PHP - Search Form Returns An Error When Entering An Apostrophie
The dreadful apostrophie problem... This search form returns an error whenever searching with an apostrophie (')
Here's the code on the form (html) <td align="center" width="135"><form method="post" action="srch_advert.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> <td align="center" width="135"><form method="post" action="srch_details.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> <td align="center" width="135"><form method="post" action="srch_artist.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> <td align="center" width="135"><form method="post" action="srch_track.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> and heres the code on srch_advert.php if ($search) // perform search only if a string was entered. { mysql_connect($host, $user, $pass) or die ("Problem connecting to Database"); $srch="%".$search."%"; $query = "select * from tvads WHERE advert LIKE '$srch' ORDER BY advert, year DESC, details ASC LIMIT 0,30"; $result = mysql_db_query("cookuk_pn", $query); if(mysql_num_rows($result)==0) { print "<h2>Your search returned 0 Results</h2>"; } else if ($result) { Similar TutorialsJust take a look at the code.. idk whats wrong. <?php // Get the search variable from URL $var = @$_GET['q'] ; $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit=10; // check for an empty string and display a message. if ($trimmed == "") { echo "<p>Please enter a search...</p>"; exit; } // check for a search parameter if (!isset($var)) { echo "<p>We dont seem to have a search parameter!</p>"; exit; } //connect to your database ** EDIT REQUIRED HERE ** mysql_connect("localhost","root",""); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("school") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "select * from the_table where 1st_field like \"%$trimmed%\" order by 1st_field"; // EDIT HERE and specify your table and field names for the SQL query $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0) { echo "<h4>Results</h4>"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; echo "<p><a href=\"http://www.google.com/search?q=" . $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a> to try the search on google</p>"; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } // get results $query .= " limit $s,$limit"; $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results"; $count = 1 + $s ; // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["1st_field"]; echo "$count.) $title" ; $count++ ; } $currPage = (($s/$limit) + 1); //break before paging echo "<br />"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< Prev 10</a>  "; } // calculate number of pages needing links $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division if ($numrows%$limit) { // has remainder so add one page $pages++; } // check to see if last page if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { // not last page so give NEXT link $news=$s+$limit; echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "<p>Showing results $b to $a of $numrows</p>"; ?> its supposed to search my databases i guess.. i need a site searcher if anyone wanna make me post it please.. or post a tut In the following code: echo $text[2] returns the word at that index but echo $key returns a blank. Can someone explain why is array-search not working in this indexed array? It does work when I test it with a key => value type array. <?php $myfile = fopen("test.txt", "r") or die("Unable to open file!"); while(!feof($myfile)) { $text[] = fgets($myfile); } fclose($myfile); $word = $_POST['word']; $key = array_search($word, $text); echo $key ?> ugh... I'm a total PHP nub and I'm having trouble with: Code: [Select] $search_by = $_POST['search_by']; $search = $_post['search']; $dbc = mysqli_connect('xx', 'artofwarsomerset', 'xx', 'artofwarsomerset') or die ('Error connecting to MySQL server'); $query = "SELECT * FROM players WHERE '$search_by' = '$search' "; $result = mysqli_query($dbc,$query) or die("Error: ".mysqli_error($dbc)); echo "<table><tr><td>Player</td><td>city</td><td>alliance</td><td>x</td><td>y</td><td>other</td><td>porters</td><td>conscripts</td><td>Spies</td><td>HBD</td><td>Minos</td><td>LBM</td><td>SSD</td><td>BD</td><td>AT</td><td>Giants</td><td>Mirrors</td><td>Fangs</td><td>ogres</td><td>banshee</td></tr>" ; while ($row = mysqli_fetch_array ($result)) { echo '<tr><td> $row['player'] </td>'; echo '<td> . $row['city']</td>'; echo '<td> . $row['alliance']</td>'; echo '<td> . $row['x']</td>'; echo '<td> . $row['y']</td>'; echo '<td> . $row['other']</td>'; echo '<td> . $row['porter']</td>'; echo '<td> . $row['cons']</td>'; echo '<td> . $row['spy']</td>'; echo '<td> . $row['hbd']</td>'; echo '<td> . $row['mino']</td>'; echo '<td> . $row['lbm']</td>'; echo '<td> . $row['ssd']</td>'; echo '<td> . $row['bd']</td>'; echo '<td> . $row['at']</td>'; echo '<td> . $row['giant']</td>'; echo '<td> . $row['fm']</td>'; echo '<td> . $row['ft']</td>'; echo '<td> . $row['ogre']</td>'; echo '<td> . $row['banshee']</td></tr></table>'; } ?> Error shows up on line 35 but I'm not sure what I've done... Also, the xx's on my dbc statement were on purpose. Current error is: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/content/64/4940664/html/artofwar/browse.php on line 35, I can't figure out where the hell it wants a ;... I know this problem comes up a lot but it can be for various reasons so after reading up I can't decide what might be wrong. I'm still learning! I have a simple php registration form (first name, second name, email address) and every time a user submits an entry a second blank record is created in the MYSQL database after it. Any help would be great. (php code in red) Code: [Select] [color=red]<? $firstname=$_POST['firstname']; $surname=$_POST['surname']; $email=$_POST['email']; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO register VALUES ('','$firstname','$surname','$email')"; mysql_query($query); mysql_close(); ?>[/color] Here is the section on the html HTML page that submits the form Code: [Select] <form action="register.php" method="post" class="BagTitle"> <table width="700" border="0" cellpadding="0" cellspacing="2"> <tr> <td width="100">First Name: </td> <td width="594"><input name="firstname" type="text" size="50" /></td> </tr> <tr> <td>Last Name: </td> <td><input name="surname" type="text" size="50" /></td> </tr> <tr> <td>E-mail:</td> <td><input name="email" type="text" size="50" /></td> </tr> <tr> <td colspan="2"> </td> </tr> <tr> <td colspan="2"><input type="Submit" value="Send" /> <input name="Reset" type="reset" value="Reset Form" /></td> </tr> </table> </form></p> Also I have no idea about securing this information. Are there any basic steps I can take? friends if an array returns results as Array() how could i set an error message to it? i tried to do a if statement to the $variable but not working httpd-2.2.17-win32-x86-openssl-0.9.8o php-5.3.6-Win32-VC9-x86 mysql-5.5.11-win32 PHP code: $dbcnx = mysql_connect('localhost','root','admin'); Error message: No connection could be made because the target machine actively refused it. I used mysql.exe -u root -p to verify that mysql is working and using the right username and password. Any ideas how to fix it? Thanks, Richard Hi I don't want errors on my page been reported on the web browser for all to see so I use this. $total2 = mysql_num_rows($qCheckUser) or die ("Error1"); however if the query returns zero rows I get Error1 on the web page. This is not an error and I don't want to taje out the or die() Any sugestions. I have an facebook application, I look in the errors log routinely to see if there are any abnormalities. Today, I found one. Code: [Select] [Sun Feb 27 15:28:25 2011] [error] [client **.***.71.38] PHP Notice: Undefined offset: 0 in *** on line 33 [Sun Feb 27 15:28:26 2011] [error] [client **.***.71.38] Exception: 121: Param pid must be a valid merged photo id Now, I have no idea what is going on here. I have found the location of where the error occurs: Code: [Select] $fb_photo = $facebook->api("me/photos", "POST", $attachement); $FQLQuery = "SELECT object_id, pid, src_big, link FROM photo WHERE object_id = " . $fb_photo['id']; $FQLResult = $facebook->api(array('method' => 'fql.query', 'query' => $FQLQuery, 'access_token'=> $session['access_token'])); $targetPhoto = $FQLResult[0]; I tried to help a user of mine get their account back on track, however, re-authenticating the application seems like the only way to fix it. It works for most of the users, however, there are a couple few who get my exception message. If anyone has any knowledge of whats going on here, that would be appreciated I am able to connect to my AD server successfully. This server serves