PHP - Help For Sorting Mysql Tabel Result From Field Named "jobnr" In Decending Ordrer
Hi - i need help with the fourth column named "jobnr" it has to be the highest number first and lowest at the bottum of the tabel.
I tried different variation like "mysql_query("SELECT * FROM tabel ORDER BY jobnr DESC")" but with no success. I attached a picture og the working script output Code: [Select] <html> <head> <meta http-equiv="Content-type" content="text/html; charset=UTF-8"/> </head> <style type="text/css"> .myclass { font-size: 8pt; font-face: Verdana; } </style> <body> <?php // Define variables $host="host"; // Host name $username="user"; // Mysql username $password="password"; // Mysql password $db_name="database"; // Database name $tbl_name="tabel"; // Table name // Connect to server and select databse mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // saetter db udtraek til UTF-8 endcoding mysql_set_charset('utf8'); // henter db data fra tabllen: jobpositons $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); // Define $color=1 $color="1"; echo '<table border=0" bordercolor="#f3f3f3" cellpadding="1" cellspacing="1">'; echo "<tr bgcolor='#00aeef'> <th>Jobtitel</th> <th>Sted</th> <th>Oprettet</th> <th>jobnr</th> </tr>"; // sortere sql db data og indsaetter i html tabel while($rows=mysql_fetch_array($result)) { // If $color==1 table row color = #ffffff if($color==1){ echo "<tr bgcolor='#ffffff'><td class='myclass'>".$rows['Jobtitel']."</td><td class='myclass'>".$rows['Sted']."</td><td class='myclass'>".$rows['oprettet']."</td><td class='myclass' align='right'>".$rows['jobnr']."</td> </tr>"; // Set $color==2, for switching to other color $color="2"; } // When $color not equal 1, use this table row color else { echo "<tr bgcolor='#f3f3f3'> <td class='myclass'>".$rows['Jobtitel']."</td><td class='myclass'>".$rows['Sted']."</td><td class='myclass'>".$rows['oprettet']."</td><td class='myclass' align='right'>".$rows['jobnr']."</td> </tr>"; // Set $color back to 1 $color="1"; } } echo '</table>'; mysql_close(); ?> </body> </html> Similar TutorialsHi guys I'm having problem on how to display my php mysql result like this.. Digitalpoint, phpfreaks, adsense here's my code Code: [Select] $result = mysql_query ("SELECT dateline, dateline_from, eventid, title FROM vb_event WHERE calendarid = 1 AND dateline_from >= '$day_before' ORDER BY dateline_from LIMIT 2"); while($row = mysql_fetch_array( $result )) { echo "<a href='/forum/calendar.php?do=getinfo&e=" . $row['eventid'] . "&c=1' title='". $row['title'] . "' style='font-size:12px;'>" . $row['title'] . "</a> " ; } the result is Digitalpoint, phpfreaks, adsense, how to remove the "," in the end? it should be like this.. Digitalpoint, phpfreaks, adsense with no , comma in the end.. Thank you.. Hey, friends. I have some trouble on the server front. My sites have been hacked, and I need to make sure I've eradicated every trace of this exploit. I'm looking for a way to search for any and all php files contained in multiple directories with specific names. For instance, I have found a commonality in relation to where these malicious files are placed, such as: Code: [Select] /some/dir/img/somename.phpor: Code: [Select] /some/dir/js/somename.php Is there a way I can easily (e.g. using ssh and the "find" command) locate all files ending in php but only found in directories named "img"? I can't seem to find anything that would allow me to do this with find, or with a combination of find and grep. I can't go directory by directory, as some of these img directories are created many levels deep, some even in .svn directories. Any and all help is appreciated. Hackers suck. Can someone please help me with an array problem i can not figure out. I need the array to be numbered from 1 to how ever many fields that are needed in the form and have a mysql field name and the title of the field also in the array. 1, "mysql_field_name", "Title of form field" 2, "", "" and so on then the form will be shown based on the array. I have the following draft code which I am working with. any suggestions on how i may do this array ? Code: [Select] <?php $options = array( '1'=> array('fieldtext'=>'option1', 'mysqlfield'=>'option1'), '2'=> array('fieldtext'=>'option2', 'mysqlfield'=>'option2'), '3'=> array('fieldtext'=>'option3', 'mysqlfield'=>'option3'), '4'=> array('fieldtext'=>'option4', 'mysqlfield'=>'option4'), ); // $options = array(1 => "option1", "option2", "option3", "option4"); // the line above works but i want to include the name of the mysql field as well. $userid = 1; ?