didy

labosanas funkcija ~ 100ls :) jocigi! (stundas darbs) "iespeeju noraadiit vai raksts ir publiks vai nee" neuzskatu ka tas butu vajadzigs!

Nav, bet doma ir ieviest sadu funkciju!

Paldies!

Kada vareetu but cena sadai lapai? http://test8.sytes.net/mtl_kl/ Ietilpst: Panelis ar rakstu pievienosanu - dzesanu!

<? require ($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php"); $connection = @mysql_connect($db_host, $db_user, $db_password); if (!$connection) { die ('Could not connect: ' . mysql_error()); } mysql_select_db("latbit_db", $connection); $query = "SELECT id, name FROM upload"; $result = mysql_query($query) or die('Error, query failed'); if(mysql_num_rows($result) == 0) { echo "Database is empty <br>"; } else { while(list($id, $name) = mysql_fetch_array($result)) { ?> <img src="index.php?id=<?=$id;?>"/><br> <?php } } mysql_close($connection); ?> un <?php require ($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php"); $connection = @mysql_connect($db_host, $db_user, $db_password); if (!$connection) { die ('Could not connect: ' . mysql_error()); } mysql_select_db("latbit_db", $connection); $result = mysql_query("SELECT * FROM upload_txt"); echo "<table border='1'> <tr> <th>id</th> <th>title</th> <th>description</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['autoID'] . "</td>"; echo "<td>" . $row['title'] . "</td>"; echo "<td>" . $row['description'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($connection); ?> Ka lie apvienoju sos divus skriptus? Neko daudz no php nezinu, tapec busu pateiciigs ja izkaidrotu cilveciski. ja pareizi domaju ta man vajag apvienot while(list($id, $name) = mysql_fetch_array($result)) ar while($row = mysql_fetch_array($result))

$data = addslashes(fread(fopen($_FILES["file"]["type"], "r"), $_FILES["file"]["size"])); echo "Upload: " . $_FILES["file"]["name"] . "<br />"; Upload: 4.jpg

Warning: fopen(image/png) [function.fopen]: failed to open stream: No such file or directory in E:\webroot\uploader.php on line 23 Warning: fread(): supplied argument is not a valid stream resource in E:\webroot\uploader.php on line 23 Field 'name' doesn't have a default value "No such file or directory" direktorija ir!

<?php // failu parbaude (extensiju parbaude + max lielums if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/jpg") || ($_FILES["file"]["type"] == "image/png") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 20000)) { // Pievienosanas DB require ($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php"); $connection = @mysql_connect($db_host, $db_user, $db_password) or die ("error connecting"); mysql_select_db($db_name, $connection); $data = addslashes(fread(fopen($_FILES["file"]["type"], "r"), $_FILES["file"]["size"])); // Faila ievietosana tabula $query = mysql_query("INSERT INTO upload_img (file) VALUES ('$data')") or die("Can't Perform Query"); mysql_query($query, $connection) or die (mysql_error()); // Augsupladeta faila info (nosaukums, lielums) echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"]) . " Kb<br />"; } // Ja vails neatbilst (izvada else) else {echo "Nepareizs fails";} ?> Warning: fopen(image/png) [function.fopen]: failed to open stream: No such file or directory in E:\webroot\uploader.php on line 23 Warning: fread(): supplied argument is not a valid stream resource in E:\webroot\uploader.php on line 23 Can't Perform Query DB izveidoju tabulu upload_img CREATE TABLE `latbit_db`.`upload_img` ( `autoID` INT( 64 ) UNSIGNED NOT NULL AUTO_INCREMENT , `name` VARCHAR( 64 ) NOT NULL , `size` INT NOT NULL , `type` VARCHAR( 64 ) NOT NULL , `file` BLOB NOT NULL , PRIMARY KEY ( `autoID` ) ) ENGINE = InnoDB Kur kluda?

query: 1 query2: query3:

izvada: 1 (ciparu 1) ja protams pareizi lietoju print_r ($query); Neparzinu tik labi php lai uzmestu kodu pats!

piemetu: mysql_query($query, $connection) or die (mysql_error()); ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

<?php if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/jpg") || ($_FILES["file"]["type"] == "image/png") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 20000)) { if ($_FILES["file"]["error"] > 0) { echo "Error: " . $_FILES["file"]["error"] . "<br />"; $mysqlp = addslashes(fread(fopen($_FILES["file"]["type"], "r"), $_FILES["file"]["size"])); } else { require ($_SERVER["DOCUMENT_ROOT"]."/folder/configs.php"); $connection = @mysql_connect($db_host, $db_user, $db_password) or die ("error connecting"); mysql_select_db($db_name, $connection); if (isset($_POST['mysqlp'])){ mysql_query("INSERT INTO img_upload (Image) VALUES ('$mysqlp')") or die("Can't Perform Query"); } echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Stored in: " . $_FILES["file"]["tmp_name"] . "<br /><br />"; } } else { echo "Invalid file"; } ?> Samonteju sadu skriptinu. Itka nekadas kludas neuzrada, bet DB nesaglabajas! KO izdariju nepareizi?

If($Picture != "none") { $PSize = filesize($Picture); $mysqlPicture = addslashes(fread(fopen($Picture, "r"), $PSize)); nomainot $Picture mainigo uz $_FILES['Picture'] - reakcija nekada!

