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PHP - "simple" Calculations Using Database Values
I was wondering how to write something like this:
if (table1.column1 + table2.column2 > table3.column3 then ..... I am confused where to put ( ) and ' Similar TutorialsCan 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> 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? 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 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! 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 cant 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']; Hi, I have set up 2 php pages page 1 - add_entry2.php In this page I have a invoice table created where I can dynamically add/delete rows. This has a View Bill button which takes me to page 2- add_entry3.php In this page it shows up the rows added in page 1 in read only format, so if the user wants to modify the data that he/she entered then he must Click on <back> that i have provided in the page 2 which will direct him to page 1 Now the problem starts here on click of Back the dynamically added rows dissappear..which is frustrating..I know its something to do with my code..but can anyone help me fix it. One more thing is that i dont want to store the data into DB till the finalise button is clicked on page 2 so that means till page 2 is submitted nothing goes to DB from Page 1. I am able to retain values if I use the code Code: [Select] <form action="add_entry2.php" name="eval_edit" method="post" format="html"> i,e if I submit back to the same page and retrieve values form $_POST but If I use the code Code: [Select] <input type="button" value="Back" onClick="history.go(-1);return true;">to get back to add_entry2.ph it looses all the values. Is there any other way to code the BACK link retaining my $_POST values(Do you think $_SESSION would work in this case?) I am getting the following error when using composer: "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? Just started. I am pretty sure composer.json wasn't changed. journalctl doesn't show anything. Maybe Doctrine related, however, nothing seems to help.
Any ideas? [michael@devserver www]$ 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]$
Hey guys, I'm new here so hello... but I'll get right to the point: I have a table named POSTS, where I store user posts. In this table, I have a row named "Popularity". Now I would like the value of "Popularity" to be a function of the number of Likes, Dislikes, Views and Date posted, all of which are stored in respective tables (linked by Primary and Secondary keys). I was wondering if it actually is possible to have the value of "Popularity" adjust automatically when a change in "Likes", "Dislikes" or "Views" occurs. And if not, what alternatives I have. The ultimate reason I need to have this value stored in a table and up to date is that I wish to sort posts by "Popularity". Many thanks !!! Hey, I wanted to know if there is any difference between catching a return value and assigning into a variable, and just catching the value when when you just simply call the function on its own. Here is what I mean below In the first example I called the return by simple calling the function. In the second example I catched the return value through a variable, and then echoed it out. Code: [Select] <?php // Example 1 function text (){ $string = "this is a string </br>"; return $string;} echo text(); // Example 2 function text (){ $string = "this is a string </br>"; return $string; } $new_value = text(); echo $new_value; } ?> This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=354981.0 This topic has been moved to Linux. http://www.phpfreaks.com/forums/index.php?topic=318175.0 Hi there, I am working on a PHP web form and I have a very simple situation I guess, I have a variable named: $MyVal When I do print_r($MyVal); To see whats inside, I get: SimpleXMLElement Object ( [0] => 8.23 ) Now I am assigning this variable into a session variable so that I can do calculations with it. So I assign: $_SESSION['MySesVal'] = $MyVal; But after assigning to session variable, when I do my calculations: $finalValue = $_SESSION['MySesVal'] * 4; I get 0. So is it because the actual $MyVal variable has some XML stuff as: SimpleXMLElement Object ( [0] => 8.23 ) So what is the right way to properly assign $MyVal variable to a session variable to do calculations. Please reply. All comments and feedbacks are always welcome. Thank you! 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 ZsereneI've been trying to simplify this code, let alone make it work. It's a loop to change the different question checkboxes to "checked" if certain degree_ids are found. All of the examples i have looked at are very strange and I'm not sure how to go about this, thought it seems to be a common question. Everyone seems to have their own way. Any advice would be greatly appreciated. for($i =1; $i <= $arraycount; $i++){ $query = "SELECT degree_id FROM `user-degree` WHERE `user_id` = '".$user_id."' AND degree_id = '".$i."'"; $result = $conn->query($query) or die(mysql_error()); $row = mysqli_fetch_array($result) ; echo $row['degree_id']; if(!