> <div style="align: center; margin: 12px; font-family:Tahoma;"> <br><br><?php if ($_POST['Update'] != "Update") { // check if form submitted yet, if not get data from mysql. $res = db_query("SELECT * FROM `users` WHERE `userid` = '" . $userid . "'"); foreach($options as $key => $value) { $_POST[$key] = mysql_result($res, 0, $value); } $ok_to_update = "no"; } elseif ($_POST['Update'] == "Update") { // check if form submitted yet, if so get POST data. // error checking // foreach($options as $key => $value) { // $_POST[$i] = ""; // } $ok_to_update = "yes"; } if ($_POST['Update'] == "Update" && $ok_to_update == "yes") { // $res = db_query("INSERT INTO `users` () VALUES ()"); // add user details to database. ?><p><br><br><br>Thank you for updating</p><?php } else { ?><form name="form1" method="post" action=""> <?php foreach($options as $key => $value) { ?><p><?php echo($value); ?>: <input type="text" name="<?php echo($key); ?>" value="<?php echo($_POST[$key]);?>"></p> <?php } ?> <input name="Update" type="submit" value="Update"> </form> <?php } ?> </div> Hey, have used a long time on this... i hope you guys can help me :D - im new in the php / mysql world. so here it is: i want to make a list on my first page that post links (list.php) list.php works fine, i get a list of links from my database, and when i click on it it goes to test.com/klub.php?id= (and the id for the tabel) fx. test.com/klub.php?id=000000018. so i think the problem is in my klub.php, here i want to select * from siteinfo1 where id = the same id as i get from clicking the link on the other site :D xD... and then post the infomation for the 'name' of the link i clicked :D well here is my code :D i think the problem is $result = mysql_query("SELECT * FROM siteinfo1 WHERE id='$id'"); for if i change the id=$id to id=000000018 it post the info for the table whit that id nr... but then i cant see all the other info i need form 000000019 and so on :D xD i hope you understand my stupidity and know how to fix my problem :D thanks... _________________________________________ list.php _________________________________________ $result1 = mysql_query("SELECT * FROM siteinfo1") or die (mysql_error()); while($row1 = mysql_fetch_array($result1)) { echo "<A href=\"klub.php?id={$row1['id']}\">{$row1['navn']}</A>"; echo "</br>"; } _________________________________________ klub.php _________________________________________ $result = mysql_query("SELECT * FROM siteinfo1 WHERE id='$id'"); while($row = mysql_fetch_array($result)) { echo "$row[id]"; echo "</br>"; echo "Navn: "; echo "$row[navn]"; echo "</br>"; } Hi, bit stuck on how to find and replace "<" and ">" with "<" and ">". I basically have a database record that outputs to screen and I need the code in the <code> tags to be rendered to the screen. I therefore need it to go through the whole array variable from the db and change the symbols just inside the code tags. Please be aware that the code tags might happen more than once here's an example below Code: [Select] <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1> </code> <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1> </code> the desired output would be: <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1 </code> <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1 </code> help on this would be great Cheers Rob Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/aci/docs/admin/hours.php on line 20 Code: [Select] <?php $user = $_SESSION['myusername']; $result = mysql_query("SELECT * hours WHERE member='$user'"); while ($row = mysql_fetch_array($result)){ echo $row['date'] . " " . $row['time'] . "<br />"; } ?