<?php if (isset($_FILES['Picture'])) { If($Picture != "none") { $PSize = filesize($Picture); $mysqlPicture = addslashes(fread(fopen($Picture, "r"), $PSize)); } require ($_SERVER["DOCUMENT_ROOT"]."/folder/configs.php"); $connection = @mysql_connect($db_host, $db_user, $db_password) or die ("error connecting"); mysql_select_db($db_name, $connection); mysql_query("INSERT INTO img_upload (Image) VALUES ('$mysqlPicture')") or die("Can't Perform Query"); } else { echo"You did not upload any picture"; } ?> Joprojam neka! Izmet joprojam else - "You did not upload any picture" Kluda "equire ($_SERVER["DOCUMENT_ROOT"]."/folder/configs.php");" nebus vainiga, ka ari "upload.php" Man nikis pirms postesanas nomainit foldera un lapas nosaukumu! Nejauta kadeel! Man vaig uploudu ar DB! Meklesu vel!

<form enctype="multipart/form-data" action=" upload.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="250000" /> Send this file: <input name="Picture" type="file" /> <input type="submit" value="Send File" /> </form> if (isset($Picture)) { If($Picture != "none") { $PSize = filesize($Picture); $mysqlPicture = addslashes(fread(fopen($Picture, "r"), $PSize)); } require ($_SERVER["DOCUMENT_ROOT"]."/fold/configs.php"); $connection = @mysql_connect($db_host, $db_user, $db_password) or die ("error connecting"); mysql_select_db($db_name, $connection); mysql_query("INSERT INTO img_upload (Image) VALUES ('$mysqlPicture')") or die("Can't Perform Query"); } else { echo"You did not upload any picture"; } ?> Kadel nesanak augsupladet datubaze ? Ko esmu nepareizi izdarijis? izvada man else - "You did not upload any picture"

Ir izveidots upload kodins. Augsupladejot bildi vina saglabajas noteiktaja mape! Jautajum: Ka lai izveidoju lai augsupledaatais fails saglabaajas un nopostojas weba! edit: PS. atradu info! Paldies, tiksu gala!

Pateicos! Izradas ka biju darijis pareizi tikai nepamanijis, ka Undefined index problema parlekusi uz nakamo $var Izlaboju kludas! Paldies!

Foruma vieta varetu ielikt saiti uz php.net un google.lv un varbuut uz wiki! Tas pats vien butu! Vareji jau njemt vera kad teicu "nevareju isti saprast ka parbaudiit ar isset vai empty" Meginaju ka paradiits isset() - nekada rezultata! Tadel jau nacu uz forumu pec palidzibas! Bet paldies! Gan jau tiku gala!

nesaubigi!

Forma ir noposteta! Ta ka esmu iesacejs php, nevareju isti saprast ka parbaudiit ar isset vai empty! Varbut var lugt isu uzmetumu!

<?php require ($_SERVER["DOCUMENT_ROOT"]."/folder/config.php"); $connection = @mysql_connect($db_host, $db_user, $db_password) or die ("error connecting"); mysql_select_db($db_name, $connection); $name = $_POST["txt_name"]; $len = strlen ($name); //Only write to database if there's a name if ($len > 0) { $email = $_POST["txt_email"]; $comment = $_POST["txt_comment"]; $date = time(); $query = "INSERT INTO guestbook (autoID, name, email, comment, date_auto) VALUES (NULL, '$name', '$email', '$comment', '$date')"; mysql_query($query, $connection) or die (mysql_error()); } ?> Problema: Notice: Undefined index: txt_name in E:\webroot\guestbook.php on line 5 ($name = $_POST["txt_name"];) kur kluda?

OS: WinXP Uzstadiju: Apache2.2.8, php5.2, mysql 5.0 Problema: nevaru piesaistiit mysql db parbaude - ieks index.php <?php $link = mysql_connect('localhost', 'lietotajvards', 'manaparole'); if (!$link) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_close($link); ?> Izmet man: Fatal error: Call to undefined function mysql_connect() in E:\folderis\index.php on line 8 meginaju citu variantu: <? require ($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php"); $connection = @mysql_connect($db_host, $db_user, $db_password) or die ("error connecting"); echo "connection made"; ?> ar papildus failu kurs norada $db_host, $db_user, $db_password Sini gadijuma neuzrada pilniba neko! KO es varetu but aizmirsis nokonfiguret? Vai varbut ko nepareizu izdarijis!

Paldies, sanaca! Problema bija ar id="textarea1

Nekad nesu lietojis JS! Esmu php iesacejs! Sit mani nost bet es nesaprotu! <form action="index.php" method="post"> <table width="390" border="0" height="10" align=center background="/style/add_style_green.png"> <tr> <td> <!-- attiecigas bildes --> <img src="/style/bloods.png" /> <img src="/style/blood.png"/> <img src="/style/underline.png" /> <img src="/style/strip.png" /> <img src="/style/gas_strip_os_83.png" /> <img src="/style/link.png" /> <img src="/style/img.png" /> <img src="/style/gas_strip_os_83.png" /> <a href="" border="0"><img src="/style/info68.png" border="0" /></a> <!-- attiecigas bildes END --> </td> </tr> </table> <table align=center width=396 border="0"> <tr> <td>Niks:<font color="red">*</font> <input type="input" name="vards" size="15" style="background-color:#FBFBFB"></td> </tr> <tr> <td colspan=2>Komentars:<font color="red">*</font><br><textarea rows=4 cols=45 name="teksts" style="background-color:#FBFBFB"></textarea></td> </tr> <tr> <td><input type=submit value="Pievienot" name="aiziet"> <input type=hidden value="<?php echo $_SERVER['REMOTE_ADDR']; ?>" name="aipii"> <input type=hidden value="<?php echo time() ?>" name="laiks"></td> </tr> </form> </table> Neizdodas panakt velamo efektu, tapec palugsu lai Jus palabojat manu source!

onclick="document.getElementById('textarea_id').value += '[smile]';" Nesanak ar onclick! Hmm.. laikam k-ko nepareizi daru!