|
PHP - Populate Dropdown On The Basis Of Another Dropdown Value In Php Using Jquery Ajax
I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time. I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.
When i load ajax.php i get 2 error mesages: This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID* is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID. Index.php
<!doctype html>
<?PHP
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass", "Database" =>
"database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
?>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>
<body>
<section id="formaT2" class="formaT2 formContent">
<div class="row">
<div class="col-md-2 col-3 row-color remove-mob"></div>
<div class="col-md-5 col-9 bg-img" style="padding-left: 0;
padding-right: 0;">
<h1>Form</h1>
<div class="rest-text">
<div class="contactFrm">
<p class="statusMsg <?php echo
!empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
<form action="connection.php" method="post">
<div>machinery</div>
<select id="machinery">
<option value="0">--Please Select Machinery--</option>
<?php
// Fetch Department
$sql = "SELECT Machinery FROM T013";
$machanery_data = sqlsrv_query($conn2,$sql);
while($row = sqlsrv_fetch_array($machanery_data) ){
$id = $row['Id'];
$machinery = $row['Machinery'];
// Option
echo "<option value='".$id."' >".$machinery."</option>";
}
?>
</select>
<div class="clear"></div>
<div>Sub Machinery</div>
<select id="sub_machinery">
<option value="0">- Select -</option>
</select>
<input type="submit" name="submit"
id="submit" class="strelka-send" value="Insert">
<div class="clear"> </div>
</form>
</div>
</div>
</div>
</div>
</section>
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#machinery").change(function(){
var machinery_id = $(this).val();
$.ajax({
url:'ajaxfile.php',
type: 'post',
data: {machinery:machinery_id},
dataType: 'json',
success:function(response){
var len = response.length;
$("#sub_machinery").empty();
for( var i = 0; i<len; i++){
var machinery_id = response[i]['machinery_id'];
var machinery = response[i]['machinery'];
$("#sub_machinery").append("<option
value='"+machinery_id+"'>"+machinery+"</option>");
}
}
});
});
});
</script>
</body>
</html>
Ajaxfile.php
<?php
$server = "server";
$options = array( "UID" => "user", "PWD" => "pass",
"Database" => "database");
$conn2 = sqlsrv_connect($server, $options);
if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
echo " ";
$machineryID = $_POST['machinery']; // department id
$sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
$result = sqlsrv_query($conn2,$sql);
$machinery_arr = array();
while( $row = sqlsrv_fetch_array($result) ){
$machinery_id = $row['ID'];
$machinery = $row['MachineID'];
$machinery_arr[] = array("ID" => $machinery_id, "MachineID" =>
$machinery);
}
// encoding array to json format
echo json_encode($machinery_arr);
?>
Edited May 6, 2019 by davidd
Similar TutorialsHello forum,
So I've been developing an app mostly in PHP, but am rather afraid of JS. Hope to fix that.
I have an AJAX dropdown using JQuery to search locations. It works great. However, I want to make it similar to what is seen on this site:
http://placefinder.com/
As you can see, the dropdown, when clicked populates a box. Then the user submits the form and the data is used in the application.
I have no clue how to make the form populate with data from the DB (I'm using mySQL) when clicked. So far, I've only been able to make it clickable as a URL (not what I want, obvioiusly!)
Is there a way to do this on a really small, simple script for starters? I'm certain their is, but don't even know where to begin.