multiple domains. Example is user1@dom1.dom.net is able to successfully bind. user2@dom2.com is not able to bind but gets error 49: invalid credentials. Using windows ldp.exe, I can connect successfully, then select bind from the connection menu, enter the username (user2), the account password, and the Domain (dom2.com) and the result indicated is successful. Using php I attempt to bind using: $adBind = ldap_bind($ad, $adUname, $ldappass); Where $ad = successful connection resource, $adUname = user1@dom1.dom.com OR user2@dom2.com, $ldappass is the account password. As user1 it is successful, with user2 it is unable to bind with error 49. Any suggestions or help is greatly appreciated. Thanks in advance! My dad loves these frozen cheeseburgers from meijer so I was gonna write a little script I can run in cron that will check Meijer's website and txt or email or something if they go on sale. Whenever I run the below script I get an Access Denied response from the server instead of the html for the cheesburger page. I'm sure I just need a CURL option or something. Thank You in Advance
can anyone see the problem here i am running php 5.3 on my server and GD seems to be enabled. and my host said it was probably something to do with the captcha files or script. <?php session_start(); /* * File: CaptchaSecurityImages.php * Author: Simon Jarvis * Copyright: 2006 Simon Jarvis * Date: 03/08/06 * Updated: 2007-08-12 by Jenny Ferenc (minor visual changes only) * Requirements: PHP 4/5 with GD and FreeType libraries * Link: http://www.white-hat-web-design.co.uk/articles/php-captcha.php * * This program is free software; you can redistribute it and/or * modify it under the terms of the GNU General Public License * as published by the Free Software Foundation; either version 2 * of the License, or (at your option) any later version. * * This program is distributed in the hope that it will be useful, * but WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the * GNU General Public License for more details: * http://www.gnu.org/licenses/gpl.html * */ class CaptchaSecurityImages { var $font = 'VeraSe.ttf'; function generateCode($characters) { /* list all possible characters, similar looking characters and vowels have been removed */ $possible = '23456789bcdfghjkmnpqrstvwxyz'; $code = ''; $i = 0; while ($i < $characters) { $code .= substr($possible, mt_rand(0, strlen($possible)-1), 1); $i++; } return $code; } function CaptchaSecurityImages($width=130, $height=40, $characters=6) { $code = $this->generateCode($characters); // font size will be 60% of the image height $font_size = $height * 0.6; $image = @imagecreate($width, $height) or die('Cannot initialize new GD image stream'); // set the colours $background_color = imagecolorallocate($image, 255, 255, 255); $text_color = imagecolorallocate($image, 20, 40, 100); $noise_color = imagecolorallocate($image, 100, 120, 180); // generate random dots in background for( $i=0; $i<($width*$height)/3; $i++ ) { imagefilledellipse($image, mt_rand(0,$width), mt_rand(0,$height), 1, 1, $noise_color); } // generate random lines in background for( $i=0; $i<($width*$height)/150; $i++ ) { imageline($image, mt_rand(0,$width), mt_rand(0,$height), mt_rand(0,$width), mt_rand(0,$height), $noise_color); } // create textbox and add text $textbox = imagettfbbox($font_size, 0, $this->font, $code) or die('Error in imagettfbbox function'); $x = ($width - $textbox[4])/2; $y = ($height - $textbox[5])/2; imagettftext($image, $font_size, 0, $x, $y, $text_color, $this->font , $code) or die('Error in imagettftext function'); // output captcha image to browser header('Content-Type: image/jpeg'); imagejpeg($image); imagedestroy($image); $_SESSION['security_code'] = $code; } } $captcha = new CaptchaSecurityImages(); ?> this seems to be the problem line he Code: [Select] $textbox = imagettfbbox($font_size, 0, $this->font, $code) or die('Error in imagettfbbox function'); [\code] ok so this problem has been laughing at me for quite some time now I tryed using some inline php in a form to pass it to my login script. The problem is that whenever the form gets submitted the variable r doesn't contain the actual microtime but it contains the inline php as a string: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Snitch</title> <link rel="stylesheet" type="text/css" href="templates/css/snitch1440x900.css" /> </head> <body> <div id="login-achtergrond"> <div id="login"> <form action="." id="loginform" name="login" method="post"> <input type="hidden" name="actie" value="Login"></input> <input type="hidden" name="r" value="<?php echo microtime();?