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PHP - Php Code For Selecting Multiple Values From Dropdown
I have a sql query similar to below in a .php file.
$sql = "select id,organisation,price from table where category = '$category';
These are the four unique values of the category column for your reference.
"A-t1"
"B-t1"
"C-t1"
"D-t1"
Now I want to create a dropdown list or listbox in .php file with option to select multiple values. If customer selects multiple values, it should fetch query for the selected categories.
E.g. dropdown list should be similar to below :
"A-t1"
"B-t1"
"C-t1"
"D-t1"
If an user selects "A-t1" and "C-t1", it should give output/query for the selected categories.
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Ok I'm trying to select the pet with the current eggid in the url. That part works fine, but when I try to select more than one color, it gets all messed up. Because I don't know how to do it and I can't find it on the internet anywhere. Code: [Select] $sql = "SELECT * FROM pets WHERE eggid='".$_GET['eggid']."' AND color='blue' OR 'green' OR 'yellow' OR 'red'"; Does anyone know how to do that? Because whenever I put this: Code: [Select] $sql = "SELECT * FROM pets WHERE eggid='".$_GET['eggid']."' AND color='blue' OR color='yellow' OR color='green' OR color='red'"; It gets all messed up. It selects ALL the colors even if the eggid isn't the current eggid. So it displays the colors of all the pets that have those colors.. :/ EDIT: OR is this the only way I can do it: Code: [Select] $sql = "SELECT * FROM pets WHERE eggid='".$_GET['eggid']."' AND color='blue' OR eggid='".$_GET['eggid']."' AND color='green' OR eggid='".$_GET['eggid']."' AND color='yellow' OR eggid='".$_GET['eggid']."' AND color='red'"; Hi, I've got a basic sign up form but I want a drop down list which will list different catergories that relate to different tables which when selected will input the sign up information into that table which was selected from the catergory drop down. This is the signup form <html><head><title>Birthdays Insert Form</title> <style type="text/css"> td {font-family: tahoma, arial, verdana; font-size: 10pt } </style> </head> <body> <table width="300" cellpadding="5" cellspacing="0" border="2"> <tr align="center" valign="top"> <td align="left" colspan="1" rowspan="1" bgcolor="64b1ff"> <h3>Insert Record</h3> <form method="POST" action="test.php"> <? print "Enter Company Name: <input type=text name=company_name size=30><br>"; print "Enter Contact Name: <input type=text name=contact_name size=30><br>"; print "Enter Telephone: <input type=text name=telephone size=20><br>"; print "Enter Fax: <input type=text name=fax size=30><br>"; print "Enter Email: <input type=text name=email size=30><br>"; print "Enter Address: <input type=text name=address1 size=20><br>"; print "Enter Address: <input type=text name=address2 size=30><br>"; print "Enter Postcode: <input type=text name=postcode size=30><br>"; print "Enter Town / City: <input type=text name=town_city size=20><br>"; print "Enter Website: <input type=text name=website size=30><br>"; print "Enter Company Type: <select name='table'> <option>stationary</option><option>reception</option></select><br>"; print "<br>"; print "<input type=submit value=Submit><input type=reset>"; ?> </form> </td></tr></table> </body> </html> This is the part which I can't figure out and is probably totally wrong! Im trying to use this script to sort the drop down list to then run the correct script to insert the form data. <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Hello!</title> </head> <body> <?php if($_POST['table']=='stationary' 'birthdays_insert_record.php') else if($_POST['table']=='reception' 'insert_reception.php') ?