Anonymní profil Cecilconrad – Programujte.com

ok zda se, ze problem vyresen, dik na spravne nasmerovani na regularni vyrazy, reseni sem nasel

preg_match("|\[$tagName\](.*)\[/$tagEnd\]|Ui", $text, $part);

a ted v $part[1] je \\1

v $part[2] je \\2

atd.

#2 KIIV
funguje me jak \\1 tak \\2

v \\1 je ulozeno URL v \\2 text odkazu

jenze nevim jak tyhle hodnoty muzu predat funkci, neco jako

$promena = "\\2";

filtr($promena);

nebo primo filtr("\\2")

byl bych vdecny za nejaky odkaz na manul, kde je o tom co je vubec \\2 zac neco napsane

Mam tento kusek kodu, ktery predelava odkazy z formatu [URL=adresa]neco[/URL] na html

$tagArray['URL=(.*)'] = array('open'=>'<a href="','close'=>'">\\2</a>');

foreach($tagArray as $tagName=>$replace)
{
         $tagEnd=preg_replace('/\W/Ui','',$tagName);
         $text = preg_replace("|\[$tagName\](.*)\[/$tagEnd\]|Ui","$replace[open]\\1$replace[close]",$text);
}

ale jedna se me jen o to \\2, o co se jedna ? funguje me to spravne, nahradi se tim ten text odkazu, ale problem je v tom, ze bych to potreboval nejak dostat do promene abych to mohl prohnat funkci, takovy filtr, ale nevim ani jak to pojmenovat a tezko takto k tomu najit nejaky manual nebo neco zkousel sem

close'=>'">'.filtr('\\2').'</a>'

ale tady uz se preda fakt jen ten retezec '\\2' a uz se nenahradi tim cim ma

zkousel sem hledat, ale vyhledavace najdou tisice odkazu ve kterem se vyskytuje dvojka, ale specialni znaky \ ignoruje

jeste jedna moznost je vytvorit si SQL dotaz pomoci PHP a pro kazdeho jezdce vygenerovat SQL dotaz a pomoci UNION vsecky spojit, vznikne hodne dlouhy dotaz, ktery se docela dlouho provadi, ale slo by tak treba zpetne se podivat kdo vedl treba po druhem zavode, takovych dotazu ale nebude moc takze by se to dalo prezit, kdyz se nekdo raz za cas bude chtit zpetne podivat tak si chvili pocka

Tak sem to nakonec vyresil pres to PHP, do tabulky Drivers jsem si udelal sloupecek TOP7 a po kazdem zavode to PHP skriptem prepocitam coz chvili sice trva, ale zato potom je uz vypis pekelne rychly, kdyz se nemusi nic pocitat. Aj kdybych to nakonec nejak vyresil v mysql tak by kazdy vypis musel znova pocitat a trvalo by to dlouho.

To by mozna slo, kdybych to nemel takto, a takto to mam proto ze ten vnoreny dotaz vraci jen 7 nejlepsich vysledku, kterych tam muze byt az 11, potrebuji cecist ale jen tech 7

SELECT SUM(Points) AS Celkem FROM (vnoreny dotaz);

Mam dotaz, ktery funguje a scita vsechny vysledky pro kazdeho jezdce:

SELECT Drivers.Name,SUM(Points) AS Celkem FROM
(
SELECT 
ZavodID AS Zavod, 
DriverID AS Driver,
Position,
IF(Position>0 AND Position<11,(SELECT Points FROM Points2011 WHERE ID=Position LIMIT 1)*(SELECT Coefficient FROM Calendar2011 WHERE ID=Zavod LIMIT 1),0) AS Points 
FROM Zavod2011
) AS Body 
INNER JOIN Drivers ON Driver=Drivers.ID
GROUP BY Driver 
ORDER BY Celkem DESC

Idea byla takovato

ten vnoreny dotaz ma jeste jeden vnoreny dotaz, ktery pro konkretniho jezdce vybere vsechny vysledky, tyto nadrazeny dotaz secte, scitat to scita spravne, kdyz misto SuperID zadam ID jednoho jezdce, jak to ale udelat aby se tam to ID stridalo pro vsechny jezdce z tabulky Drivers, tal abych to potom podle tohoto souctu mohl seradit nevim