>"></input> <input type="text" id="gebruikersnaam" name="gebruikersnaam" value="" style="opacity:0.7;filter:alpha(opacity=70)"></input> <input type="password" id="paswoord" name="paswoord" value="" style="opacity:0.7;filter:alpha(opacity=70)"></input> <input id="aanmelden" type="submit" name="submit" value="" style="opacity:0;filter:alpha(opacity=0)"> </form> </div> <div id="registreer"> </div> </div> </body> </html> the $_POST['r']; contains "<?php echo microtime();?>" instead of for example: 0.53192500 1305636839 does someone know what I am doing wrong here? I also tryed doing it like this: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Snitch</title> <link rel="stylesheet" type="text/css" href="templates/css/snitch1440x900.css" /> </head> <body> <div id="login-achtergrond"> <div id="login"> <form action=".r=<?php echo microtime();?>" id="loginform" name="login" method="post"> <input type="hidden" name="actie" value="Login"></input> <input type="text" id="gebruikersnaam" name="gebruikersnaam" value="" style="opacity:0.7;filter:alpha(opacity=70)"></input> <input type="password" id="paswoord" name="paswoord" value="" style="opacity:0.7;filter:alpha(opacity=70)"></input> <input id="aanmelden" type="submit" name="submit" value="" style="opacity:0;filter:alpha(opacity=0)"> </form> </div> <div id="registreer"> </div> </div> </body> </html> both $_POST and $_GET contains a string instead of the actual microtime single quotes didn't work eather Hi folks, I am a complete n00b at php and mysql. I am teaching myself from books and the WWW, but alas I am stuck... the error I get is: Parse error: syntax error, unexpected T_STRING in X:\xampp\htdocs\search.php on line 7 here is the code: <?php mysql_connect ("localhost", "user", "password") or die (mysql_error()); mysql_select_db ("it_homehelp_test") or die (mysql_error()); $term = $_POST['term']; $sql = $mysql_query(select * from it_homehelp_test where ClientName1 like '%term%'); <<<------this is line 7 while ($row = mysql_fetch_array($sql)){ echo 'Client Name:' .$row['ClientName1']; echo 'Address:' .$row['Address1']; echo 'Phone:' .$row['Tel1']; } ?> Any help you can offer would be great. I can also post the ".html" file that creates the search bar if it is needed. Thanks hi there, i am fairly new to OOPs in php, i get an error when i declare the argument type (as object) in a function and pass the same type (object). class eBlast { public static function getEmail(object $result) { return $result->email; } } $r = mysql_fetch_object($query); eBlast::getEmail($r); echo gettype($r); // outputs: object error is : Code: [Select] Catchable fatal error: Argument 1 passed to eBlast::getEmail() must be an instance of object, instance of stdClass given, called in C:\wamp\www\integra\client\pl_eblast\admin\send_emails.php on line 145 and defined in C:\wamp\www\integra\client\pl_eblast\app\app.eBlast.php on line 8 if i remove the type declaration in the function it works, but just would like to know why it shows error when pass the same type, also isnt mysql_fetch_object is the instance of stdclass? thanks in advance! I'm not sure why, but once I added a search form in my nav menu, it made my other forms on the website such as login and signup form take them to where the search button would take them. any ideas??? Error: Warning: mysql_query() expects parameter 1 to be string, resource given in C:\Program Files\xampp\htdocs\NEW\search.php on line 35 Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\Program Files\xampp\htdocs\NEW\search.php on line 36 Results Sorry, your search: "mod" returned zero results Click here to try the search on google 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 'Resource id #3 limit 0,10' at line 1 My PHP code <?php // Get the search variable from URL $var = @$_GET['q'] ; $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit=10; // check for an empty string and display a message. if ($trimmed == "") { echo "<p>Please enter a search...</p>"; exit; } // check for a search parameter if (!isset($var)) { echo "<p>We dont seem to have a search parameter!