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PHP - Multiple Dropdown List To Load Content
(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 Similar TutorialsHi , I have one question .. Can I split showing of content of dynamic list in 2 parts , when I echo list in code .. Code: [Select] <?php // Run a select query to get my letest 8 items // Connect to the MySQL database include "../connect_to_mysql.php"; $dynamicList = ""; $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 8"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $product_name = $row["product_name"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamicList .= '<table width="100%" border="2" cellspacing="2" cellpadding="2"> <tr> <td width="17%" valign="top"><a href="product.php?id=' . $id . '"><img style="border:#666 1px solid;" src="inventory_images/' . $id . '.jpg" alt="' . $product_name . '" width="77" height="102" border="2" /></a></td> <td width="83%" valign="top">' . $product_name . '<br /> $' . $price . '<br /> <a href="product.php?id=' . $id . '">View Product Details</a></td> </tr> </table>'; } } else { $dynamicList = "We have no products listed in our store yet"; } mysql_close(); ?> Code: [Select] <p><?php echo $dynamicList; ?><br /> </p> It works ok, and putting my files, everything works, but when I put 8 pictures with price and other details, it just show one image with details and another image below with details, and the third image below and so on .. Can I split dynamic list to show 4 images with details on the left side and 4 on the right side? Thank you in advance for help , if is possible I understand that is might be something that is already answered and I apologize if it is, I could not find it.
What I need to do is build a simple form that has two options, they will be dropdown options. Dropdown A and Dropdown B then a Submit button. This part I understand in HTML, although it may be easier in php or javascript.
Then I need it to take the two options and create a "if/then" statement that loads a specific pdf that matches the two options selected.
Example.
If someone selects Option 1 from Dropdown A and Option 2 From Dropdown B then it loads 12.pdf If someone selects Option 5 from Dropdown A and Option 3 From Dropdown B then it loads 53.pdf If someone selects Option 2 from Dropdown A and Option 1 From Dropdown B then it loads 21.pdf and so on... It does not have to be the exact thing just some way to take both inputs and have it equal a specific pdf. Here is the form I built but I don't know what to put in the form_action.php file in order to make it work <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <center> <h1> Get Directions</h1> <form action="form_action.php" method="get" name="directions" target="_new"> <select name="startpoint" size="1"> <option value="north">North Tower Entrance</option> <option value="south">South Tower Entrance</option> <option value="moba">MOB A Entrance</option></select> -----> <select name="endpoint" size="1"> <option value="onco">Oncology</option> <option value="radio">Radiology</option> <option value="pulm">Pulmanary</option></select> <br /><br /> <input type="submit" value="Submit" /> </form> </center> </body> </html>Any help is appreciated, thanks. I have 10 tables in my database. I want load these 10 table names into a drop-down list. If a user selects a value form the drop-down list, a form should be displayed with related to the table. (form fields should be appropriate table's column headers) When user fill the form and submitted, data should be saved in the selected table. Give me a code example to do this. Hi All,
Am hoping someone can point out my perhaps obvious issue here.
I've a series of links that are dynamically created:
<a class="team-link" rel="<?php the_ID(); ?>" href="#">When a link is clicked, I'm trying to get content to load into this DIV: <div id="single-post-container"></div>I have the following jquery:
<script>
$(document).ready(function(){
$.ajaxSetup({cache:false});
$(".team-link").click(function(){
var post_id = $(this).attr("rel");
$.ajax({
url : "http://www.domain.co.uk/wp-content/themes/name/loop.php",
data : {"the_team": post_id },
type : "POST",
dataType : "html",
success: function(response) {
$("#single-post-container").html(response);
}
return false;
});
});
</script>
This is my loop.php file:
require_once('../../../wp-load.php');
// Our variables
$the_team = (isset($_GET['post_id'])) ? $_GET['post_id'] : 0;
echo $the_team;
query_posts(array(
'post_type' => 'work',
'meta_key' => 'author_name',
'meta_value' => $the_team
));
// our loop
if (have_posts()) {
while (have_posts()){
the_post();
?>
<h1><?php the_title(); ?></h1>
<?php the_content(); ?>
<hr />
<?php
}
}
wp_reset_query();
However, when you click the link, nothing happens. I know loop.php works as I can specify a value and manually browse to it to see the content.