Any help appreciated
Hi freaks, I'm new to php first of all. I'm dynamically binding a dropdownlist with mysql database . After the user selects an item from it , I want to match that item with another table so as to populate another database. The code I'm using to populate dropdown: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if(!$con) { die ('Can not connect to : '.mysql_error()); } mysql_select_db("ims",$con); $result=mysql_query("select cat_id,cat_name from category"); echo "<select name=cat>"; while($nt=mysql_fetch_array($result)) { echo "<option value=$nt[cat_id]> $nt[cat_name] </option>"; } echo "</select>"; mysql_close($con); ?> Now after the user selects any one of the item , I want to bind another dropdown on the same page using such query like $result=mysql_query("select subcategory.sc_id,subactegory.sc_name from subcategory,category where subcategory.sc_id=$nt[cat_id]"); Please anyone tell me the logic and code to do it. Also tell me do I need an intermediate page to post the 1st dropdown value and then continue with 2nd dropdown. I couldn't figure out the concept anyhow. Help on this will be highly appreiable . (Tell me if I'm not clear with my question) 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> This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=351349.0 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> i skipped solving this the other day with an easier way but now i am stuck with this stumble again in another area and i have no way out.... so here it goes, i have an ajax dropdown box...i need to get the value that is selected by the user when it is clicked and then pass this value to a new pop window by appending to its url....any suggestions? This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=328006.0 I am not a developer but I can modify code to work for me. The following code works on my test machine (Windows 10, IIS, PHP 7.4) but doesn't work on my website (cPanel, Some version of Linux, PHP 7.4). The two dropdowns are for State and City. You are supposed to be able to select the state and then select a city from that state then bring up a report for craft breweries in the city. When selecting State from the first dropdown, the page refreshes, the URL is correct with the reports.php?cat=<STATE> so $cat is being set, but the first dropdown no longer has the state selected and the second dropdown is populated with All cities and not just one ones from the selected state. Any ides why this is working fine on one machine and not the other? Selected code from reports.php
<?php
?>
/////// for second drop down list we will check if State is selected else we will display all the cities/////
echo "<form method=post action='brewerylistbycity.php'>";
}
////////// Starting of second drop downlist /////////
//// End Form /////
hi, I was going through this tutorial: http://www.electrictoolbox.com/json-data-jquery-php-mysql/ here is my HTML Code: [Select] <!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> <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js"></script> <script language="javascript" type="text/javascript"> function populateFruitVariety() { $.getJSON('search-by.php', {fruitName:$('#fruitName').val()}, function(data) { var select = $('#fruitVariety'); var options = select.attr('options'); $('option', select).remove(); $.each(data, function(index, array) { options[options.length] = new Option(array['variety']); }); }); } $(document).ready(function() { populateFruitVariety(); $('#fruitName').change(function() { populateFruitVariety(); }); }); </script> </head> <body> <form> Search by: <select name="name" id="fruitName"> <option>Please Select</option> <option id="Town" value="Town">Town</option> <option id="County" value="County">County</option> </select> Variety: <select name="variety" id="fruitVariety"> </select> </form> </body> </html> here is my PHP Code: [Select] $dbhost = "xxx"; $dbname = "xxx"; $dblogin = "xxx"; $dbpass = "xxx"; function dbConnect() { global $dbhost; global $dbname; global $dblogin; global $dbpass; $db = mysql_connect($dbhost, $dblogin, $dbpass) or die("could not connect to database: ".mysql_error()); mysql_select_db($dbname) or die("could not select database"); return $db; } function dbClose($db) { mysql_close($db); } // basis code ^^ $db = dbConnect(); $rows = array(); if(isset($_GET['Town'])) { $query = "SELECT DISTINCT rsCounties FROM pubs"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo json_encode($rows); } } if(isset($_GET['County'])) { $stmt = $pdo->prepare("SELECT rsCounty FROM pubs"); $stmt->execute(array($_GET['rsCounty'])); $rows = $stmt->fetchAll(PDO::FETCH_ASSOC); } echo json_encode($rows); Can someone please help? I have the JQuery working but not the PHP?! here is a link to what I have at the moment: http://www.mypubspace.com/dashnew/index.html thanks Ran the following but got a segfault as it doesn't seem to support php7.4. ./vendor/bin/concrete5 c5:install -i So I changed ./vendor/bin/concrete5 from: #!