SELECT SUM(Points) AS Celkem FROM
(
SELECT
Calendar2011.ID AS Zavod,
IF(Position>0 AND Position<11,(SELECT Points FROM Points2011 WHERE ID=Position LIMIT 1)*(SELECT Coefficient FROM Calendar2011 WHERE ID=Zavod LIMIT 1),0) AS Points
FROM Zavod2011
INNER JOIN Calendar2011 ON Zavod2011.ZavodID=Calendar2011.ID
INNER JOIN Drivers ON Zavod2011.DriverID=Drivers.ID
WHERE Drivers.ID=SuperID ORDER BY Points DESC LIMIT 7
) AS Jeden

ID kazdeho jezdce, kdyz to prepisu bez toho vnoreneho dotazu

SELECT ID AS SuperID, (vnoreny) AS Celkem FROM Drivers ORDER BY Celkem DESC;

a v tom vnorenem dotazu scitam tech 7 nejlepsich vysleku pro kazdeho jezdce, ty hledam podle SuperID

Problem asi bude ze SuperID je pouze alias, zkousel jsem to i tak, ze jsem si ID jezdce ulozil do promene a vyhledaval podle ni, ale vracelo to same NULL zaznamy.

struktura:

CREATE TABLE `Calendar2011` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `Name` varchar(100) NOT NULL,
  `Coefficient` float NOT NULL DEFAULT '1',
  PRIMARY KEY (`ID`)
);

CREATE TABLE `2011Points` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `Points` int(11) NOT NULL,
  PRIMARY KEY (`ID`)
);

CREATE TABLE `Zavod2011` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `DriverID` int(11) NOT NULL,
  `TeamID` int(11) DEFAULT NULL,
  `ZavodID` int(11) DEFAULT NULL,
  `Position` int(11) DEFAULT NULL,
  PRIMARY KEY (`ID`)
);

CREATE TABLE `Drivers` (
  `ID` int(11) NOT NULL,
  `Name` varchar(100) NOT NULL,
  PRIMARY KEY (`ID`)
);

Mam databazi zavodu a v tabulce mam seznam jezdcu a jejich vysledku (pozici a vypocitam si kolik ziskali bodu), potrebuji z teto tabulky pro kazdeho jezdce vybrat 7 nejlepsich vysledku a ty secist a seradit

Povedlo se me udelat dotaz ktery tohle umi pro jednoho konkretniho jezdce

SELECT SUM(Points) AS Celkem FROM
(
SELECT
Calendar2011.ID AS Zavod,
IF(Position>0 AND Position<11,(SELECT Points FROM Points2011 WHERE ID=Position LIMIT 1)*(SELECT Coefficient FROM Calendar2011 WHERE ID=Zavod LIMIT 1),0) AS Points
FROM Zavod2011
INNER JOIN Calendar2011 ON Zavod2011.ZavodID=Calendar2011.ID
INNER JOIN Drivers ON Zavod2011.DriverID=Drivers.ID
WHERE Drivers.ID=54453 ORDER BY Points DESC LIMIT 7
) AS Jeden

kdyz se ale snazim z druhe tabulky kde mam seznam jezdcu spoustet tento prikaz pro kazde ID tak mam problem

SELECT ID AS SuperID,
(
SELECT SUM(Points) AS Celkem FROM
(
SELECT 
Calendar2011.ID AS Zavod,
IF(Position>0 AND Position<11,(SELECT Points FROM Points2011 WHERE ID=Position LIMIT 1)*(SELECT Coefficient FROM Calendar2011 WHERE ID=Zavod LIMIT 1),0) AS Points
FROM Zavod2011
INNER JOIN Calendar2011 ON Zavod2011.ZavodID=Calendar2011.ID
INNER JOIN Drivers ON Zavod2011.DriverID=Drivers.ID
WHERE Drivers.ID=SuperID ORDER BY Points DESC LIMIT 7
) AS top7
) AS Celkem
FROM Drivers ORDER BY Celkem DESC

#1054 - Unknown column 'SuperID' in 'where clause'

Jedine co me napada je udelat si tabulku prubeznych souctu, v PHP bych mel script ktery by ten muj dotaz spustil pro kazdeho jezdce, ale radeji bych to primo v sql.

