I have 2 selectboxes
<h3>Results</h3>
<select id="register_form" name="sport" />
<option value="Rugby">Rugby</option>
<option value="Cricket">Cricket</option>
<option value="Football">Football</option>
</select>
<?php
echo'<select name="match">';
echo'<option value="'.$row['event_id'].'">'.$row['team1'].' VS '.$row['team2'].'</option>';
echo'</select>';
?>
<input id="register_form" type="submit" value="Display" name="submit" />
User searches for a result by:
selecting sport type in 1st selectbox and then in 2nd selectbox option values are populated based on sport type.
Is it possible to do this in PHP without the user having to first press submit to get the $_POST value of sport type?
What is my best option here?
PHP always need to reload the page to refresh your informations, so, as anant kumar singh said, you need to use AJAX for that. And as yak613 said, jQuery will help you to use AJAX easily
1.Ajax is the only option what you asked for that(without page refresh)
When you use php it's only possible with page refresh. but with ajax without page refresh it's possible.
helping links are:-
Use jQuery to change a second select list based on the first select list option
https://www.daniweb.com/web-development/php/threads/372228/php-and-ajax-auto-populate-select-box
https://remysharp.com/2007/01/20/auto-populating-select-boxes-using-jquery-ajax
You can use this Multiple Select Dropdawn lists: http://coursesweb.net/ajax/multiple-select-dropdown-list-ajax_t , it can be used for multiple sets of Select lists.
I've faced with the same problem in my project. But the needed functionality was higher - not two dependent selectboxes and bigger number. I've written a simple function to load my selectboxes:
//formId - form where selectbox is
//name - attribute "name" of selectbox
//dataSourceUrl - url to PHP-file
//affectingField - string with value that filters the selecbox's data
function loadSelectbox( formId, name, dataSourceUrl, affectingField ){
//console.log('Loading data to selectbox name="'+name+'":');
var selectbox = $('#'+formId+' select[name="'+name+'"]');
if(selectbox){
//console.log("Selecbox found");
if(affectingField != null){
var affectingValue = $('#'+formId+' [name="'+affectingField+'"]').val();
dataSourceUrl += '?affectingValue='+affectingValue;
}
var options = selectbox.find('option');
var jqxhr = $.ajax({
url: dataSourceUrl,
dataType: 'text'
})
.done(function(data) {
//console.log(data);
if(data != ""){
var optionsObject = JSON.parse(data);
var i = 0;
console.log(optionsObject);
var options = [];
$(optionsObject).each(
function(){
options[i] = '<option value="'+$(this)[0]['val']+'">'+$(this)[0]['text']+'</option>';
i++;
}
);
selectbox.html(options);
if(urlParamsSet[name] == false){
setParamFromUrl(name);
}
}
else{
selectbox.html('<option value="">Все</option>');
}
})
.fail(function() {
alert("Problems with server answer");
})
selectbox.prop("disabled", false);
}
else{
console.log("No selectbox with such name");
}
}
Not saying that this code is perfect, but it works. PHP-file must return the values to selecbox in JSON format (convert from with structure: array(index, value, text) ).
Related
I have a page with a select list (gets successfully populated from mysql) and a text box. The text box has to be populated with a value from mysql based on the item selected in the list. But the ajax call to php is not working and i can not figure out what the issue is. I am just learning ajax and php, so a novice.. Please help. i am stuck with this for a long time.