I feel I'm missing something obvious as I'm new to AJAX.
Your help/patience is very much appreciated
Hi, I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code: mysql: Code: [Select] create table years ( yearID integer auto_increment, year varchar(30), primary key (yearID) ); insert into years (yearID, year) values ('1', '2007-2008'); insert into years (yearID, year) values ('2', '2008-2009'); insert into years (yearID, year) values ('3', '2009-2010'); insert into years (yearID, year) values ('4', '2010-2011'); insert into years (yearID, year) values ('5', '2011-2012'); insert into years (yearID, year) values ('6', '2012-2013'); PHP: Code: [Select] <?php require_once('../Connections/connection.php'); ?> <?php $result = @mysql_query( "select yearID, year, from sss.years"); print "<p>Select a year:\n"; print "<select name=\"yearID\">\n"; while ($row = mysql_fetch_assoc($result)){ $yearID = $row[ 'yearID' ]; $year = $row[ 'year' ]; print "<option value=$yearID>$year\n"; } print "</select>\n"; print "</p>\n"; ?> Thank you! Hi, I am trying to load a block of code for all operating systems except Linux. I have the following code. Does it look ok?- I am not too sure what to put in the IF statement.
Thanks! Alright, so I have an xml file differences.xml that is being parsed in XML. This is what the xml looks like: <item code="lM" name="dog"> <cost>5000</cost> <Start>12/15/2010</Start> <End>01/13/2011</End> </item> <item code="lF" name="cat"> <cost>5000</cost> <Start>04/15/2010</Start> <End>04/23/2011</End> </item>[/ I want to have the item names (dog, cat) show in a dropdown menu so that I can select these items for editing before storing in my mysql database. This is the php code I have so far: <?PHP $xml = simplexml_load_file("differences.xml"); $object = $xml->xpath("//item"); echo '<SELECT name=object>'; foreach ($object['name'] as $key => $value) { echo '<OPTION value='.$value.'> '.$value; } echo '</select>'; ?> I do have a dropdown list but there are no values inside it (it is empty). Can anyone help me figure out why? I do have this code that does work which lists the items in plaintext (not in a dropdown) so hopefull this will help us out: <?PHP $xml = simplexml_load_file("differences.xml"); $object = $xml->xpath("//item"); $count = count($object); $i = 0; while($i < $count) { echo '<h1>'.$object[$i]['name'].'</h1>'; $i++; } ?> I have the following code currently: Code: [Select] <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php print $item['view'] ?> <?php } ?> I would like to make the list a drop down with a link so that when a user selects, he goes to a new page. I tried the following: Code: [Select] <select name="select"> <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php $url = $node->field_buy_at[0]['url']; $store = $item['view']; ?> <? echo "<option value='$url'>$store</option>";?> <?php } ?> </select> I'm pretty sure it's this "$url = $node->field_buy_at[0]['url'];" that I don't have correct. Hey Guys, I know it may seem pretty simple, but im having trouble populating a drop down list. Here is my code at the moment, but what it's doing is displaying the names all in one value, where it should be in separate select values. *Note that i have only done it to the first one. See attachment. 'AntonMatt' are next to each other, they should be separate select values. Code: [Select] <? $id = $_GET['id']; $selectplayers="SELECT * FROM players WHERE club='$club' AND team='$team'"; $player=mysql_query($selectplayers); ?> <table class='lineups' width="560" cellpadding="5"> <tr> <td colspan="2">Starting Lineup</td> <td colspan="2">On the Bench</td> </tr> <tr> <td width="119"> </td> <td width="160"> </td> <td width="69"> </td> <td width="160"> </td> </tr> <tr> <td>Prop</td> <td><select name="secondary" style="width: 150px"> <option value='' selected="selected"><? while($rowplayer = mysql_fetch_array($player)) { echo $rowplayer['fname']; } ?></option> </select></td> <td>16.</td> <td><select name="secondary16" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> <tr> <td style="padding-top: 8px;">Hooker</td> <td style="padding-top: 8px;"><select name="secondary2" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> <td style="padding-top: 8px;">17.