PHP - How To Multiply(or Sum) The Dropdown Box Value That Trieved From Mysql Database
hi. I am stuck on the dropdown box value calculation. anyone can help me on this problem ? I appreciate it.
I have two dropdown boxes on my webpage, and the values that in the dropdown boxes are retrived from the mysql database. dropdown box "A" displays two differnt values, which are "Chassise_Name" and "Chassise_price", and The dropdown box "B" displays the quantity of the item. Dropdown box "A" Dropdown box "B" SuperServer 7036A-T(Black) $90 3 "Superserver 7036A-T(Black)" is retrieved from the Chassis_Name ; "$90" is retrieved from the chassis_Price; "3" is retrieved from the Quantity in other table. Question: how can I take only the price on the Dropdown box "A" multiply the number in the Dropdown box "B"? Here is my code: Code: [Select] echo "<table border='1'>"; echo "<tr><td>"; $result = mysql_query("SELECT Chassis_Name, Chassis_Price FROM Chassis where Chassis_Form_Factor='ATX'") or die(mysql_error()); echo '<select name="Chassis" onChange="change()">'; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo '<option Name="Chassis">',$row['Chassis_Name'],' ',$row['Chassis_Price'],' ','</option>'; } echo '</select>'; echo "</td>"; echo "<td>"; $result= mysql_query("SELECT Number FROM Quantity") or die(mysql_error()); echo '<select name="Quantity" Value="Quantity" onChange="change()" >'; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result)) { // Print out the contents of each row into a table echo '<option Name="Quantity">',$row['Number'],'</option>'; } echo '</select>'; echo "</td></tr>"; echo "</table>"; ?> Similar TutorialsHeja I have a MySQL database which contains information of members and the current balance of there accounts (eg: $1000) it also shows there current gain (eg: 1.2%) what i have been doing is individually updating each clients percent and balance. What i'm trying to work out is how to add a percent gain for the week and both the percent and balance of the clients are updated according to the submitted gain for example: Client 1 has $1,000 and has a current gain of 1.5% Client 2 has $,2000 and has a current gain of 2.5% I want to update both clients by 1.5% and then both clients be updated like so: Client 1 would now have $1,010.50 and now has a gain of 3.0% Client 2 would now have $2,020.50 and now has a gain of 4.0% I was looking around and could find much information on the internet or exactly has this can be achieved so i started to work on a addition and try atleast add the % however it dont go accordingly, i suppose im very rusty at my php. I started working around something like this: Code: [Select] mysql_select_db('mem') or trigger_error("SQL", E_USER_ERROR); $a1 = mysql_query("SELECT profits FROM account"); $a= $a1; $b=$b1; /* html <input type="text" name="b1" id="b1"> the % i want to update*/ $add=($a+$b); $sql = "UPDATE mem SET balance = balance +'$add'"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } This obviously does not work but it could possibly give you a better understanding of maybe what im trying to do here. Any help would be appreciated. Thanks At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. Hello, im trying to make a dropdown show up for every single row gotten from the database table. Instead it shows only one for the first row. Here is the code: <!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> <title>Example Form</title> <link rel="stylesheet" type="text/css" href="dd.css" /> <script type="text/javascript" src="jquery-1.3.2.min.js"></script> <script type="text/javascript" src="jquery.dd.js"></script> </head> <body> <?php mysql_connect("localhost", "", "")or die("cannot connect"); mysql_select_db("test")or die("cannot select DB"); $tbl_name="test_mysql"; $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); $count=mysql_num_rows($result); ?