/usr/bin/env php <?php ... to #!/usr/bin/env php72 <?php ... I am sure I will forget I did this and would rather configure either the directory or the user to use php72. Is this possible? Also, should I be making any other changes? For instance, maybe:
"config": {"platform": {"php": "7.2.30"}}
Thanks Hello, I decided I would make my websites load less by implementing ajax/jQuery.. I however stumbled upon a problem.. This is the code I use to direct myself to the registeration page... Code: [Select] function register(){ $.ajax({ type: "POST", url: "user/registerConfirm.php", data: "regUsername=" + document.getElementById("regUsername").value + "®Password=" + document.getElementById("regPassword").value + "&myPassCheck=" + document.getElementById("myPassCheck").value + "®Email=" + document.getElementById("regEmail").value, success: function(html){ $('#response').html(html); alert("success..."); }, complete: function(html){ $('#response').html(html); alert("completed.."); } }); } When running - it shows me the "alert("completed..");", and in fact- the code used to insert the user in my database is working. However, I never seem to get the 'html' variable that they use in tutorials I found. (meaning I can't show some sort of response) The php code used: Code: [Select] <?php session_start(); ob_start(); require("../widgets/functions.php"); connectToDB(); // Check email if (!filter_var(clean($_POST['regEmail']), FILTER_VALIDATE_EMAIL)) { $error = 'Invalid email!'; } else { if ( clean($_POST['regPassword']) == clean($_POST['myPassCheck']) && clean($_POST['regPassword']) != NULL && clean($_POST['regUsername']) != NULL && clean($_POST['regEmail']) != NULL ) // Register can be allowed { // Check if their already is a user with the same name.. $sql="SELECT * FROM `users` WHERE username='".clean(trim($_POST['regUsername']))."'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); if($count==1){ // Their was already a user with this name! $error = 'Registration failed - username already taken..'; } else if ( $count == 0 ){ // Registration allowed, no user found with the same name // Encrypt password $encryptPass = md5($_POST['regPassword']); $subject = "Confirmation registration"; $message = ' <html> <head> <title>Registration at codexplained</title> </head> <body> <p>Hello '.clean($_POST['regUsername']).',</p> <p>Thank you for registering at our website! </p> <p>If you wish, you can now edit your profile, to change your display options and/or to upload your own profile image!<br /> We hope to see you commenting on our articles soon!<br /> If you wish to receive more information - Please contact us, or message any of our moderators.</p> <hr /> <p>- Current moderators - <br /> Ruud<br /> Willem<br /> Quint </p> <hr /> </p> - Contact details - <br /> Codexplained.tk<br /> Codexplained@gmail.com </p> </body> </html> '; $from = "Codexplained@admin.com"; $headers = 'From: Codexplained'."\r\n"; $headers .= 'MIME-Version: 1.0' . "\r\n"; $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n"; $to = clean($_POST['regEmail']); if ( mail($to,$subject,$message,$headers) ) { // Success } else { // Failed } // Insert data $query = "INSERT INTO `users` ( `Id` ,`Username` ,`Password` ,`Rank`,`E-mail` ,`PostAmount`, `ProfileImage`, `Ip`, `LastIP` ) VALUES ( NULL , '".clean(trim($_POST['regUsername']))."' , '".$encryptPass."' , 'member', '".clean($_POST['regEmail'])."' , '0', 'none', '".$_SERVER['REMOTE_ADDR']."','".$_SERVER['REMOTE_ADDR']."' ) "; mysql_query($query)or die(mysql_error()); $error = 'Registration completed!'; } } else { if ( clean($_POST['regPassword']) != clean($_POST['myPassCheck']) ) { $error = 'Passwords do not match...'; } else { $error = 'Registration failed - not enough data..'; } } } echo $error; exit; mysql_close(); //header("location:../index.php?page=register"); ?