<script>
$(document).ready(function() {
$('.selectpicker').on("change", function(){
var selected_data = $(this).find("option:selected").val();
alert(selected_data);
$.ajax ({
type: "POST",
data: { selected_data: selected_data },
url: "getoldcharity.php",
dataType: "json",
success: function(res) {
$('#charity_new').val(data.charity_new);
}
});
});
});
</script>
<form id="assign-fundraiser_form" class="form-horizontal" action="" method="post">
<div class="form-group">
<div class="col-md-3">
<select class="selectpicker form-control" id="fundraiser" name="fundraiser" required>
<option value="" selected disabled>Select a Fundraiser</option>
<?php
include('session.php');
$result1 = mysqli_query($db,"select concat(f_firstname,' ',f_lastname) fundraiser from fundraiser where f_company in (select contractor_name from contractor where company_name = '$_SESSION[login_user]') and f_status = 'Active' order by concat(f_firstname,' ',f_lastname)");
while ($rows = mysqli_fetch_array($result1))
{
echo "<option>" .$rows[fundraiser]. "</option>";
}
?>
</select>
</div>
</div>
<input type="text" name="charity" id="charity_new" />
</form>
<?php
include "session.php";
if (ISSET($_POST['.selectpicker'])) {
$ref = $_POST['.selectpicker'];
$query = $db->query("select f_charity charity_new from fundraiser limit 1");
$row = $query->fetch_assoc();
$charity_new = $row['charity_new'];
$json = array('charity_new' => $charity_new);
echo json_encode($json);
}
$db->close();
?>
There are a few problems that I've spotted from quick glance, so I've separated them below.
PHP
In your AJAX request, you are using data: { selected_data: selected_data } which means the PHP code will be expecting a POSTed key named selected_data but you're looking for .selectpicker. You seem to have mixed up a couple of things, so instead of:
$_POST['.selectpicker']
it should be:
$_POST['selected_data']
JavaScript
As Ravi pointed out in his answer, you also need to change your success function. The parameter passed through to this function is res not data, so instead of:
$('#charity_new').val(data.charity_new);
it should be:
$('#charity_new').val(res.charity_new);
MySQL
It also appears as though your query itself is invalid - you seem to be missing a comma in the column selection.
select f_charity charity_new from fundraiser limit 1
should be:
select f_charity, charity_new from fundraiser limit 1
or, seeing as you're not using the f_charity column in the results anyway:
select charity_new from fundraiser limit 1
You aren't using the value that is being POSTed either, meaning that whatever option is selected in the dropdown makes no difference to the query itself - it will always return the first record in the database.
Other
One other thing to be aware of is you're using a class selector on your change function. This means if you have multiple dropdowns with the same class name in your HTML, they will all be calling the same AJAX function and updating the textbox. I don't know if this is what you're aiming for, but from your code posted, you only have one dropdown in the form. If you only want that one dropdown to be calling the AJAX function, you should use an ID selector instead:
$('#fundraiser').on("change", function() {
// ...
}
I think, it should be
$('#charity_new').val(res.charity_new);
instead of
$('#charity_new').val(data.charity_new);
I'm using Select2 3.4.5 for create select boxes,
I use this code for creatre a Multi-Value Select Boxe and everything is fine.
<select id="e1" name="mydata" multiple>
<option value="D1">Data1</option>
<option value="D2">Data2</option>
<option value="D3">Data3</option>
</select>
...
<script>
$("#e1").select2();
</script>
For get multiple selected values of select box in php I have to modify name="mydata" by name="mydata[]", and in PHP I get values by this code:
<?php
foreach ($_POST['mydata'] as $names) {
print "You are selected $names<br/>";
}
?>
But my question: How can I send selected values of select box to PHP as string to recover in php like this : 'D1,D2,D3' , and thanks.
Edit:
I want to send the data as string, not receive it as an array then
change it as string
Server-side with PHP
Ideally you would do this with PHP once the value is sent. To convert the selected items just want to implode the array
$names=implode(',', $_POST['mydata']);
Where $_POST['mydata'] is an array
[0]=>'D1',
[1]=>'D2',
[2]=>'D3'
implode(',', $_POST['mydata']) would be 'D1,D2,D3'
Client-side with jQuery
Your title says "send selected values of select box to PHP as string". You would do that in JS by catching the submit event of the form and using .join() to change the value of that field.
$('#formid').on('submit', function(){
$('#e1').val($('#e1').val().join(','));
});
Your form (not given) would need an id <form id="formid" ...>
If you want a client-side solution, try getting the val() and calling join():
$('#e1').val().join()
http://jsfiddle.net/gwgLV/
You can do it with javascript.
