SkyD
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Posts posted by SkyD
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Kristabs, tnx...
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tas kods man ir jaliek ieksh, piem fails.php failaa?? .. iebaazu tur ieksh vinju, bet izmet: "Parse error: parse error, unexpected T_STRING in d:\lapa\m.php on line 3"
3line - create database reg;
Varbūt var kautkā sīkāk šitās štelles paskaidrot?? Būtu baig lab=]
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Vnk testējot ir šāda padarīšana:
<? $link = mysql_connect("localhost", "root", "") or die ("<b>Nevar savienoties ar datu baazi!</b>"); mysql_select_db ("reg") or die ("<b>Taadu datu baazi neatradu;pPP</b>"); ?> <?php // This could be supplied by a user, for example $firstname = 'fred'; $lastname = 'fox'; // Formulate Query // This is the best way to perform a SQL query // For more examples, see mysql_real_escape_string() $query = sprintf("SELECT firstname, lastname, address, age FROM friends WHERE firstname='%s' AND lastname='%s'", mysql_real_escape_string($firstname), mysql_real_escape_string($lastname)); // Perform Query $result = mysql_query($query); // Check result // This shows the actual query sent to MySQL, and the error. Useful for debugging. if (!$result) { $message = 'Invalid query: ' . mysql_error() . "\n"; $message .= 'Whole query: ' . $query; die($message); } // Use result // Attempting to print $result won't allow access to information in the resource // One of the mysql result functions must be used // See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc. while ($row = mysql_fetch_assoc($result)) { echo $row['firstname']; echo $row['lastname']; echo $row['address']; echo $row['age']; } // Free the resources associated with the result set // This is done automatically at the end of the script mysql_free_result($result); ?>Un izmet šo:
Invalid query: Table 'reg.friends' doesn't exist Whole query: SELECT firstname, lastname, address, age FROM friends WHERE firstname='fred' AND lastname='fox'
Kā ir jaizveido tā reg.friends tabula?? nu reg man ir tā pati mape, kas atrod MySQL datos.. Plz pastāstiet, jo gribu apgūt to lietu ar MySQL tabulām..
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Izmet tieshi to pashu... !! Kas par probleemu?
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Hey.. Uzinstaleeju MySQL 4.0.18 versiju.. Veru valjaa ../bin/winmysqladmin.exe failu, tur piedavaa ievadiit username un passwordu, ievadu viss OK..
Veru valjaa PHP editoru, rakstu iekshaa
$link = mysql_connect("localhost", "Logins", "Parole") or die ("Could not connect to MySQL"); mysql_select_db ("test") or die ("Could not select database");saglabaaju index.php ... eju http://localhost izmet
Warning: mysql_connect(): Access denied for user: 'Logins@localhost' (Using password: YES) in d:\lapa\index.php on line 2
Could not connect to MySQL
Kas par probleemu??
... Lūdzu palīdziet ...
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a kaadam nav kaads vienkaarsh style skripts?? Vajag, taa: ja izveelas STILS 1, lai uzstaada kaa default 1.css, ja izveelas STILS 2, lai uzstaada kaa default 2.css .. PLZ kaads uzstaisat..
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Tagad ir shitaa, bet vienalga neiet:
style.php
<? $style = $_GET['style']; switch($style) { case '1': $style = 'styles/1.css'; setcookie("1", $style, time()+3600); break; case '2': $style = 'styles/2.css'; setcookie("2", $style, time()+3600); break; } ?> <a href="?style=1">Stils 1</a> <a href="?style=2">Stils 2</a><br>index.php
<link rel="stylesheet" href="styles/'.echo $_COOKIE['$style'].'.css" type="text/css">
Kas vareetu buut par kljuudu, plz notesteejiet uz sevim, ja zinat kur kljuuda tad izlabojiet.. =]
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Tag viss shtokos.. bet plz pasakat kapeec man neiet shitas kods??
style.php
<? session_start(); $style = $_GET['style']; switch($style) { case 1: $_SESSION['style'] = 1; break; case 2: $_SESSION['style'] = sarkans; break; } ?>index.php
<link rel="stylesheet" href="styles/'.$SESSION['style'].'1.css" type="text/css">
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nu taa jau izdariju, bet vinjsh neiet vienalga..