> And here is the form code: Code: [Select] <?php session_start(); ob_start(); connectToDB(); $query = "SELECT * FROM `users`"; $result = mysql_query($query); $num = mysql_num_rows($result); echo ' <div id="registerHolder"> <h1> <strong> Registration </strong> </h1> <em>Currently... we have '.$num.' registered members! Join them by filling in the form below!</em><br /> <p> Please fill in these boxes and press confirm to register: <br /> <form id="registerForm" method="post" action=""> <table> <tr> <td style="width:200px"> Username: </td> <td> <input name="regUsername" type="text" id="regUsername" style="width:300px" class="box" value="" /> </td> </tr> <tr> <td> Password: </td> <td> <input name="regPassword" type="password" id="regPassword" style="width:300px" class="box" value="" /> </td> </tr> <tr> <td> Confirm Password: </td> <td> <input name="myPassCheck" type="password" style="width:300px" id="myPassCheck" value="" /> </td> </tr> <tr> <td> E-mail: </td> <td> <input name="regEmail" type="text" id="regEmail" style="width:300px" class="box" value="" /> </td> </tr> <tr> <td> <input type="submit" name="Submit" class="btn" value="Confirm" onclick="register()" /> </td> </tr> </table> <br /> </form> </p> <div id="response"> </div> </div> '; ?> I am wondering what I missed, and hoped someone out here could assist me. Thanks in advance. Edit: Added register.php code (form) This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=314220.0 Hi Everyone,
I am struggling with a pdf.
On our server I get the pdf as a byte array from the database, which I then build up in a StreamingOutput object and return as a pdf file. This is done using java using JAX-RS service.
Now when I open the path/link in my browser(chrome/firefox) the pdf opens in the browser. So this means my backend is working.
On my webapp I want to use $.ajax to download the pdf and view it in an Iframe or embed it in an <embed> or <object> tag. I need to do this as the service call is actually under username and password, and we do not create a session between the browser and the server - there is a long explanation for this.
I saw the option to use http://[username]:[password]@www.myserver.com/... but this was dropped by most browsers. This is why I am trying ajax.
Now in my ajax call I do receive the pdf in the success function as:
%PDF-1.4....This open the browser's pdfviewer correctly and every thing, but the pdf only returns blank pages. The Base64 is a java object that I found on the internet. But I have tried it without the base64 but still now luck. Here is my success function from my ajax call.
success: function(data, status, xhr) {
var pdfText = Base64.encode(xhr.responseText);
var url = "data:application/pdf;base64," + escape(pdfText);
var html = '<embed width=100% height=600'
+ ' type="application/pdf"'
+ ' src="' + url + '">'
+ '</embed>';
$("div.inner").html("");
$("div.inner").append(html);
},
Hi there, I am using the jQuery, Ajax PHP code which is given at http://roshanbh.com.np/2008/04/ajax-login-validation-php-jquery.html The form I am using, which is in index.html, is: Code: [Select] <form method="post" action="" name="login" id="login_form"> <div class="field_row"> <div class="label_container"> <label>Email</label> </div> <div class="field_container"> <input type="text" placeholder="login with your email address..." name="email_address" id="email_address" value="" class="large" /> </div> <div class="clear"><span class="nodisp"> </span></div> </div> <div class="field_row"> <div class="label_container"> <label>Password</label> </div> <div class="field_container"> <input type="password" placeholder="...and password" name="password" id="password" value="" class="large" /> </div> <div class="clear"><span class="nodisp"> </span></div> </div> <div class="final_row"> <input type="image" src="images/login_blue.gif" id="user_login_button" name="user_login_button" value="login" id="submit" class="submit_button" /> <div class="final_row_text_container" > <a href="/login/forgot_password" style="color: #008ee8;" class="small_text">Forgot your Password?