> Code: [Select] CREATE TABLE `hours` ( `hoursID` int(5) NOT NULL auto_increment, `member` varchar(20) NOT NULL default '', `date` date NOT NULL default '0000-00-00', `time` time NOT NULL default '00:00:00', PRIMARY KEY (`hoursID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ; -- -- Dumping data for table `hours` -- INSERT INTO `hours` VALUES (4, 'aci', '2010-11-30', '14:31:39'); INSERT INTO `hours` VALUES (2, 'aci', '2010-11-30', '14:31:08'); INSERT INTO `hours` VALUES (3, 'aci', '2010-11-30', '14:31:23'); INSERT INTO `hours` VALUES (5, 'aci', '2010-11-30', '14:31:40'); Hi guys I am a newbie with PHP and I have been trying to solve this problem for 3 days.. Ive set up a booking form on my website to be sent directly to my email once the client clicked on the send button.. The problem I am having while checking these pages on wamp is that the booking form is ok but when I click on the send button the following message error appears "Warning: mail() [function.mail]: Failed to connect to mailserver at "localhost" port 25, verify your "SMTP" and "smtp_port" setting in php.ini or use ini_set() in C:\wamp\www\fluffy_paws\booking_form2.php on line 27" When I check online I have this message "500 - Internal server error. There is a problem with the resource you are looking for, and it cannot be displayed." Are my codes wrong??? My webhosting is Blacknight.com , Im on Windows Vista, my internet is a dongle with O2 ireland Thanks for your help! An insert function is not inserting. I keep getting "Could not store data, please try again." After some probing, it seems that "$result = $conn->query($user_degree_insert_query); " is where the problem is. Variables before it seem to pass testing. I can't figure out why this line is messing thins up though. Any idea? Thank you in advance for your help! The "//$degree_id = mysql_fetch_array($degree_id_query, MYSQL_BOTH) ;" is commented out because it was giving me trouble, i replaced it with the line above that $conn = db_connect(); $arraycount = count($degree_Array); //loop through the categories chosen and add them each into member-category for($i =0; $i < $arraycount; $i++){ $degree_id_query = "SELECT degree_id FROM degree WHERE degree_type ='".$degree_Array[$i]."'"; $result = $conn->query($degree_id_query); //place the id found into the array, which automatically cancatonates it. $degree_id = $result->fetch_array(); //$degree_id = mysql_fetch_array($degree_id_query, MYSQL_BOTH) ; $user_degree_insert_query = "INSERT INTO user-degree ( `user_id` , `degree_id` ) VALUES ( '".$user_id."', '".$degree_id[0]."' )"; $result = $conn->query($user_degree_insert_query); if($result) { header("Location: survey2.php"); }else { echo "<p>Could not store data, please try again.</p>"; exit; Hi guys I'm struggling a bit, I need to replace a word that occurs multiple times in text with an array("up","down","forward","backwards") of words. $find = "left"; $replace = array("up","down","forward","backwards"); $text = "left left left left"; echo str_replace($find,$replace,$text); The Output is: array array array array Did try this with a foreach statement as well, but no luck. Is there a better way of doing this? Thanks I have a table with a list of users and an edit button and delete button. When the edit button is pressed on a site it passes the user_id as p_id to the page that catches it and displays the data. The problem is when I click on the "update user" button, I get the following error:
Warning: Undefined variable $the_user_id in C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php on line 10 The weird thing is I had another update user page with a table I created that ran the query to update the table in the database just fine. But as I created it, it didn't look all that great so I recreated the page and used a bootstrap table because of the much cleaner look. Both pages have the exact same PHP code, the only difference is the bootstrap table I added in. So I'm really at a loss with this. Other than the table and PHP code, there is a script at the bottom of the page for the table itself to allow for searching within the table, i'll include that as well. The PHP code is as follows:
<?php //THE "p_id" IS BROUGHT OVER FROM THE EDIT BUTTON ON VIEW_ALL_USERS if (isset($_GET['p_id'])) { $the_user_id = $_GET['p_id']; } // QUERY TO PULL THE SITE INFORMATION FROM THE p_id THAT WAS PULLED OVER $query = "SELECT * FROM users WHERE user_id = $the_user_id "; $select_user = mysqli_query($connection,$query); //SET VALUES FROM ARRAY TO VARIABLES while($row = mysqli_fetch_assoc($select_user)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_email = $row['user_email']; $user_phone = $row['user_phone']; //$user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id']; } THE UPDATE QUERY CODE....................................................................................................................