> <form name="form1" method="post" action=""> <tr> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td align="center"><strong>Row</strong></td> <td align="center"><strong>Month Date</strong></td> <td align="center"><strong>Message</strong></td> <td align="center"><strong>Title</strong></td> <td align="center"><strong>Icon</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td align="center"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td> <td align="center"><input name="month[]" MAXLENGTH="3" size="3" type="text" id="month" value="<?php echo $rows['month']; ?>"> <input name="date[]" MAXLENGTH="2" size="2" type="text" id="date" value="<?php echo $rows['date']; ?>"> </td> <td align="center"><input name="message[]" size="50" type="text" id="message" value="<?php echo $rows['message']; ?>"></td> <td align="center"><input name="title[]" size="50" type="text" id="title" value="<?php echo $rows['title']; ?>"></td> <td align="center"> <select name="test[]" style="display:none; width:200px" class="mydds"> <option value="icon_phone.gif" title="icon/icon_phone.gif">Phone</option> <option value="icon_sales.gif" title="icon/icon_sales.gif">Graph</option> </select> </tr> <?php } ?> <tr> <td colspan="4" align="center"><br><input type="submit" name="Submit" value="Submit"></td> </tr> </table> </td> </tr> </form> <script language="javascript" type="text/javascript"> function showvalue(arg) { alert(arg); //arg.visible(false); } $(document).ready(function() { try { oHandler = $(".mydds").msDropDown().data("dd"); oHandler.visible(true); //alert($.msDropDown.version); //$.msDropDown.create("body select"); $("#ver").html($.msDropDown.version); } catch(e) { alert("Error: "+e.message); } }) </script> <hr> <?php if (isset($_POST['Submit'])) { for($i=0;$i<$count;$i++){ $month = $_POST['month']; $date = $_POST['date']; $message = $_POST['message']; $title = $_POST['title']; $monthday = $month[$i]."<br>".$date[$i]; $sql1="UPDATE $tbl_name SET monthday='$monthday', month='$month[$i]', date='$date[$i]', message='" . mysql_real_escape_string($message[$i]) . "', title='" . mysql_real_escape_string($title[$i]) . "' WHERE id='$id[$i]'"; //$sql1="UPDATE $tbl_name SET monthday='$monthday', month='$month[$i]', date='$date[$i]', message='$message[$i]', title='$title[$i]' WHERE id='$id[$i]'"; $result1 = mysql_query($sql1); } header("location:update2.php"); } ?> </body> </html> <BR> I have no idea if it has to do with the PHP, HTML, or JAVASCRIPT. So if someone can tell me what is going on I would be happy. Here is how it looks: Youl see it shows the fields but not the dropdown... Please help and thank You! Hi, I'm trying to get data from one field in a table (database). But I get undesirable result: Here is my code -> <?php $result2 = mysql_query("SELECT DISTINCT theme FROM mytable ") or die(mysql_error()); while($row2 = mysql_fetch_array( $result2 )) { ?> <form method="post" action='<?php echo $_SERVER["PHP_SELF"]; ?>'> <select name='themes'"> <?php $arr= array($row2['theme']); foreach($row2 as $value) { echo "<option value='$value'><b>". $value."</b> </option><br> "; } } ?> The attached image file show the result that I don't wont. (It's not a dropdown). Is there anyone who may help me, I spent a lot of time to find out but I can't. Thanks a lot for your help mysql.php <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ckeditor",$con); ?> --------------------------------------------------------- add.php <?php include("mysql.php"); if(isset($_POST["button2"])) { $sql="INSERT INTO cktext (section) VALUES ('$_POST[select2]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } ?> ----------------------------------------------------------------- home.php <form id="form1" name="form1" method="post" action="add.php"> <tr> <td>Section:</td> <td><select name="select2" id="select2"> <option selected="selected" value="MALE">Male</option> <option selected="" value="FEMAIL">FEMAIL</option> </select></td> </tr> <input type="submit" name="button2" id="button2" value="Upload" /> </form> In DATABASE :- cktext table attribute "section" is varchar type. BUT IT RETURN ME BLANK OUTPUT. Hi all, Does anyone know if it's possible to generate a dropdown list filled with database information. I use the following to get the information from the database: $query = "SELECT Name FROM city WHERE