> </body> </html> This is the script which works! that inserts the form data into a specific table <html><head><title>Birthdays Insert Record</title></head> <body> <? /* Change db and connect values if using online */ $company_name=$_POST['company_name']; $contact_name=$_POST['contact_name']; $telephone=$_POST['telephone']; $fax=$_POST['fax']; $email=$_POST['email']; $address1=$_POST['address1']; $address2=$_POST['address2']; $postcode=$_POST['postcode']; $town_city=$_POST['town_city']; $website=$_POST['website']; $db="myflawlesswedding"; $link = mysql_connect('localhost', 'root' , ''); if (! $link) die(mysql_error()); mysql_select_db($db , $link) or die("Select Error: ".mysql_error()); $result=mysql_query("INSERT INTO reception (company_name, contact_name, telephone, fax, email, address1, address2, postcode, town_city, website) VALUES ( '$company_name', '$contact_name', '$telephone', '$fax', '$email', '$address1', '$address2', '$postcode', '$town_city', '$website')") or die("Insert Error: ".mysql_error()); mysql_close($link); print "Record added"; ?> <form method="POST" action="birthdays_insert_form.php"> <input type="submit" value="Insert Another Record"> </form> <br> <form method="POST" action="birthdays_dbase_interface.php"> <input type="submit" value="Dbase Interface"> </form> </body> </html> I hope somebody can help me out here! or can point me in a better way to sort this problem! Thanks for any advice! I am trying to create an autocomplete form for 'city,state zip' I want to be able to search by either a distinct zip code that will show 'city, state zip' or by distinct city 'city, state' Can anyone tell me how to fix my script? $sql = "SELECT DISTINCT zip,city,state FROM `residential` WHERE `zip` LIKE '$input%' OR `city` LIKE '$input%' OR `state` LIKE '$input%' UNION SELECT DISTINCT city,state,zip FROM `residential` WHERE `zip` LIKE '$input%' OR `city` LIKE '$input%' OR `state` LIKE '$input%' UNION SELECT DISTINCT state,city,zip FROM `residential` WHERE `zip` LIKE '$input%' OR `city` LIKE '$input%' OR `state` LIKE '$input%' LIMIT $limit"; $result = mysql_query($sql); if (!$result || !mysql_num_rows($result)) exit; include_once "headers.php"; echo "<response>"; while ($row = mysql_fetch_array($result)) { $keywords = "$row[city], $row[state] $row[zip]"; echo "<keywords>". $keywords ."</keywords>"; } while ($row = mysql_fetch_array($result)) { $keywords = "$row[city], $row[state]"; echo "<keywords>". $keywords ."</keywords>"; } echo "</response>"; I want to query the five heighest values in a coloum but the coloum is not the primarykey. i.e: i have a coloum on the end of my table (articles) called views which get updated when ever an article is viewed by adding 1. On anther page i want to be able to view the 5 most viewed articles. hi i have a table like this rank xp 1 0 2 50 3 100 4 200 say my username has 75xp how would a select which rank i am not sure how i can do this? :S cheers matt I am running a query to get a certain student's quiz results from all attempts and print one after the other in a simple table for now. The student is designated by an email address. I need results from all of the rows containing this particular email. The issue I think I am having is that when I use "WHERE" and specify an email address that exists many times in the table it isn't storing the variable correctly. Here is code below that I cannot get working right. When I try to print these variables as a test to see if they are working I get nothing. No errors either. I have checked all names, spellings, and capitalization about 5 times. Thanks for looking at it. Code: [Select] $QuizQuery = mysql_query("SELECT * FROM Quiz_Results WHERE Email = '".$_SESSION['UserEmail']."' AND Quiz_Name = 'M1'") or die(mysql_error()); $i=1; while ($QuizResults = mysql_fetch_array($QuizQuery)){ $UserEmail = $QuizResults['Email']; $Score = $QuizResults['Score']; $Date = $QuizResults['Date_Taken']; echo "<table width='650' border='1'><tr><td>"; echo "Attempt #".$i."<br></td><td>Sco ".$Score."</td><td>"; if($QuizResults['Pass'] == "PASSED"){ echo "Passed"; } else { echo "Failed"; } echo "</td><td>Date Taken: ".