<?php if(isset($_POST['update_user'])) { $user_id = $_POST['user_id']; $user_firstname = $_POST['user_firstname']; $user_lastname = $_POST['user_lastname']; $username = $_POST['username']; $user_email = $_POST['user_email']; $user_phone = $_POST['user_phone']; //$user_image = $_POST['user_image']; $user_title_id = $_POST['user_title_id']; $user_role_id = $_POST['user_role_id'];
$query = "UPDATE users SET "; $query .= "user_id = '{$user_id}', "; $query .= "user_firstname = '{$user_firstname}', "; $query .= "user_lastname = '{$user_lastname}', "; $query .= "username = '{$username}', "; $query .= "user_email = '{$user_email}', "; $query .= "user_phone = '{$user_phone}', "; //$query .= "user_image = '{$user_image}', "; $query .= "user_title_id = '{$user_title_id}', "; $query .= "user_role_id = '{$user_role_id}' "; $query .= "WHERE user_id = '{$the_user_id}' "; $update_user = mysqli_query($connection,$query); if(! $update_user) { die("QUERY FAILED" . mysqli_error($connection)); } } ?> THE "UPDATE USER" BUTTON THE USER CLICKS ON TO UPDATE....................................................................................................................
<div class="col-1"> <button class="btn btn-primary" type="submit" name="update_user">Update User</button> </div>
Any Help is Greatly Appreciated! Edited March 23 by Zserenecant work out this mysql syntax error "operation":"medupdate","medid":"","name":"ibo","medyear":"5","medmonth":"21","medday":"1","recuser":1,"SuccFail":"fail","SuccFailMessage":"error occured 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 '''WHERE med_id=' at line 2","ResultData":""} Code: [Select] $sql="update metodology set med_description='".$req['name']."', med_year='".$req['medyear']."', med_month='".$req['medmonth']."', med_day='".$req['medday']."', med_recorddate= now(), med_recorduserid= 1'"; $sql.= " WHERE med_id=".$req['medid']; Forgive me if I am using the word "globally" incorrectly... I want to reset the content of a particular field in all rows of a table to a specific value, I was thinking of nesting a query inside a query for the same table, but was not sure that this would work: ... $query = "SELECT * FROM music WHERE picked >'0'"; $result = mysql_query($query) or die ("Couldn't execute query."); /* reset the picked field to zero (0) */ while ($row = mysql_fetch_array($result)) { $picked = 0; $query= "UPDATE music SET picked='$picked' WHERE title='$title'"; $selectresult = mysql_query($query) or die ("Problem with the query: $query<br>" . mysql_error()); } ... Is this the correct approach or is there a better way of doing this? Folks, I am not able to find any function that will sort an array elements by its element's length. Does such function exist? I am getting the following error when using composer: "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? Just started. I am pretty sure composer.json wasn't changed. journalctl doesn't show anything. Maybe Doctrine related, however, nothing seems to help.
Any ideas? [michael@devserver www]$ php -v PHP 7.3.4 (cli) (built: Apr 2 2019 13:48:50) ( NTS ) Copyright (c) 1997-2018 The PHP Group Zend Engine v3.3.4, Copyright (c) 1998-2018 Zend Technologies with DBG v9.1.9, (C) 2000,2018, by Dmitri Dmitrienko [michael@devserver www]$ composer -V Composer version 1.4.2 2017-05-17 08:17:52 [michael@devserver www]$ yum info composer Loaded plugins: fastestmirror Loading mirror speeds from cached hostfile * base: mirror.us.oneandone.net * epel: mirror.rnet.missouri.edu * extras: mirror.us.oneandone.net * remi-php73: mirror.bebout.net * remi-safe: mirror.bebout.net * updates: mirror.us.oneandone.net Installed Packages Name : composer Arch : noarch Version : 1.8.4 Release : 1.el7 Size : 1.8 M Repo : installed From repo : epel Summary : Dependency Manager for PHP URL : https://getcomposer.org/ License : MIT Description : Composer helps you declare, manage and install dependencies of PHP projects, : ensuring you have the right stack everywhere. : : Documentation: https://getcomposer.org/doc/ [michael@devserver www]$ composer update Loading composer repositories with package information Updating dependencies (including require-dev) [ErrorException] "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? update [--prefer-source] [--prefer-dist] [--dry-run] [--dev] [--no-dev] [--lock] [--no-custom-installers] [--no-autoloader] [--no-scripts] [--no-progress] [--no-suggest] [--with-dependencies] [-v|vv|vvv|--verbose] [-o|--optimize-autoloader] [-a|--classmap-authoritative] [--apcu-autoloader] [--ignore-platform-reqs] [--prefer-stable] [--prefer-lowest] [-i|--interactive] [--root-reqs] [--] [<packages>]... [michael@devserver www]$