member_ID = '$member_ID'"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($result)) { echo $row['Name'] . '<br>'; } Thnx Ryflex Hey Everyone... First off, I am only a young web developer and i'm working on a school project and am making a text-based game online... Now what i'm having trouble with... I want a drop-down list that has a list of characters classes Clubber Mixer Sauceror Tamer And I want whatever is selected to be placed into the database along with the username/password (THIS ALL WORKS FINE JUST NOT THE DROP DOWN LIST) All help appreciated Hi, I have a MySQL database called "2011_database" that has a table called "2011_list." In that table I have fields, amongst others, called "name" and "district." I need to find way to get the data from the table and put them into a drop down list on other PHP page. But they need to be listed as "name - district" on one line. I am PHP beginner and if I understand it correctly there need to be two references to get all the data in all records, a third reference to merge them together with " - " in between; and what eludes me the most, putting them in a drop down menu. Any help is greatly appreciated Thanks I am very new new php (wrote my first PHP script 5 Days ago) and am trying to give myself a crash course but I have hit a pit-stop which is killing me a little! I hope that title makes sense.... Basically I created PHP script to take data from a database and display in, I then wrote some code to use a drop down menu to order that data. That all worked ok until I tried to utilise some pagination. I can make the pagination work, and I can make the ordering work, but not at the same time! At the moment the code that I have will allow me to order the list and almost paginate it. There are 40 results and I want to display 10 at a time. When not using the ordering code I can paginate it perfectly but when I try to intergrate the two bits of code it will only display the first 10 results and not give me an option to go to the next page to see the rest! The code: if (!isset($_GET['start'])) { $_GET['start'] = 0; } $per_page = 10; $start = $_GET['start']; if (!$start) $start = 0; $sort = @$_POST['order']; if (!empty($sort)) { $get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn FROM booktable ORDER BY ".mysql_real_escape_string($_POST['order'])." ASC LIMIT $start, $per_page"); } else { $get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn FROM booktable ORDER BY bookname ASC LIMIT $start, $per_page"); } $record_count = mysql_num_rows($get); ?> <?php if (isset($_GET['showerror'])) $errorcode = $_GET['showerror']; else $errorcode = 0; ?> wont include all the html rubbish and the ordering menu! <div id="mid"> <?php echo "<table>"; echo "<tr>"; echo "<th>"; echo "</th>"; echo "<th>"; echo "Book Title"; echo "</th>"; echo "<th>"; echo "Book Author"; echo "</th>"; echo "<th>"; echo "Book Publisher"; echo "</th>"; echo "<th>"; echo "Book ISBN"; echo "</th>"; echo "<th>"; echo "</th>"; echo "</tr>"; while ($row = mysql_fetch_assoc($get)) { // get data $bookname = $row['bookname']; $bookauthor = $row['bookauthor']; $bookpub = $row['bookpub']; $bookisbn = $row['bookisbn']; echo "<tr>"; echo "<td>"; echo "<a href='addtolist.php?bookname=".$bookname."&bookauthor=".$bookauthor."&bookpub=".$bookpub."&bookisbn=".$bookisbn."'>Add to basket</a>"; echo "</td>"; echo "<td>"; echo $bookname; echo "</td>"; echo "<td>"; echo $bookauthor; echo "</td>"; echo "<td>"; echo $bookpub; echo "</td>"; echo "<td>"; echo $bookisbn; echo "</td>"; echo "</tr>"; } echo "</table>"; $prev = $start - $per_page; $next = $start + $per_page; if (!($start<=0)) echo "<a href='products.php?start=$prev'>Prev</a> "; //set variable for first page number $i=1; //show page numbers for ($x = 0; $x < $record_count; $x = $x + $per_page) { if ($start != $x) echo "<a class='pagin' href='products.php?start=$x'> $i </a>"; else echo "<a class='pagin' href='products.php?start=$x'><b> $i </b></a>"; $i++; } //show next button if (!($start >= $record_count - $per_page)) echo "<a class='pagin' href='products.php?start=$next'> Next </a>"; ?