</p>"; exit; } //connect to your database ** EDIT REQUIRED HERE ** mysql_connect("localhost","root",""); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("school") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = mysql_query("SELECT * FROM main WHERE username LIKE '%".$trimmed."%' ORDER BY username") or die(mysql_error()) ; // EDIT HERE and specify your table and field names for the SQL query $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0) { echo "<h4>Results</h4>"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; echo "<p><a href=\"http://www.google.com/search?q=" . $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a> to try the search on google</p>"; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } // get results $query .= " limit $s,$limit"; $result = mysql_query($query) or die(mysql_error()) ; // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results"; $count = 1 + $s ; // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["1st_field"]; echo "$count.) $title" ; $count++ ; } $currPage = (($s/$limit) + 1); //break before paging echo "<br />"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< Prev 10</a>  "; } // calculate number of pages needing links $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division if ($numrows%$limit) { // has remainder so add one page $pages++; } // check to see if last page if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { // not last page so give NEXT link $news=$s+$limit; echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "<p>Showing results $b to $a of $numrows</p>"; ?> Lines: Code: [Select] 35 == $numresults=mysql_query($query); 36 == $numrows=mysql_num_rows($numresults); 59 == $result = mysql_query($query) or die(mysql_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 'Resource id #3 limit 0,10' at line 1 Help Please.. I made a search engine to search my database for specific users ( well I used a tutorial ). I'm not getting any errors atleast when loading the page which is a first for me. However, now I am tackled by something new. When entering the keywords and submitting the search I get: Access forbidden! You don't have permission to access the requested object. It is either read-protected or not readable by the server. If you think this is a server error, please contact the webmaster. Error 403 127.0.0.1 04/30/07 11:09:21 Apache/2.2.16 (Win32) PHP/5.3.3 Which is odd for me because I am using the same config.php with connection code to connect to the db which I used with all my other pages. And in those pages I have been able to add/delete/edit and display records from the database tables. Does this have something to do with it being a search engine? Here is my code: Code: [Select] <html> <head> </head> <body> <h2>Search</h2> <form name="search" method="post" action="<?=$PHP_SELF?>"> Seach for: <input type="text" name="find" /> in <Select NAME="field"> <Option VALUE="fname">First Name</option> <Option VALUE="sname">Last Name</option> <Option VALUE="sno">Profile</option> </Select> <input type="hidden" name="searching" value="yes" /> <input type="submit" name="search" value="Search" /> </form> </body> </html> <?php //This is only displayed if they have submitted the form $searching ='searching'; if ($searching =="yes") { echo "<h2>Results</h2><p>"; //If they did not enter data then they receive an error if ($find == "") { echo "<p>You forgot to enter a search term"; exit; } // Otherwise connect to database include 'includes/config.php'; include 'includes/functions.php'; connect(); // filtering $find = strtoupper($find); $find = strip_tags($find); $find = trim ($find); // search for term in specified field $data = mysql_query("SELECT * FROM student WHERE upper($field) LIKE'%$find%'"); // display results while($result = mysql_fetch_array( $data )) { echo $result['fname']; echo " "; echo $result['sname']; echo "<br>"; echo $result['sno']; echo "<br>"; echo "<br>"; } //Counts number or results and display message if there we're non $anymatches=mysql_num_rows($data); if ($anymatches == 0) { echo "Sorry, but we can not find an entry to match your query<br><br>"; } //And we remind them what they searched for echo "<b>Searched For:</b> " .$find; } ?> Any ideas? Thanks in advance Getting the following error message: Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in This error occurs when I first navigate to the search page. When I put a search term in the box and click search it works fine and the error is gone. I've been trying to figure this out for days with no luck, any help you can offer would be greatly appreciated. Code: [Select] <?php require('config.php'); ?