This topic has been moved to Linux. http://www.phpfreaks.com/forums/index.php?topic=318175.0 When I put "hello" in a text field it disappears, when the html page is submitted it dont like "" for example <input name="FirstName" type="text" id="FirstName" value="<?php if (isset($_POST['FirstName'])) echo ($_POST['FirstName']); ?>" maxlength="20" /> I have a varchar field called date that looks like this "2/06/2011 TUE" or like this "10/05/2011 WED". When I do an Order BY date Desc in php it does Not sort properly. Thanks for your assistance Steve I have a form where people will be putting in specific info such as their names and other info. When the form is posted I need to add a "." between the fields on the output depending on if it is filled out or not if field is blank do not need a "." Example if they put in their name Bob Marley on the form I need it to appear on the output as Bob.Marley. Here is the code using currently which also excludes blank fields echo $_POST["First"]; if(isset($_POST["Middle"])){ echo $_POST["Middle"]; } echo $_POST["Last"]; I am trying to add another "required field" to my PHP form, but I'm not sure what I'm doing wrong. The form and "required fields" work fine, but I am now trying to make the "Select skill level which is an "input type="radio" as a required field, but it is not working. As you can see below, I just tried adding the new "required" field the same way as the other 4 were set up, but it is not working (I think I have to do something different because the new "required field" is a "radio" type and the rest are "text"). The relevant parts of the code are below. Any assistance would be appreciated Code: [Select] <script type="text/JavaScript"> function verify() { var Skill = document.indiv_reg.skill.value; var Fname = document.indiv_reg.name.value; var Hphn = document.indiv_reg.home_num.value; var Email = document.indiv_reg.email.value; var Tname = document.indiv_reg.tname.value; if ((Fname =="") || (Skill =="") || (Hphn =="") || (Email =="") || (Tname =="")) { alert('Please Fill Out ALL Required Fields'); } else { document.indiv_reg.submit(); } } </script> <?php $url = "https://www.paypal.com/cgi-bin/webscr?cmd=_s-xclick&hosted_button_id=6Y76L7TX6FVRW"; $sport = $_GET['sport']; $league = $_GET['league']; $visitor = $_GET['name']; $skill = $_GET['skill']; $tname = $_GET['tname']; $tcap = $_GET['tcap']; $homephn = $_GET['home_num']; $cellphn = $_GET['cell_num']; <?php * Denotes a Required Field<br /> <br /> <form name="indiv_reg" method="GET" action="tourney_mailRedirect.php"> Sport:<br /> <input type="text" name="sport" value="'.$sport.'" readonly="readonly" /><br /> <br /> Tournament:<br /> <input type="text" size="90" name="league" value="'.$league.'" readonly="readonly" /><br /> <br /> * Please select preferred skill level:<br /> <input type="radio" name="skill" value="recreational" /> Recreational<br /> <input type="radio" name="skill" value="competitive" /> Competitive<br /> <br /> * Name:<br /> <input type="text" name="name" /><br /> <br /> * Team Name:<br /> <input type="text" name="tname" /><br /> <br /> Team Captain:<br /> <input type="text" name="tcap" /><br /> <br /> * Home Phone:<br /> <input type="text" name="home_num" /><br /> <br /> Cell Phone:<br /> <input type="text" name="cell_num" /><br /> <br /> * Email:<br /> <input type="text" name="email" /><br /> <br /> ?> Hi, I currently have an if, elseif, else program that starts off with Code: [Select] if( $v_name == "" || $v_msg == "" ) echo "something" how do I turn this into a switch? but the very next elseif I have Code: [Select] elseif( strcspn( $_REQUEST['msg'], '0123456789' ) == strlen( $_REQUEST['msg'] ) ) echo "something" is it possible to turn this into a switch with 2 different strings like that to evaluate? |