> Thank you so much for reading! Hi, I'm trying to create a dropdown box that is filled by a mysql column. I found this example, which fills the dropdown box with a list of columns in the database, but I want to fill the drop down box with the content in a column. Any guidance on how to do this would be appreciated. Thank you! Example: $connection = mysql_connect("localhost","username","password"); $fields = mysql_list_fields("database", "table", $connection); $columns = mysql_num_fields($fields); echo "<form action=page_to_post_to.php method=POST><select name=Field>"; for ($i = 0; $i < $columns; $i++) { echo "<option value=$i>"; echo mysql_field_name($fields, $i); } echo "</select></form>"; Hey guys. I am new to PHP and I have figured out how to populate a dropdown box with data from my database but now I want to create another dropdown box with data from my database but I want it to be data that depends on the data from the first drop down box. I am making code for a supply order for a website I am making for tafe and I filled in the first drop down box with the supplier names and now I want a second drop down box of all the products that that supplier is supplying us. I cannot for the life of me work out how to do this and I have no idea what to search to get this. Any help will be greatly appreciated. I will show you the code for my first dropdown box. Code: [Select] <?php $db_host = "localhost"; $db_username = "root"; $db_password = ""; $db_name = "ocs"; @mysql_connect("$db_host","$db_username","$db_password") or die("Could not connect to MySQL"); @mysql_select_db("$db_name") or die("Cannot find database"); $query = "SELECT ProductID FROM products"; $result = mysql_query($query) or die(mysql_error()); $dropdown = "<select name='products'>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['ProductID']}'>{$row['ProductID']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> Thanks a lot -Teammuffin This is actually a page to edit the drama details. Database- 3 tables 1. drama dramaID drama_title 1 friends 2. drama_genre drama_genreID dramaID genreID 1 1 2 2 1 1 3 1 3 3. genre genreID genre 1 comedy 2 romance 3 family 4 suspense 5 war 6 horror <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = ''; $dbname = 'drama'; $link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $id = $_GET['dramaID']; $link2 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql2 = "SELECT * from drama , drama_genre , genre WHERE drama.dramaID='".$id."' AND drama_genre.dramaID='".$id."' AND drama.dramaID = drama_genre.dramaID AND drama_genre.genreID = genre.genreID"; $status2 = mysqli_query($link2,$sql2) or die (mysqli_error($link2)); $link3 = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or trigger_error('Error connecting to mysql'); $sql3 = "SELECT * from genre "; $status3 = mysqli_query($link3,$sql3) or die (mysqli_error($link3)); <form method='Post' action='doUpdate.php' enctype="multipart/form-data"> <?php while ($row2 = mysqli_fetch_assoc($status2)) { ?> <td height="1"></td> <td><select name="gdrop"> <option value="<?php echo $row2['genreID'];?>"><?php echo $row2['genre'];?></option> <?php while ($row3 = mysqli_fetch_assoc($status3)) { ?> <option value="<?php echo $row3['genreID'];?>"><?php echo $row3['genre'];?></option> <?php } mysqli_close($link3); ?> <input type='hidden' id='drama_genreID' name='drama_genreID' value = "<?php echo $row2['drama_genreID']; ?>"/> </select> <input type="hidden" id="dramaID" name="dramaID" value = "<?php echo $row['dramaID'];?