$Date."</td></tr></table>"; $i++; } i am making a comparison site for products say i have a table ! A ! B ! C ! ---------------------------------------------------------------------- product 1 ! 1.34 ! 2.78 ! 7.90 ! ---------------------------------------------------------------------- product 2 ! 1.56 ! 3.90 ! 8.00 ! ---------------------------------------------------------------------- product 3 ! 1.22 ! 2.99 ! 9.00 ! ---------------------------------------------------------------------- product 4 ! 1.09 ! 3.00 ! 8.50 ! ---------------------------------------------------------------------- say i wanted to buy product 1 and 2 from the cheapest company A/B or C how would i go about doing this ? any guidance would be helpful cheers matt Hi, My company has 240+ locations and as such some users (general managers) cover multiple sites. When I run a query to pull user information, when the user has multiple sites to his or her name, its adds the second / third sites to the next columns, rather than wrapping it inside the same table cell. It also works the opposite way, if a piece of data is missing in the database and is blank, its pull the following columns in. Both cases mess up the table and formatting. I'm extremely new to any kind of programming and maybe this isn't the forum for this question but figured I'd give it a chance since I'm stuck. The HTML/PHP code is below: <table id="datatables-column-search-select-inputs" class="table table-striped" style="width:100%"> <thead> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> <th>Actions</th> </tr> </thead> <tbody> <?php //QUERY TO SELECT ALL USERS FROM DATABASE $query = "SELECT * FROM users"; $select_users = mysqli_query($connection,$query);
// SET VARIABLE TO ARRAY FROM QUERY while($row = mysqli_fetch_assoc($select_users)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_phone = $row['user_phone']; $user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id'];
// POPULATES DATA INTO THE TABLE echo "<tr>"; echo "<td>{$user_id}</td>"; echo "<td>{$user_firstname}</td>"; echo "<td>{$user_lastname}</td>"; echo "<td>{$username}</td>"; echo "<td>{$user_phone}</td>";
//PULL SITE STATUS BASED ON SITE STATUS ID $query = "SELECT * FROM sites WHERE site_manager_id = {$user_id} "; $select_site = mysqli_query($connection, $query); while($row = mysqli_fetch_assoc($select_site)) { $site_name = $row['site_name']; echo "<td>{$site_name}</td>"; } echo "<td>{$user_title_id}</td>"; echo "<td>{$user_role_id}</td>"; echo "<td class='table-action'> <a href='#'><i class='align-middle' data-feather='edit-2'></i></a> <a href='#'><i class='align-middle' data-feather='trash'></i></a> </td>"; //echo "<td><a href='users.php?source=edit_user&p_id={$user_id}'>Edit</a></td>"; echo "</tr>"; } ?>
<tr> <td>ID</td> <td>FirstName</td> <td>LastName</td> <td>Username</td> <td>Phone #</td> <td>Location</td> <td>Title</td> <td>Role</td> <td class="table-action"> <a href="#"><i class="align-middle" data-feather="edit-2"></i></a> <a href="#"><i class="align-middle" data-feather="trash"></i></a> </td> </tr> </tbody> <tfoot> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> </tr> </tfoot> </table>
Hi All ,
I have a small table with 4 fields namely Day_ID, Dues, Last_Visit, Points. where Day_ID is an auto-increment field. The table would be as follows:
Day_ID -- Dues --- Last_Visit --- Points.
1 --------- 900 -------- 1/12 -------- 6
2 --------- 700 -------- 4/12 -------- 7
3 --------- 600 -------- 7/12 -------- 5
4 --------- 600 -------- 9/12 -------- 6
5 --------- 600 -------- 10/12 ------- 6
6 --------- 600 -------- 14/12 ------- 6
So this is the record of a person's visit to say a club. The last row indicates the last date of his visit to the club. His points on this date are 6. Based on this point value of 6 in the last row I want to retrieve all the previous BUT adjoining all records that have the same Points i.e. 6.