</td> <td style="padding-top: 8px;"><select name="secondary17" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> </table> </div> </div> If I can get this fixed, I will have completed all but the admin login for this project - my first php/mysql project. Here is what I need. I have a list_records.php that list all the records in the table 'links' and the category each entry is in from the table 'categories'. Here are my table structures. Code: [Select] -- Table structure for table `categories` -- DROP TABLE IF EXISTS `categories`; CREATE TABLE IF NOT EXISTS `categories` ( `id` int(11) NOT NULL AUTO_INCREMENT, `categories` varchar(37) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ; -- -------------------------------------------------------- -- -- Table structure for table `links` -- DROP TABLE IF EXISTS `links`; CREATE TABLE IF NOT EXISTS `links` ( `id` int(4) NOT NULL AUTO_INCREMENT, `catid` int(11) DEFAULT NULL, `name` varchar(255) NOT NULL DEFAULT '', `url` varchar(255) NOT NULL DEFAULT '', `content` varchar(255) NOT NULL DEFAULT '', PRIMARY KEY (`id`), KEY `catid` (`catid`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ; On the update.php file, I have a form that lets me make changes to the record. Here is the codes for update.php Code: [Select] <? include "menu.php" ?> <? include "db.php" ?> <?php $id=$_GET['id']; $sql = "select * from links where id =$id"; $query = mysql_query($sql); while ($row = mysql_fetch_array($query)){ $id = $row['id']; $catid = $row['catid']; $name = $row['name']; $url = $row['url']; $content = $row['content']; //we will echo these into the proper fields } mysql_free_result($query); ?> <table width="65%" align="center"> <tr><td align="left"> <form action="updated.php" method="post"> <input type="hidden" value="<?php echo $id; ?>" name="id"/> <br> <b>Website Name:</B><br> Change the name of the website listing.<br> <input type="text" value="<?php echo $name; ?>" name="name"/> <br> <br> <b>URL:</b><br> Change the URL of the website listing.<br> <input type="text" value="<?php echo $url; ?>" name="url"/> <br> <br> <b>Description:</b><br> Change the description of the website listing.<br> Limit 255 characters.<br/> <textarea name="content" cols="45" rows="4" wrap="soft"><?php echo($content);?></textarea> <br> <?php $result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $id = $rows["id"]; $categories=$row["categories"]; $options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>'; }; ?> <SELECT NAME=catid> <OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION> </SELECT> <?php mysql_close(); ?> <div align="center"> <input type="submit" value="submit changes"/> </div> </form> <br> </td></tr></table> The part of he code I need help with is Code: [Select] <?php $result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $id = $rows["id"]; $categories=$row["categories"]; $options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>'; }; ?> <SELECT NAME=catid> <OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION> </SELECT> I want it to default to the category that the entry is in. If you look, you will see in the select portion that I I have Code: [Select] <$ echo $catid; ?> which echos the proper category ID, but if I use Code: [Select] <? echo $categories; ?> it echos Writing, which is the last category in the list. Yet, the $options echo the catid and it corresponding category. How can I get the default option to echo BOTH the catid and category name while also listing all the other categories so that the records can be moved to a new category is needed? Any help will be appreciated. Thank you in advance. Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. 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. I'm trying to sort this dropdown box. It reads from a directory, and lists the file name in the dropdown box. Here's the tricky part... the filename is listed differently in the dropdown than in the directory by using explode(). I want to sort it though since it's still being sorted by the directory listings... For example: Filename starts out as: 123_abc_567.pdf then gets listed as abc_123_567.pdf in the dropdown, but it's still getting sorted as if it were 123_abc_567.pdf How can I do that? Here's my code: // Define the full path to folder from root $path = "C:/Work_Orders/"; // Open the folder $dir_handle = @opendir($path) or die("Unable to open $path"); echo "<form method=\"POST\" action='".