>"/> <input type="submit" Value="Update"/> The result was there were 3 dropdown menus but only the first dropdown menu has all 6 genres from my database and also the genre that belongs to the drama. I'm also wondering how I can bring all the drama_genreIDs to my save(doupdate.php) page and update all 3 of them because it seems like only the last dropdown menu's data is saved. And also how can I display only 6 genres instead of 7 with the genre that belongs to the drama , being set as the default selection. i have been trying to get this code to get a list of usernames from a database and i have now got that to work but when i try and save it it saves all the usernames from the drop down list and not just the one i have selected how can i get it to just use the one i have selected Code: [Select] <?php include "connect.php"; //connection string include("include/session.php"); print "<link rel='stylesheet' href='style.css' type='text/css'>"; print "<table class='maintables'>"; print "<tr class='headline'><td>Post a message</td></tr>"; print "<tr class='maintables'><td>"; // Write out our query. $query = "SELECT username FROM users"; // Execute it, or return the error message if there's a problem. $result = mysql_query($query) or die(mysql_error()); $dropdown = "<select name='username'>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>"; } $dropdown .= "\r\n</select>"; if(isset($_POST['submit'])) { $name=$session->username; $yourpost=$_POST['yourpost']; $subject=$_POST['subject']; $to=$dropdown; if(strlen($name)<1) { print "You did not type in a name."; //no name entered } else if(strlen($yourpost)<1) { print "You did not type in a post."; //no post entered } else if(strlen($subject)<1) { print "You did not enter a subject."; //no subject entered } else { $thedate=date("U"); //get unix timestamp $displaytime=date("F j, Y, g:i a"); //we now strip HTML injections $subject=strip_tags($subject); $name=strip_tags($name); $yourpost=strip_tags($yourpost); $to=strip_tags($to); $insertpost="INSERT INTO forumtutorial_posts(author,title,post,showtime,realtime,lastposter,name) values('$name','$subject','$yourpost','$displaytime','$thedate','$name','$to')"; mysql_query($insertpost) or die("Could not insert post"); //insert post print "Message posted, go back to <A href='forum.php'>Forum</a>."; } } else { print "<form action='newtopic.php' method='post'>"; print "Your name:<br>"; print "$session->username<br>"; print "User to send to:<br>"; print "$dropdown"; print "Subject:<br>"; print "<input type='text' name='subject' size='20'><br>"; print "Your message:<br>"; print "<textarea name='yourpost' rows='5' cols='40'></textarea><br>"; print "<input type='submit' name='submit' value='submit'></form>"; } print "</td></tr></table>"; ?> MOD EDIT: Changed PHP manual link [m] . . . [/m] tags to [code] . . . [/code] tags. Hi. I am using this script to populate a dropdown list box from sql, it works but does anyone know how to sort the list in alphabetical order? $sql="SELECT * FROM Fish WHERE ***** = '".$_GET['stocktype']."'"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $ID=$row["ID"]; $Stock=$row["Commonn"]; $Options.="<OPTION VALUE=\"$ID\">".$Stock; } <SELECT NAME='stock1'> <OPTION VALUE='$Options'>$Options</option> </SELECT> Hi Guys, I am a typical newbie with a twist, i program in VB, recently moved to PHP. I need help, preferable an example, but first let me explain what i have. I have a .tpl page that has 3 drop down boxes, one of these is populated from within the .tpl page itself. The other 2 i waant to pupulate from a SELECT on a database from a .php file. The .TPL file i have so far is:- Code: [Select] <style type="text/css"> <!-- .style7 { color: #006600; font-weight: bold; font-size: 18px; } .style14 {color: #003300; font-weight: bold; } .style16 { color: #000066; font-style: italic; } .style25 { color: #FF0000; font-weight: bold; font-size: 12px; } .style28 { font-size: 12px; color: #330000; } .style29 { font-size: 12px } .style31 {color: #003300; font-weight: bold; font-size: 16px; } .style33 { color: #000099; font-weight: bold; font-style: italic; font-size: 18px; } .style34 { color: #FF0000; font-weight: bold; font-size: 24px; } .style35 {font-size: 12px; color: #003300; } .style36 { color: #FF0000; font-weight: bold; font-size: 14px; font-style: italic; } .style38 {color: #003300; font-weight: bold; font-size: 16px; font-style: italic; } --> </style> <script src="Scripts/AC_RunActiveContent.js" type="text/javascript"></script> <h1 align="center" class="style34">Scotbirds Alertz - Rare and Scarce</h1> <h3 align="center" class="style36">Soon Only VIP Members will be able to Access this Page</h3> <h3 align="center" class="style14">If you have seen something unusual / rare then please call it in on -- <span class="style33">Hotline: 0333 5772473</span></h3> <p align="center" class="style14"> <table width="90%" border="0" align="center" cellpadding="5"> <tr> <?php print $_SERVER['PHP_SELF']; $the_date_filter = $_GET["DATE_FILTER"];?