Tak vyresene, tabulky sem si exportoval, vymazal je z databaze pak je tam znova vlozil a uz to funguje. Zajimavy problem, diky moc za rady.

*tak to funguje tak jak ma

No je v tom prikazu par syntaktickych chyb, nekde jsou navic carky nekde chybi uvozovky, u druhe tabulky sem nechal puvodni nazvy je to tim jak sem to menil aby tam nebyly prebytecne data a sloupce ktere nejsou potreba, ale co je zajimavejsi, ze kdyz jsem si ten prikaz spustil na teto zjednodusene verzi tak to tak jak ma. Pritom moje pracovni tabulka je v podstate uplne stejna.

Stejny vysledek, nahodim sem svoje data

CREATE TABLE IF NOT EXISTS `Tabulka` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `Hodnota` int(11) NOT NULL,
  `ZaznamID` int(11) NOT NULL,
  PRIMARY KEY (`ID`),
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=642 ;

CREATE TABLE IF NOT EXISTS `Zaznamy` (
  `ID` int(11) NOT NULL AUTO_INCREMENT,
  `Nazev` varchar(100) NOT NULL,
  PRIMARY KEY (`ID`),
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=12 ;

INSERT INTO `Tabulka` (`ID`, `Hodnota`, `ZaznamID`) VALUES
(1, 10, 1),
(2, 6, 1),
(3, 8, 1),
(4, 5, 1),
(5, 6, 1),
(6, 5, 1),
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(18, 9, 1),
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(590, 6, 11),
(591, 7, 11),
(592, 8, 11),
(593, 3, 11),
(594, 5, 11),
(595, 1, 11),
(596, 13, 11),
(597, 6, 11),
(598, 11, 11),
(599, 16, 11),
(600, 7, 11),
(601, 5, 11),
(602, 9, 11),
(603, 9, 11),
(604, 8, 11),
(605, 5, 11),
(606, 5, 11),
(607, 1, 11),
(608, 7, 11),
(609, 8, 11),
(610, 8, 11),
(611, 10, 11),
(612, 5, 11),
(613, 6, 11),
(614, 5, 11),
(615, 5, 11),
(616, 5, 11),
(617, 3, 11),
(618, 5, 11),
(619, 29, 11),
(620, 16, 11),
(621, 6, 11),
(622, 8, 11),
(623, 8, 11),
(624, 3, 11),
(625, 8, 11),
(626, 3, 11),
(627, 16, 11),
(628, 10, 11),
(629, 2, 11),
(630, 6, 11),
(631, 6, 11),
(632, 13, 11),
(633, 6, 11),
(634, 6, 11),
(635, 24, 11),
(636, 17, 11),
(637, 9, 11),
(638, 10, 11),
(639, 5, 11),
(640, 1, 11),
(641, 10, 11);

INSERT INTO `Kalendar` (`ID`, `Jmeno`) VALUES
(1, 'a'),
(2, 'b),
(3, 'c'),
(4, 'd'),
(5, 'e'),
(6, 'f'),
(7, 'g'),
(8, 'h'),
(9, 'i'),
(10, 'j'),
(11, 'k');

poustim to na localu, zkusim to jeste na nejakem jinem serveru, treba tam to pojede, ale jestli moze byt chyba v nejakem nastaveni serveru ktere sem ale nemenil netusim