$y){ $y = 0; } if ($row['degree_id'] == 1){ $q4[0] = 'checked'; } if ($row['degree_id'] == 2){ $q4[1] = 'checked'; } if ($row['degree_id'] == 3){ $q4[2] = 'checked'; } if ($row['degree_id'] == 4){ $q4[3] = 'checked'; } if ($row['degree_id'] == 5){ $q4[4] = 'checked'; } if ($row['degree_id'] == 6){ $q4[5] = 'checked'; } if ($row['degree_id'] == 7){ $q4[6] = 'checked'; } if ($row['degree_id'] == 8){ $q4[7] = 'checked'; } if ($row['degree_id'] == 9){ $q4[8] = 'checked'; } } "BACK" or REFRESH: Preventing database interaction / code execution how to prevent database interaction / code execution when user presses back or refresh button? can i detect? can i disable back/refresh? I am currently creating a form and I want to populate a drop down selection menu with data from two fields in a form. For example, I want it to pull the first and last name fields from a database to populate names in a drop down menu in a form. I want the form to submit to the email address of the person selected in the drop down. Is this possible to do? The email is already a field in the record of the person in the database. Can anyone give me some pointers or advice on how I should go about setting up the "Select" box drop down? I am not sure how to code it to do what I am wanting. Any links to relevant help would be appreciated too. Thanks in advance! I have a simple form to add details to the database. I want to use the same form layout to call back details for editting and updating. The main page is showroomedit.php which lists (in an array) the records in the database. I have added an edit.php but cant seem to work the code out. Can some one point me in the right direction. here is edit.php: <?PHP include('dbconnect.php') ?> <HTML> <HEAD> <TITLE>Barry Ottley Motor Co - Quality Used Motors - Stanford-Le-Hope, Essex</TITLE> </HEAD> <BODY LINK="#000000" ALINK="#000000" VLINK="#000000"> <CENTER><IMG SRC="logo.jpg"></CENTER> <CENTER> <TABLE WIDTH="840" CELLPADDING="0" CELLSPACING="0" BORDER="0"> <TR> <TD WIDTH="186" valign="top"> <BR> <CENTER><A HREF="index.html"><IMG SRC="homepage.jpg" alt="Homepage" border="0"></A></CENTER> <CENTER><A HREF="showroom.php"><IMG SRC="showroom.jpg" alt="Our Online Showroom" border="0"></A></CENTER> <CENTER><A HREF="location.html"><IMG SRC="ourlocation.jpg" alt="Our location" border="0"></A></CENTER> <CENTER><A HREF="aboutus.html"><IMG SRC="aboutus.jpg" alt="About Us" border="0"></A></CENTER> <CENTER><A HREF="contactus.php"><IMG SRC="contactus.jpg" alt="Contact Us" border="0"></A></CENTER> </TD> <TD WIDTH="30" valign="top"> </TD> <TD valign="top"> <FONT SIZE="2" FACE="ARIAL"> <CENTER><A HREF="addcar.html"><IMG SRC="vehicleadd.jpg" border="0"></A><A HREF="showroomconfig.php"><IMG SRC="vehicleedit.jpg" border="0"></A><A HREF="showroomconfig.php"><IMG SRC="vehicledelete.jpg" border="0"></A></CENTER> <CENTER><B>Update a Vehicle</B></CENTER> <BR> <?php $query1="SELECT cars FROM CarName='$carname', CarTitle='$CarTitle', CarPrice='$carprice', CarMiles='$carmiles', CarDescription='$cardesc'"; mysql_query($query1); mysql_close(); ?> <form action="editvehicle.php" method="post"> <CENTER>Vehicle Name:</CENTER> <CENTER><input type="text" name="CarName" value="<?php echo $carname; ?>"></CENTER> <br> <CENTER>Vehicle Type:</CENTER> <CENTER><input type="text" name="CarTitle" value="<?php echo $cartitle; ?>"></CENTER> <br> <CENTER>Vehicle Price:</CENTER> <CENTER><input type="text" name="CarPrice" value="<?php echo $carprice; ?>"></CENTER> <br> <CENTER>Vehicle Mileage:</CENTER> <CENTER><input type="text" name="CarMiles" value="<?php echo $carprice; ?>"></CENTER> <br> <CENTER>Vehicle Description:</CENTER> <CENTER><textarea name="CarDescription" rows="10" cols="30" value="<?php echo $cardesc; ?>"></textarea></CENTER> <br> <CENTER><input type="Submit"></CENTER> </form> </TD> </TR> </TABLE> <BR><BR><BR><BR> <CENTER><FONT FACE="VERDANA" SIZE="1"> This website was designed and is hosted by <A HREF="http://www.IRCDirect.co.uk">IRC Direct Website Design™</A> 2010 </CENTER> <BR> <CENTER><A HREF="admin.php">Employee Area</A> </BODY> </HTML> That is not even showing me the table details at all. Im lost . There is an "Update" link in each array in showroom.php in which when the user clicks it opens edit.php . They update the form as they wish... then click update and it updates the database and sends them back to showroomedit.php Much help would be appreciated ! Cheers paul if your out there still Hello I code by hand (HTML & CF -14 years now), and some of my customers want the ability to do updates to their sites themselves using Adobe Contribute, and I can build any app they need that Contribute can't handle in CF if they host on my server - however, in this particular case, the site is hosted by GoDaddy which is no longer supporting CF. The site is PHP-enabled, and the client was using WordPress to update his entire site. Because of the blogging-centric features of Wordpress, a user can enter a "headline" then an associated story (commentary) - publish, and the headline ends up on one page linked to the page where the headline and story live. So, I need a simple form handler written in PHP that allows the client to enter his "headline" (which is a hard-coded link to the "rest of the story" page), the headline showing up above the story content (using a second form field) - not necessarily targeting <a href="#end-up-here"> - just simply land on the page where the content gets dynamically displayed. Thanks in advance for your patience and be kind---- I have never written a line of PHP- just thousands of lines of CF..... Norman Hello All, I have a custom session.set_save_handler to handler sessions for my application. It seems to work fine. The database table is of type innodb. I see some errors in my log file of the nature, "(1205) Lock wait timeout exceeded; try restarting transaction". I have a class which contains the session handler callbacks and have made it a singleton class. session_set_save_handler( array($this, "db_open"), array($this, "db_close"), array($this, "db_read"), array($this, "db_write"), array($this, "db_destroy"), array($this, "db_gc") ); In my "db_read" method, I am doing a "SELECT FOR UPDATE" on the row with the session id. So, while performing load testing on my app. which has a lot of ajax calls, I noticed the "lock wait timeout" errors. How do I resolve this problem? Any suggestions? 1) Should I catch these errors and issue a explicit "commit" in the read or write methods? 2) Should I specify a higher timeout value for "innodb_lock_wait_timeout"? Thanks |
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