$_SERVER['PHP_SELF']."' name='selectworkorder'><select name='ordernumber2'>"; // Loop through the files while ($file = readdir($dir_handle)) { //Remove file extension $ext = strrchr($file, '.'); if($ext !== false) { $file = substr($file, 0, -strlen($ext)); } if($file == "." || $file == ".." || $file == "index.php" ) continue; //explode file name $changedordernumber = explode("_",$file); //put in new order $changedordernumber = $changedordernumber[1]."_".$changedordernumber[0]."_".$changedordernumber[2]; $changedordernumber=trim($changedordernumber,"_"); //list options echo "<option name='$file' value='$file'>$changedordernumber</option>\n"; } echo "</select><input type='submit' value='Change' name='submit'/></form></div>"; // Close closedir($dir_handle); 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> This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=316599.0 Hey, I have the following coding: Quote <? $dbuser="*******"; $dbpass="*******"; $dbname="virtuda_db"; //the name of the database $chandle = mysql_connect("localhost", $dbuser, $dbpass) or die("Connection Failure to Database"); mysql_select_db($dbname, $chandle) or die ($dbname . " Database not found. " . $dbuser); $mainsection="license"; $query1="select name from license"; $result = mysql_db_query($dbname, $query1) or die("Failed Query of " . $query1); //do the query while($thisrow=mysql_fetch_row($result)) { $i=0; while ($i < mysql_num_fields($result)) { $field_name=mysql_fetch_field($result, $i); echo $thisrow[$i] . " "; //Display all the fields on one line $i++; } echo "<br>"; //put a break after each database entry } ?> How would I set up this so that instead of just "listing" them out on new lines, it would list the results into a drop down list? Thanks! hi, im new here i hope to learn and help in the way i can since i'm a newbie with php but here it goes. I have some old databases i need to test, i want to mail these old users from a website i had but i need to know if the database contains invalid e-mails and remove them so i dont mess up my server's good ip reputation. So i found the checkdnsrr function: Code: [Select] function checkEmail($email) { // checks proper syntax if(preg_match("/^([a-zA-Z0-9])+([a-zA-Z0-9\._-])*@([a-zA-Z0-9_-])+([a-zA-Z0-9\._-]+)+$/" , $email)) { // gets domain name list($username,$domain)=split('@',$email); // checks for if MX records in the DNS if(!checkdnsrr($domain, 'MX')) { return false; } return true; } return false; } so everyone knows this function, i just wanted to run it to multiple e-mail addresses like from a text file. when i say multiple, i mean arround 500.000 or 50k at the time. i just want to output the bad e-mails to a text file. hope someone can help me on this one. thanks in advance hello! in my header.php file i have the following line of code: Code: [Select] <link rel="stylesheet" href="<?php echo $pageCSS; ?>"> in the document proper, i have the following at the top of the page: Code: [Select] <?php $pageCSS="_css/compare.css"; ?> this works as expected and it's clean and neat. what i want to know is how i can adapt this to pass MULTIPLE JavaScript files through it. would it be best to pass an array of files and then step through the array? would be nice to be able to pass a simple comma-delimited list. any help would be appreciated. WR! hi all,i have a php form that i would like to populate from a drop down list,my problem is that i cant place the drop down list at the correct place.what i want is that where the franchisee is the input should be the drop down,please help....this is the code that i am using Code: [Select] <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post"> <p>Add a Bus in the Bus table.</p> <p> Fleet Number:<br /> <input type="text" name="fleet_number" size="6" maxlength="10" value="" /> </p> <p> Registration Number:<br /> <input type="text" name="registration_number" size="10" maxlength="20" value="" /> </p> <p> Model:<br /> <input type="text" name="model" size="10" max length="15" value="" /> </p> <p> Franchisee :<br /> <input type="select" name="franchisee_id" size="10" max length="15" value="" /> <select name=dropdown_list> <?php while($row = oci_fetch_array($stid)) { echo "<option value='".$row['FRANCHISEE_NAME']."'>"; echo $row['FRANCHISEE_NAME']; echo "</option> "; } ?> </select> </p> <p> <input type="Submit" name="submit" value="Submit !" /> </p> </form> |
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