Vraci me to uplne to stejne co ten muj, tedy:

iid | col_pocet
1  |       10
2  |        3
3  |        3
4  |        3
5  |        3
6  |        3
7  |        3
8  |        3
9  |        3
10 |       3
11 |       3

kdezto kdyz to poustim po jednom timto prikazem s tim ze menim ZaznamID u WHERE

SELECT `Hodnota` FROM `Tabulka` WHERE `ZaznamID`=1 GROUP BY `Hodnota` ORDER BY COUNT(`Hodnota`) DESC LIMIT 1;

tak to vraci pro ZaznamID

1 -> 10

2 -> 11

3 -> 9

4 -> 10

5 -> 9

6 -> 10

7 -> 12

8 -> 11

9 -> 13

10 -> 12

11 -> 11

Takze prvni radek je spravne a pak ostatni jsou vsecky stejne a spatne, nedokazu si to nijak vysvetlit, ale kazdopadne dik za snahu. :)

Takze mam tabulku `Tabulka` se sloupcema `ZaznamID`  a `Hodnota` typu integer
tento prikaz vypise ktera hodnota se zaznamem s ID 1 se v tabulce vyskytuje nejcasteji
SELECT `Hodnota` FROM `Tabulka` WHERE `ZaznamID`=1 GROUP BY `Hodnota` ORDER BY COUNT(`Hodnota`) DESC LIMIT 1;

Druha tabulka `Zaznam` obsahuje sloupec `ID`
tento prikaz by mel udelat statistiku nejcastejsiho vyskytu hodnot pro vsechny zaznamy
SELECT `ID`,
(
SELECT `Tabulka.Hodnota` FROM `Tabulka` WHERE `Tabulka.ZaznamID`=`Zaznam.ID` GROUP BY `Tabulka.Hodnota` ORDER BY COUNT(`Tabulka.Hodnota`) DESC LIMIT 1
) AS `Hodnota`
From `Zaznamy`;

Prvni radek vrati jeste spravne, ale vsechny ostatni jsou vsecky stejne a vetsinou spatne, pritom pokud ten prvni prikaz poustim po jednom se stejnym ID tak vraci vzdy prvni spravnou hodnotu, nepremyslejte nad tim jestli neni spatne navrzena struktura tabulek, ve skutecnosti je tam mnohem vic sloupcu a dat, ale pro rychlejsi pochopeni jsem to cele hodne zjednodusil. V cem moze byt chyba ?

Prosím vás, už mám ten kód, bohužel mám někde chybu, protože při kontrole ze souboru mi to hodí vždy, že slovo není ve slovníku :(

#include <stdio.h>

#include <stdlib.h>

#include <string.h>



int kontrola (char *slovo_s)

{

	FILE *f;

	f = fopen ("slova.txt", "r");

	char slovo[256];

	

	int i;

	while (fgets (slovo, sizeof (slovo), f) != NULL)

	{

		int znak = 0;

		for (i = 0; i < sizeof(slovo); i++)

		{

			if (slovo[i] == ';')

			{

				break;

			}

			

			znak++;

		}

	

	char slovo_t[znak];

	for (i = 0; i < znak; i++)

	{

		slovo_t[i] = slovo[i];

	}

	

	if (strcmp (slovo_s, slovo_t) == 0)

	{

		fclose(f);

		return 1;

	}

	}

	fclose (f);

	return 0;

}



int main (int argc, char** argv)

{

	char slovo[50], zac[10];

	char p;

	int error = 0;

	

	printf ("Slovni fotbal\n");

	printf ("Zadej pismeno, jimz se bude zacinat:");

	scanf ("%s", zac);

	

	p = zac[0];

	

	int hrac = 1;

	do

	{

		printf ("Hrac %d: Zadej slovo:", hrac);

		scanf ("%s", slovo);

		if (p != slovo[0])

		{

			printf ("Slovo nezacina znakem '%c' \n.", p);

			error++;

		}

		else if (kontrola (slovo) == 0)

		{

			printf ("Slovo neni ve slovniku.\n");

			error++;

		}

		

		if (error > 0)

		{

			printf ("Vitezem je hrac %d.\n", hrac ? 2:1);

			break;

		}

		else

		{

			p = slovo[strlen(slovo)-1];

			if (hrac == 1)

			{

				hrac++;

			}

			else

			{

				hrac = 1;

			}

		}

	} while (error <= 0);

	return (EXIT_SUCCESS);

}