work Theres 2 spaces at the beginning.

[redarrow]

use trim() function.........

[sean123]

Just tried it didn't work:

 

$txt = ereg_replace( '  ', '', $txt);

[doni49]

Just tried it didn't work:

 

$txt = ereg_replace( '  ', '', $txt);

 

That's not the trim function.  And run it on the car BEFORE the code above runs.

[sean123]

Doesn't work after I do it all it outputs is the name.

 

$daStr = trim($txt);

[thebadbad]

Wouldn't it be easier to just scrape the data properly, assigning the vars from the start? If that's what you're doing, that is.

[phpSensei]

Just tried it didn't work:

 

$txt = ereg_replace( '  ', '', $txt);

 

Mine Works, just that you need to insert the user data into the format i showed you.

 

<?php
$stats = '25;1.54;3266th;73761;9977;674534632';
            //username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632 

$break = explode(';',$stats);
$name = "username";
$level = str_replace($break[0]);
$ratio = str_replace($break[1]);
$rank = str_replace($break[2]);
$kill = str_replace($break[3]);
$death = str_replace($break[4]);
$exp = str_replace($break[5]);
?>

[sean123]

Can't use that code And is it possible to get what line the stat is on then store it in a variable? I've got it so it's like this:

 

  webmonster 

Level:

25

Ratio:

1.54

Rank:

4266 th

Kills:

73761

Deaths:

9977

Experience:

674534632

 

So if I wanted to put the number of kills in a variable I would put the line number somewhere and it'll store the number 73761.

[laffin]

use preg match

 

<?php
  header('Content-type: text/plain');
  $str="username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632";

  preg_match('@([^\s]+)\s*Level: (\d+)\s+Ratio:\s+([\d.]+)+\s+Rank:\s+(\d+).+?Kills:\s+(\d+)\s+Deaths:\s+(\d+)\s+Experience:\s+(\d+)$@',$str,$matches);
  print_r($matches);
?>
/* Output:
Array
(
[0] => username   Level: 25 Ratio: 1.54 Rank: 3266 th Kills: 73761 Deaths: 9977 Experience: 674534632
[1] => username
[2] => 25
[3] => 1.54
[4] => 3266
[5] => 73761
[6] => 9977
[7] => 674534632
)
*/

 

as u can see everything is captured

well, item 0 can be ignored, as that is just the matched string.

[sean123]

That doesn't work because of the 3 spaces at the start I tried:

 

$txt = str_replace("  ", "", $txt);

 

But it didn't work.

[laffin]

preg_match('@\s*([^\s]+)\s*Level: (\d+)\s+Ratio:\s+([\d.]+)+\s+Rank:\s+(\d+).+?Kills:\s+(\d+)\s+Deaths:\s+(\d+)\s+Experience:\s+(\d+)$@',$str,$matches);

 

\s represents white spaces (spaces tabs newlines)

 

[sean123]

I tried:

 

preg_match('@\s*\s*([^\s]+)\sLevel: (\d+)\s+Ratio:\s+([\d.]+)+\s+Rank:\s+(\d+).+?Kills:\s+(\d+)\s+Deaths:\s+(\d+)\s+Experience:\s+(\d+)$@',$str,$matches); //print_r($matches);

 

Didn't work If I send you a private message with the full source can you try it?