So my query should retrieve for me, based on the column value of Points of the last row (i.e. Day_ID - 6 ), as follows:
4 --------- 600 -------- 9/12 -------- 6
5 --------- 600 -------- 10/12 ------- 6
6 --------- 600 -------- 14/12 ------- 6
This problem stated above had been completely resolved, thanks to a lot of help from Guru Barand by this following query :-
$query = "SELECT cv.day_id, cv.dues, cv.last_visit, cv.points FROM clubvisit cv WHERE last_visit >= ( SELECT MAX(last_visit) FROM clubvisit WHERE points <> ( SELECT points as lastpoints FROM clubvisit JOIN ( SELECT MAX(last_visit) as last_visit FROM clubvisit ) as latest USING (last_visit) ) )";I am using this and it works perfectly except that now there is a slight change in the table because the criteria for points is now dependent on more than one column cv.points and is more like cv.points1, cv.points2, cv.points3 etc. So now I need to make a selection based on each of these cv.points columns. As of now I can still get the results by running the query multiple times for each of the cv.points columns ( seperately for cv.points1, cv.points2, cv.points3) and it works correctly. However I am wondering if there is a better way to do this in just one go. This not only makes the code repetitive but also since the queries are interconnected, involves the use of transactions which I wish to avoid if possible. The values that I require for each of the cv.point columns is 1. day_id of the previous / old day on which the cv.points value changed from the current day value, and 2. cv.points on that old/ previous day. So for example if the table is as below: Day_ID -- Dues --- Last_Visit --- Points1 --- Points2. 1 --------- 900 -------- 1/12 ----------- 9 ------------ 5 2 --------- 600 -------- 4/12 ----------- 6 ------------ 6 3 --------- 400 -------- 7/12 ----------- 4 ------------ 7 4 --------- 500 -------- 9/12 ----------- 5 ------------ 8 5 --------- 600 -------- 10/12 ---------- 6 ------------ 8 6 --------- 600 -------- 11/12 ---------- 6 ------------ 8 7 --------- 600 -------- 13/12 ---------- 6 ------------ 7 8 --------- 500 -------- 15/12 ---------- 5 ------------ 7 9 --------- 500 -------- 19/12 ---------- 5 ------------ 7 Then I need the following set of values : 1. day_id1 -- Day 7, points1 ---- 6, days_diff1 -- (9-7 = 2) . // Difference between the latest day and day_id1 2. day_id2 -- Day 6, points2 ---- 8, days_diff2 -- (9-6 = 3) 3. day_id3 -- .... and so on for other points. Thanks all ! Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. I don't even know what I'm trying to do.. but I can't figure out how to do it.. or how I should do it. I have a table with a bunch of rows in it.. let's say 10 rows. I want to output two fields from each one of those rows wherever I want on my page. So if I do it this way, I can't use: Code: [Select] while($result = mysql_fetch_array($query)){ I can't use that because that assumes I want to repeat each line. However, I don't want to use a new query each time because the actual number I'll be using is closer to 100. So if that makes sense.. does anyone have any ideas? Folks, I have a dropdown, once values are selcted, these values should be put in a URL structure so that it matches with the one MoD_rewrite rule in my .htaccess. Here is my Mod_Rewrite Rule: RewriteRule ^(.*)/([^/]*)\.html$ search.php?q=$1&sc=$2 [QSA,L] Here is my Dropdown Code: <div id="search" > <form id="searchform" method="get" action="search.php"> <label>Search By Brand/ Manufacturer: </label> <select name="q"> <option value="SelectBrand">Select Brand</option> <?php if(isset($this->search->options)): ?> <?php foreach($mfg as $lolachild): ?> <option value="<?php echo $lolachild; ?>"><?php echo ucwords($lolachild); ?></option> <?php endforeach; ?> <?php endif; ?> </select> <select name="sc"> <option value="All"><?php eprint(LangAll); ?></option> <?php if(isset($this->search->options)): ?> <?php foreach($this->search->options as $cat): ?> <option value="<?php echo $cat->value; ?>"><?php echo $cat->name; ?></option> <?php endforeach; ?> <?php endif; ?> </select> <input type="submit" value="<?php eprint(LangSearch); ?