> <td width="1%"><td width="5%"><td width="2%"><td width="2%"><td width="2%"><form action="alertz_VIP.php" method="post"> <td width="2%"><td width="2%"><td width="5%"><span class="style38">Date</span><td width="3%"></td> <td width="12%"><span class="style14"> <select name=DATE_FILTER size="1" id=DATE_FILTER onchange="this.form.submit()"> <option value="All Dates">All Dates</option> <option value="Today">Today</option> <option value="Last 48hrs">Last 48hrs</option> <option value="Last week">Last Week</option> <option value="last month">Last Month</option> </select> </span></td> <td width="1%"> </td> <td width="7%"><span class="style16"><span class="style14"><span class="style31">Region</span></span></span></td> <td width="22%"><span class="style16"><span class="style14"> <select name=REGION_FILTER size="1" id=REGION_FILTER onchange="this.form.submit()"> <option value="AllRg">All Regions</option> <option value={REGION_FILTER}></option> </select> </span></span></td> <td width="1%"> </td> <td width="8%"><span class="style16"><span class="style14"><span class="style31">Species</span></span></span></td> <td width="20%"><span class="style16"><span class="style14"> <select name=SPECIES_FILTER size="1" id=SPECIES_FILTER onchange="this.form.submit()"> <option value="AllSpec">All Species</option> <option value={SPECIES_FILTER}></option> </select> </span></span></td> <td width="9%"><input type="Submit" name="submit2" value="Search" method="get" action="alertz_VIP.php" /></td> </table> <p> </p> <table width="870" border="1" align="center" cellpadding="0" cellspacing="0" bordercolor="#99CC99"> <tr> <th width="178" class="style31" scope="col">Region</th> <th width="184" class="style31" scope="col">Species</th> <th width="96" class="style31" scope="col">Date</th> <th width="96" class="style31" scope="col">Time</th> <th width = "304" class="style31" scope="col">Comments</th> </tr> <!-- BEGIN alerts --> <tr> <th class="style7 style29" scope="col">{alerts.REGION}</th> <th class="style25" scope="col">{alerts.SPECIES}</th> <th class="style35" scope="col">{alerts.DATE} </th> <th scope="col"><span class="style35">{alerts.TIME} </span></th> <th scope="col"><span class="style28">{alerts.COMMENTS} </span></th> </tr> <!-- END alerts --> </table> <p align="center" class="style14"> </p> Now from here i expect the user to select a date from the date box, ( Ideally i would like the other 2 options to only become visible once a selection on the date has been made, i would also like to retain the selected in the date box once the form has been submited. Now the code i have for the PHP so far is quite long winded as my PHP skills are not so good, although i am learning fast. Code: [Select] <?php /*************************************************************************** * Alertz.php * ------------------- * begin : 30/10/04/10 * copyright : (C) 2010 Andy Guppy * email : webmaster@scotbird.co.uk * * ***************************************************************************/ define('IN_ICYPHOENIX', true); if (!defined('IP_ROOT_PATH')) define('IP_ROOT_PATH', './'); if (!defined('PHP_EXT')) define('PHP_EXT', substr(strrchr(__FILE__, '.'), 1)); // Include files include(IP_ROOT_PATH . 'common.' . PHP_EXT); include_once(IP_ROOT_PATH . 'includes/functions_groups.' . PHP_EXT); include(IP_ROOT_PATH . '/alerts/alertsconfig.'