</a> <br /> <span id="msgbox" style="display:none"></span> </div> </div> <div class="clear"><span class="nodisp"> </span></div> </form> The Javascript, which is situated in the head of index.html. is: Code: [Select] <script language="javascript"> $(document).ready(function() { $("#login_form").submit(function() { //remove all the class add the messagebox classes and start fading $("#msgbox").removeClass().addClass('messagebox').text('Validating....').fadeIn(1000); //check the email address exists or not from ajax $.post("login_ajax.php",{ email_address:$('#email_address').val(),password:$('#password').val(),rand:Math.random() } ,function(data) { if(data=='yes') //if correct login detail { $("#msgbox").fadeTo(200,0.1,function() //start fading the messagebox { //add message and change the class of the box and start fading $(this).html('Logging in.....').addClass('messageboxok').fadeTo(900,1, function() { //redirect to secure page document.location='secure.php'; }); }); } else { $("#msgbox").fadeTo(200,0.1,function() //start fading the messagebox { //add message and change the class of the box and start fading $(this).html('Your login details are incorrect.').addClass('messageboxerror').fadeTo(900,1); }); } }); return false; //not to post the form physically }); //now call the ajax also focus move from $("#password").blur(function() { $("#login_form").trigger('submit'); }); }); </script> And the PHP, in login_ajax.php, is: <?php session_start(); $host = "localhost"; $user = "bford"; $pass = "bford"; $db = "bford"; $link = mysql_connect($host, $user, $pass); if (!link) { die('<strong>Error(s) occured:</strong> Could not connect: ' . mysql_error()); } $db_selected = mysql_select_db($db, $link); if (!db_selected) { die ('<strong>Error(s) occured:</strong> Cant use bford: ' . mysql_error()); } //get the posted values $email_address=$_GET['emailaddress']; $pass=$_GET['password']; //now validating the username and password $sql="SELECT * FROM users WHERE email_address='".$email_address."'"; $result=mysql_query($sql); $row=mysql_fetch_array($result); //if username exists if(mysql_num_rows($result)>0) { //compare the password if($row["password"],$pass)==1 { echo "yes"; //now set the session from here if needed $_SESSION["user_name"]=$userID; } else echo "no"; } else echo "no"; //Invalid Login ?> I have been working on this for days now, changing around the form names, database table names, php variables, allsorts! I still cannot get it functioning properly. When I input a correct email_address and password combination, the 'Your login details are incorrect.' message still appears. Help would be much appreciated. Ben. I need some help adjust a search form that uses PHP/MySQL/AJAX/JQuery. I am not sure what area needs to be adjusted, but I think it's in the Query. The search from searches across the MySQL database of sports team info and works fine. the only problem is in the return of one key word when I needed the exact match of two or more words. For instance you type in 'Miami Heat' and the results returned are for 'Miami Heat' and 'Miami Dolphins', so anything with the key word Miami is returned. Here is the PHP code <?php $dbhost = "localhost"; $dbuser = "username"; $dbpass = "pw"; $dbname = "db_name"; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); mysql_select_db($dbname); if(isset($_GET['query'])) { $query = $_GET['query']; } else { $query = ""; } if(isset($_GET['type'])) { $type = $_GET['type']; } else { $query = "count"; } if($type == "count") { $sql = mysql_query("SELECT count(category_id) FROM players WHERE MATCH(sport,first_name,last_name,MVP,team) AGAINST('$query' IN BOOLEAN MODE)"); $total = mysql_fetch_array($sql); $num = $total[0]; echo $num; } if($type == "results") { $sql = mysql_query("SELECT sport,first_name,last_name,MVP,team FROM players WHERE MATCH(sport,first_name,last_name,MVP,team) AGAINST('$query' IN BOOLEAN MODE)"); echo "<table>"; echo "<tr>"; echo "<th>First Name</th><th>Last Name</th><th>Team</th><th>MVP</th>"; while($array = mysql_fetch_array($sql)) { $sport = $array['sport']; $first_name = $array['first_name']; $last_name = $array['last_name']; $team = $array['team']; $MVP = $array['MVP']; echo "<tr><td>" . $first_name . "</td>\n<td>". $last_name ."</td>\n<td>" .$team."</td>\n<td> . $MVP .</td>\n</tr>\n"; } echo "</table>";=\ } mysql_close($conn); ?> Hi, I have a dropdown box, and I want it so that if dropdownbox.value='selected:' then show error message. The checkbox is dynamically populated. Pleaee Advise, - stuart Hi all. I kinda need some from a more advanced PHP expert. I have a table that displays time slots that are available to be booked. As you can see I have a table with two columns, The 1st column displays the times and the 2nd column has a link that says 'Available'. I want to be able to put all this in a select dropdown box to save space, but how can I do this?