>" /> </form> </div> Problem is, upon Submit, it goes to this link structu http://mydomain.co.uk/search.php?q=fisher&sc=302 It should rather be: http://au2.co.uk/fisher/302.html What am i missing or doing wrong? Cheer Natasha (not sure if this should be in php or javascript section) I am trying to create a dropdown list that loads content according to the option chosen.. The idea is that you select from 4 or so dropdown boxes, which then selects the appropriate content that the user picked Pretty much like http://www.cooldiamonds.com/ (after you click enter) This is what I've got so far: Option.html Code: [Select] <script src="http://code.jquery.com/jquery-latest.js" type="text/javascript"></script><br> <script type="text/javascript"> $(document).ready(function(){ $('.select').change(function() { var option = $(this).val(); $.get('option.php', {select:option}, function(data) { $('#result').html(data).hide().fadeIn(1000); }); }); }); </script> </head> <body> <p> Selection 1</p> <select name="select" id="select" class="select"> <option value="">Select</option> <option value="option1">Option 1</option> <option value="option2">Option 2</option> </select> <p> Selection 2</p> <select name="select2" id="select2" class="select"> <option value="">Select</option> <option value="option3">Option 3</option> <option value="option4">Option 4</option> </select> <p> Selection 3</p> <select name="select3" id="select3" class="select"> <option value="">Select</option> <option value="option5">Option 5</option> <option value="option6">Option 6</option> </select> <p> Selection 4</p> <select name="select3" id="select3" class="select"> <option value="">Select</option> <option value="option7">Option 7</option> <option value="option8">Option 8</option> </select> <div id="result" style="border:1px solid #000;padding:10px;color:#ff0000;display:none;"></div> option.php Code: [Select] <?php // Multiple Selections if ($_GET['select'] == 'option1' && $_GET['select2'] == 'option3') { echo 'the option you have chosen is 1 and 3';} elseif ($_GET['select2'] == 'option4' && $_GET['select3'] == 'option5' && $_GET['select4'] == 'option7') { echo 'the option you have chosen is 4, 5 and 7';} // and so on //Selection 1 elseif($_GET['select'] == 'option1') { echo 'the option you have chosen is 1';} elseif($_GET['select'] == 'option2') { echo 'the option you have chosen is 2';} //Selection 2 elseif($_GET['select2'] == 'option3') { echo 'the option you have chosen is 3';} elseif($_GET['select2'] == 'option4') { echo 'the option you have chosen is 4';} //Selection 3 elseif($_GET['select3'] == 'option5') { echo 'the option you have chosen is 5';} elseif($_GET['select3'] == 'option6') { echo 'the option you have chosen is 6';} // Selection 4 elseif($_GET['select4'] == 'option7') { echo 'the option you have chosen is 7';} elseif($_GET['select4'] == 'option8') { echo 'the option you have chosen is 8';} ?> You can view it he http://adamwatkin.com/option.html As you can see selection 1 works fine, but the other selections do not work and neither does multiple selections. Its almost certainty theres errors in the code, does anyone have any idea what the problem is? Any help will be appreciated Cheers Is is possible to give multiple values in a single checkbox? i need to have a checkbox that holds multiple values. Help Anyone.. Hi Guys, I don't know how to do this but basically I want the value of $postcode3 to equal the value of $postcodep1 aswell as $postcodep2 like this... else {$postcode3 = $postcodep1 $postcodep2;} That doesn't work... Say $postcodep1 = LS14 $postcodep2 = 2AB Then $postcode3 = LS14 2AB How do I write this in the above php code?! Thanks, S I'm sorting out a simple PHP email form, and I would like, when the email arrives in my inbox, for it to display both the email of the sender AND their name as specified in the form on my website. So far I have $from = $_POST['from']; with the 'from' being the client's email address. I would also like their name ('name') to show up next to their email in the header. I have tried such things like $from = $_POST['from' . 'name']; //or $from = $_POST['from']; $_POST['name']; and a few more, but I'm terrible at PHP coding in general. I have searched, but I really am at a loss as to what to search for. Can someone please give me an idea as to what I should do? Thanks, Sacha. I have a form that has a list of checboxes, each checkbox has multiple values. IE. Checkbox1 has Date 1 and Cost1. Checkbox2 has Date2 and cost2 and so on. This is what I have in form. echo "<input type=checkbox name=service_id[] value=".$id."><label>".$description."</label> Date:<input type=date name=date[] value=".date('Y-m-d').">Date:<input type=number name=cost[] value=".$cost."><br>"; When i run the for loop to insert the values into a table i have this.
foreach($_POST['service_id'] as $key => $value){
The problem is that when I select certain boxes (ie, checkbox #2), it inserts Date1 and Cost1 but does use the correct service id (ie checkbox) 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. |
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