<table style="width: 15%;"> <? $f = file('fails.txt'); $i = 0; foreach($f as $rinda) { echo '<tr style="color: ' . ($i ? 'gray' : 'red') . '">'.trim($rinda).'</tr>'; $i = ($i+1)&1; } ?> </table> -
mjaa... es nepareizi iisteniibaa izteicos.. mana vaina.. Man vajag, lai taas rindinjas iekraasojas tabulaa..
Vareetu jau taa:
<table> <tr class="nepara"><td>...</td></tr> <tr class="para"><td>...</td></tr> <tr class="nepara"><td>...</td></tr> <tr ... utt ...</tr> </table>
..bet man nav noteikts skaits ar rindinjaam..
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Sveiki! Man ir txt fails kuraa iekshaa ir vairaakas rindinjas, teiksim:
aabols
apelsiins
banaans
cukurs
sviests
.. Taatad, man vajag taa, izvadot sho txt failu lapaa, lai iekraasojas katra rindinja savaa kraasaa, piem paara rindinjas zaljaa, nepaara zilaa.. Kaa to izdariit??
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Nu paša neuzmanība, izskatās ka ir pati lielākā kļūda :D~
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parrymason ja tev veel daudz tutotriāļi priekšā, tad pasaki kādu no tiem man arī ;)
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a kaada shitam saistiiba ar CTRL + C, un CTRL + V? :rolleyes:
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nu ir taada probleema, uzzinastaleeju jau nokonfigureetu apache_php_mysql ...
Ierubiju apaci, paraadaas ->
Don't close this windows, while Apache is running
Fenster nicht schliessen, solange Apache laeuft
If you have installed APACHE in an other folder change this entry please.
[Wed Mar 23 15:29:34 2005] [warn] pid file c:/server/apache/logs/httpd.pid overw
ritten -- Unclean shutdown of previous Apache run?
Apache/1.3.29 (Win32) PHP/4.3.4 running... <-- peectam rakstu http://localhost, bet neiet, kas pa lietu?
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ar switch($id) - arii laikam var.. ;)
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tieš tā šitas te ir normālākais forums.. nu okey, pats kko bīdīšu, ja vajadzēs kkas help tad teikšu :)
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Nu baigi žēl :P~~
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Hey.. Iedomājamies ka man ir komentaaru forma ar name, email, text.. nju a vinju var ja saglabaat txt failaaa, un tad izvadīt vinju virs tās komentāru formas.. Bet vajadzētu arī tā, ja nav komenti lai pasaka, ka to nav:)~ varētu man uztaisīt tāda veida? Nju formu es pats uzmestu
Admin sistema
in PHP skripti
Posted
Ir shitas kods, kursh ieraksta iekshaa Loginu un paroli:
<? $hostname="localhost"; $mysql_login="root"; $mysql_password=""; if ($db = mysql_connect("$hostname", "$mysql_login" , "$mysql_password")) { echo "<br>"; } else { echo "<br>"; } $q = mysql_query("create database admin"); if($q){ echo "<br>"; } if (!(mysql_select_db("admin",$db))) { echo "nav kaartiibaa ar DB<br>"; } else { $q= mysql_query("create table admini ( `logins` varchar (250) NULL , `parole` varchar (250) NULL )")or die(mysql_error()); if($q){ echo "<br>";} $q == mysql_query("INSERT INTO admini (logins,parole) VALUES ('$logins','$parole')")or die(mysql_error()); if($q) { echo "Pievienots!"; } } ?>Šitas kods ir pareizs???
Nu tag man ir http://localhost/admin un ieksh tur ir divi lauki, loginam un parolei.. Vajag taa, nospiezhot uz pogas "Pievienoties", lai kods no MySQL izvelk loginu un paroli un pasaka vai tam lietotaajam ir pareiza vai nepareiza .. Ja nepareiza izmet: "Nepareizi", ja pareiza iet uz main.php
... Plss, paliidzat!