.PHP_EXT); // Page Authorise $cms_page_id = 'scotalertz'; $cms_page_nav = (!empty($cms_config_layouts[$cms_page_id]['page_nav']) ? true : false); $cms_global_blocks = (!empty($cms_config_layouts[$cms_page_id]['global_blocks']) ? true : false); $cms_auth_level = (isset($cms_config_layouts[$cms_page_id]['view']) ? $cms_config_layouts[$cms_page_id]['view'] : AUTH_ALL); check_page_auth($cms_page_id, $cms_auth_level); // Obtain Select Criteria $temp_region_filter = $_REQUEST["REGION_FILTER"]; $temp_species_filter = $_REQUEST["SPECIES_FILTER"]; $temp_date_filter = $_REQUEST["DATE_FILTER"]; // Filter characters if required $region_filter =str_replace(" & ", "&", $temp_region_filter); $species_filter = $temp_species_filter; $date_filter = $temp_date_filter; // standard session management $userdata = session_pagestart($user_ip); // Check to see if user is logged in if ((!$userdata['session_logged_in']) ) // No he isnt { redirect(append_sid(LOGIN_MG . '?redirect=alerts.' . PHP_EXT)); } else // Yes they are { init_userprefs($userdata); // set page title $page_title = "ScotBird alerts - VIP's ONLY !! "; // standard page header include(IP_ROOT_PATH . 'includes/page_header.'.PHP_EXT); // Connect to the database $db = new sql_db($alerts_mysql_host,$alerts_mysql_username,$alerts_mysql_password,$alerts_mysql_db,false); if(!$db) { die("Database Connection Failed:- Please Contact Site Admin" . mysql_error()); } $template->set_filenames(array('body' => 'alertz_VIP.tpl')); if (!isset($_REQUEST['DATE_FILTER'])) { // if date is NOT set then perform this echo 'Date has not been set'; $sql2 = "SELECT * FROM alerts ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts GROUP BY region"; echo $sql2; } else { // if date is set then perform this $year =date("Y"); $month = date("m"); $day = date("d"); $theenddate = $year . '-' .$month . '-' . $day ; echo 'the choice selected was :- '.$_REQUEST['DATE_FILTER']; switch ($date_filter) { case "All Dates": $sql2 = "SELECT * FROM alerts ORDER BY Date DESC, time DESC "; break; case "Today": $sql2 = "SELECT * FROM alerts WHERE Date = '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date = '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date ='".$theenddate ."' GROUP BY region"; break; case "Last 48hrs": $year =date("Y"); $month = date("m"); $day = date("d")-1; $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; case "Last week": $year =date("Y"); $month = date("m"); $day = date("d")-7; $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; case "last month": $year =date("Y"); $month = date("m")-1; $day = date("d"); $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; case "last 3 months": $year =date("Y"); $month = date("m")-3; $day = date("d"); $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; break; default: // Default is the last 48hrs $year =date("Y"); $month = date("m"); $day = date("d")-1; $thestartdate = $year . '-' .$month . '-' . $day ; $sql2 = "SELECT * FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' ORDER BY Date DESC, time DESC "; $sql3 = "SELECT DISTINCT species FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY species"; $sql = "SELECT DISTINCT region FROM alerts WHERE Date BETWEEN '".$thestartdate ."' AND '".$theenddate ."' GROUP BY region"; } } $result2 = $db->sql_query($sql2); if (!($result2 = $db->sql_query($sql2))) { die("Database Query Failed" . mysql_error()); } $i = 1; while ( $row2 = $db->sql_fetchrow($result2) ) { $template->assign_block_vars('alerts', array( 'POS' => $i , 'REGION' => str_replace("&", " & ", $row2['region']), 'SPECIES' => str_replace("%", "'",$row2['species']), 'DATE' => $row2['Date'], 'TIME' => $row2['time'], 'COMMENTS' => str_replace("'", "%",$row2['comments']), ) ); $i++; } $template->assign_vars(array( 'USERNAME' => htmlspecialchars($userdata[username]), 'REGION_FILTER' => str_replace("&", " & ", $region_filter_options), 'SPECIES_FILTER' => $species_filter_options, ) ); $result = $db->sql_query($sql); if (!($result = $db->sql_query($sql))) { die("Database Query Failed" . mysql_error()); } // This is where you would add a new VARS Array if you intend to use your own custom VARS. while ($row = $db->sql_fetchrow($result)) { $region_filter_options .= '<option value="' . $row['region'] . '">' . $row['region'] . '</option>'; } $result = $db->sql_query($sql3); if (!($result = $db->sql_query($sql3))) { die("Database Query Failed" . mysql_error()); } // This is where you would add a new VARS Array if you intend to use your own custom VARS. while ($row = $db->sql_fetchrow($result)) { $species_filter_options .