<?php
$doc = JFactory::getDocument();
$doc->addStyleSheet(JURI::root(false)."components/com_pbbooking/user_view.css");
?>
<style>
table#pbbooking td, table#pbbooking th {padding: 0em;}
</style>
<h1><?php echo JText::_('COM_PBBOOKING_DAY_VIEW_HEADING').' '.Jhtml::_('date',$this->dateparam->format(DATE_ATOM),JText::_('COM_PBBOOKING_DAY_VIEW_DATE_FORMAT'));?></h1>
<table id="pbbooking">
<!-- Draw header row showing calendars across the top....-->
<tr>
<th></th> <!-- first column left blank to display time slots -->
<?php foreach ($this->cals as $cal) :?>
<th><?php echo $cal->name;?></th>
<?php endforeach;?>
</tr>
<!-- draw table data rows -->
<?php while ($this->day_dt_start <= $this->dt_last_slot) :?>
<?php $slot_end = date_create($this->day_dt_start->format(DATE_ATOM),new DateTimezone(PBBOOKING_TIMEZONE));?>
<?php $slot_end->modify('+ '.$this->config->time_increment.' minutes');?>
<tr>
<th><?php echo Jhtml::_('date',$this->day_dt_start->format(DATE_ATOM),JText::_('COM_PBBOOKING_SUCCESS_TIME_FORMAT'));?></th>
<?php foreach ($this->cals as $cal) :?>
<td class="pbbooking-<?php echo (!$cal->is_free_from_to($this->day_dt_start,$slot_end)) ? 'free' : 'busy';?>-cell">
<?php if ($this->day_dt_start>date_create("now",new DateTimeZone(PBBOOKING_TIMEZONE)) && !$cal->is_free_from_to($this->day_dt_start,$slot_end)) :?>
<a href="<?php echo JRoute::_('index.php?option=com_pbbooking&task=create&dtstart='.$this->day_dt_start->format('YmdHi').'&cal_id='.$cal->cal_id);?>">
<?php echo (!$cal->is_free_from_to($this->day_dt_start,$slot_end)) ? JText::_('COM_PBBOOKING_FREE') : JText::_('COM_PBBOOKING_BUSY');?>
</a>
<?php else :?>
<?php echo JText::_('COM_PBBOOKING_BUSY');?>
<?php endif;?>
</td>
<?php endforeach;?>
</tr>
<?php $this->day_dt_start->modify('+ '.$this->config->time_increment.' minutes');?>
<?php endwhile;?>
<!-- end draw table data rows-->
</table>
I'm trying to have it shows the data from my array as each option will have a value of the user's id and the text for the option will be their name. As of right now it only shows the last person. Code: [Select] <?php echo form_label('Recipient', 'recipient'); ?> <?php $data = array( 'name' => 'to', 'class' => 'required' ); <?php echo form_label('Recipient', 'recipient'); ?> <?php $data = array( 'name' => 'to', 'class' => 'required' ); $options = array(); foreach($users AS $user) { $options = array ( $user->user_id => $user->first_name.' '.$user->last_name ); } ?> <?php echo form_dropdown($data, $options); ?> Hi, I'm wondering how i would go about making a drop down menu in php to insert the option into users > Maint of my database, if anyone has any tutorials or examples it would be great, Thanks a lot. |
|||||||