> <?php $conn = mysql_connect(SQL_HOST, SQL_USER, SQL_PASS) or die('Could not connect to MySQL database. ' . mysql_error()); mysql_select_db(SQL_DB, $conn); $result = NULL; if (isset($_GET['keywords'])) { $sql = "SELECT id, hostname, ip, clust_name, clust_ip, node_name, ". "node_ip, os, patch " . "FROM serv_main " . "WHERE MATCH (hostname,ip) " . "AGAINST ('" . $_GET['keywords'] . "' IN BOOLEAN MODE) " . "ORDER BY MATCH (hostname,ip) " . "AGAINST ('" . $_GET['keywords'] . "' IN BOOLEAN MODE) DESC"; $result = mysql_query($sql) or die('Could not perform search; ' . mysql_error()); } ?> <html> <body> <form method="get" action="search.php"> <p class="head">Type in the Server name below to search for an SLA</p> <p> <input id="searchkeywords" type="text" name="keywords" <?php if (isset($_GET['keywords'])) { echo ' value="' . htmlspecialchars($_GET['keywords']) . '" '; } ?> > <input id="searchbutton" class="submit" type="submit" value="Search"> </p> <?php echo "<h1>Search Results</h1>\n"; if ($result and !mysql_num_rows($result)) { if (isset($_GET['keywords'])) { $k = $_GET['keywords']; } echo "<p>No SLA for <b><font size=\"6\" color=#990000>" . $k . "</b></font> found.</p>\n"; } else { $bg = ''; //Error occurs right here********************* while ($row = mysql_fetch_array($result)) { //($row['id'], TRUE); $bg = ($bg=='F2F2FF'?'99CCCC':'F2F2FF'); $table = "<table border=\"0\" cellpadding=\"5\">"; $table .= "<tr bgcolor=\"#" . $bg . "\">" . "<td><a href=\"servedit.php?s=" . $row['id'] . "\">" . $row['hostname'] . "</a></td><td>" . $row['ip'] . "</td><td align=\"center\">" . $row['clust_name'] . "</td><td>" . $row['clust_ip'] . "</td><td>" . $row['node_name'] . "</td><td>" . $row['node_ip'] . "</td><td>" . $row['os'] . "</td><td>" . $row['patch'] . "</td></tr>"; } $table .= "</table>"; echo $table; } ?> </form> </body> </html> MOD EDIT: [code] . . . [/code] tags added. Hi I am finding that I get a random LDAP search fail the initial time i run a script against 2008R2 when I refresh all is ok ? PHP 5.2.5 get the same thing with the latest version of PHP. any ideas ? its as if the DC is sleeping until you give it a nudge, running on 2008R2 running on vmware. I'm working on the following assignment and keep coming up with error messages.
"Warning: require(mysqli_connect_registrar.php): failed to open stream: No such file or directory in /home/students/r/b.richards/public_html/cis231/labs/sectionresults.php on line 3 I've included the code I have and any help would be appreciated. Thank you. Assignment Directions: Introduction This lab will demonstrate how to create a dynamic form that is created from database contents. And then that form will be used to pull results from the database, based on the user's selection.
STEP 1 - Create the Database Connection File In your labs directory, define a mysql_connect.php file for the registrar database (with the student | student credentials).
STEP 2 - Create the Form Define the form as sectionsearch.php. Create a dynamic form with a drop-down menu listing available courses by title. (Note: the courseID will be the value that is posted to the results page.)
Step 3 - Create the Results page Define the results page as sectionresults.php. Present the results of the query for sections (unless there are none—then show an appropriate message). If there are results, use an HTML table (dynamically-generated) to display available sections with the following information: Faculty last name Room number Class meeting day(s) Start time End time The information should be sorted by the faculty member's last name. Note: You will need to consult the structure of the database tables to determine which fields need to be drawn from which tables and how to join the appropriate tables needed for the query. Sectionsearch.php code:
<!doctype html>
<head> <body>
<form id="form1" name="form1" method="post" action="sectionresults.php">
<p>Courses
<p>
</form> mysql_connect.php code:
<!doctype html>
//Set database connection information
//Database connection
//Set the encoding sectionresults.php code:
<?php
$cat = "";
$q = "SELECT lastame, FROM faculty WHERE registrar = $cat ORDER BY lastname ASC"; |