= '<option value="' . $row['species'] . '">' . $row['species'] . '</option>'; } $template->assign_vars(array( 'USERNAME' => htmlspecialchars($userdata[username]), 'REGION_FILTER' => str_replace("&", " & ", $region_filter_options), 'SPECIES_FILTER' => $species_filter_options, ) ); // Build the page $template->pparse('body'); // standard page footer include(IP_ROOT_PATH . 'includes/page_tail.'.PHP_EXT); } ?> Now my questions are these:- 1) In the .tpl file hows can i submit from any of the dropdown boxes and retain the selection after submission 2) How can i have it so that the second and third drop down boxes are only visible after the previous one, ie the first box ( date ) has to have a selection before the Region ( 2nd one ) is visible and so 3) How can i reduce the amount of code to cover all options in building a select statement or is Switch - case the best way. I would greatly appreciate help with this and even more so for some example of what i am asking for so i can learn from them. Hi guys,
Does anyone know why my code isn't working?:
$servername="127.0.0.1"; $username="root"; $password=""; $dbname="testdb"; $connection = new mysqli($servername,$username,$password); $result = "SELECT 'Country' from 'w'"; echo '<select>'; foreach($result as $res) { echo '<option value="'.$res['Country'].'">' . $res['Country'] . '</option>'; } echo '</select>'; }There's not a credential issue, the table is called 'w' and the collumn I'm trying to get the values from is 'Country' Any help is appreciated. Cheers, Broken Hi, I don know how to explain this in words too well, so i have create 2 images to show what I need. What I need to do is use whats in the first image (table.jpg)(the database) to create whats in the second image (dropdown.jpg)(the drop-down menu) The results of the form will be entered into the database. category_id will be the value thats inserted. Many kind regards Eoin [attachment deleted by admin] Hello,
I have been trying to figure this for a while now and reading the tutorials are not helping, I think I'm a little over my head on this one and was hoping someone could help me out with this issue.
I am making a User Edit page and would like to have the access level part of the form show the users access current access level thats set in the database when the page loads, and if it needs to be changed you can press the dropdown box and select a new access level. I can't figure out how to show the current access level as default and populate the drop down box with the other access levels in my table.
My Tables look like this
Users table (users):
---------------------------------------------------------------------------------------------------
| id | username | password | flag | realfirst | reallast | dept |
---------------------------------------------------------------------------------------------------
1 loderd 9 test guy Service
Auth Table (auth):
--------------------------------------------
| id | auth_level | descrip |
--------------------------------------------
1 1 Service Tech
2 2 Office Staff
3 9 Super Admin
My SQL Query looks like this
$users = $db->fetch_all_array("SELECT users.id, users.username, users.password, users.realfirst, users.reallast, users.dept, users.flag, auth.auth_level, auth.descrip, auth.id FROM users LEFT JOIN auth ON users.flag = auth.auth_level WHERE users.id = '".$_GET['id']."'");I can't seem to figure out how I can do this for the Access Level dropdown box. <tr> <td width="19%" align="right" valign="top">Access Level :</td> <td width="1%" align="left"> </td> <td width="80%" align="left" valign="top"> <?php echo "<select name='flag' id='flag'>"; foreach ($users as $row){ if($row[auth_level]==$row[auth_level]){ echo "<option value=$row[auth_level] selected>$row[auth_level] - $row[descrip]</option>"; }else{ echo "<option value=$row[auth_level]>$row[auth_level] - $row[descrip]</option>"; } } echo "</select>"; ?> </td> </tr>Any help would be greatly appreciated This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=351349.0 |