<select id="e1" name="mydata" multiple>
<option value="D1">Data1</option>
<option value="D2">Data2</option>
<option value="D3">Data3</option>
</select>
<button id="but" onclick="now()">Show selected values</button>
javascript code
function getSelectValues(select) {
var result = [];
var options = select && select.options;
var opt;
for (var i=0, iLen=options.length; i<iLen; i++) {
opt = options[i];
if (opt.selected) {
result.push(opt.value || opt.text);
}
}
return result;
}
function now(){
var el = document.getElementsByTagName('select')[0];
var x = getSelectValues(el);
alert(x);
}
Demo here
Instead of alert store in a variable and send it along with the rest of the form data. Or you can use join (as mentioned in other answers ) to send it over post to php.
So, here's the deal. I have an html table that I want to populate. Specificaly the first row is the one that is filled with elements from a mysql database. To be exact, the table is a questionnaire about mobile phones. The first row is the header where the cellphone names are loaded from the database. There is also a select tag that has company names as options in it. I need to trigger an onChange event on the select tag to reload the page and refill the first row with the new names of mobiles from the company that is currently selected in the dropdown list. This is what my select almost looks like:
<select name="select" class="companies" onChange="reloadPageWithNewElements()">
<?php
$sql = "SELECT cname FROM companies;";
$rs = mysql_query($sql);
while($row = mysql_fetch_array($rs)) {
echo "<option value=\"".$row['cname']."\">".$row['cname']."</option>\n ";
}
?>
</select>
So... is there a way to refresh this page with onChange and pass the selected value to the same page again and assign it in a new php variable so i can do the query i need to fill my table?
<?php
//$mobileCompanies = $_GET["selectedValue"];
$sql = "SELECT mname FROM ".$mobileCompanies.";";
$rs = mysql_query($sql);
while ($row = mysql_fetch_array($rs)) {
echo "<td><div class=\"q1\">".$row['mname']."</div></td>";
}
?>
something like this. (The reloadPageWithNewElements() and selectedValue are just an idea for now)
Save the value in a hidden input :
<input type='hidden' value='<?php echo $row['cname'] ?>' id='someId' />
in your JavaScript function use the value from this hidden input field:
function reloadPageWithNewElements() {
var selectedValue = document.getElementById('someId').value;
// refresh page and send value as param
window.location.href = window.location + '?someVal='+ selectedValue;
}
Now again in your PHP file retrieve this value from url for use as:
$someVal = null;
if (isset($_GET['someVal']) {
$someVal = $_GET['someVal'];
}
see if this works!!!
The best option would be using AJAX.
reloadPageWithNewElements() is a function which calls a page of your own site which will return the data you would like to put in your table.
If you are using JQuery, AJAX is very easy to implement:
$.ajax({
url: '/yourPage',
data: { selectedCompany: $('.companies').val() },
success: function(result) {
//delete your tablerows
$(".classOfTable tr").remove();
//put the result in your html table e.g.
$('.classOfTable').append(result);
},
dataType: html
});
The browser will send a request to "/yourPage?selectedCompany=Google" or something
All you have to do is let this page print out only html (maybe even easier is to print only the tablerow (<tr>).
If you have any further questions, please ask.
I would use jQuery to do it.
first You need to add 'id' attribute to every option tag
<option id="option1">
<option id="option2">
and so on...
then with jQuery:
$('<option>').change(function() {
var id=$(this).attr('id');
...save data here (i.e: with ajax $.post(url, { selected_id: id}, callback }
});
I'm still trying to learn jquery so bear with me. I have a dual select box that only works if I select all the results of the second select box after I move them there. What I want is when the first box transfers values to the second second select box, it doesn't require highlighting the options, but posts that second select box on form submit. Here is what
I have:
HTML:
<span id="dualselect1" class="dualselect">
<select name="select1[]" multiple="multiple" size="10">
<?php
$c='0';
foreach($lp_name as $lpn){
echo '<option value="'.$lp_id[$c].'">'.$lpn.' ('.$lp_url[$c].')</option>';
$c++;
}
?>
</select>
<span class="ds_arrow">
<span class="arrow ds_prev">«</span>
<span class="arrow ds_next">»</span>
</span>
<select name="select2[]" multiple="multiple" size="10">
<option value=""></option>
</select>
</span>
JQUERY:
<script type="text/javascript">
jQuery(document).ready(function(){
var db = jQuery('#dualselect1').find('.ds_arrow .arrow'); //get arrows of dual select
var sel1 = jQuery('#dualselect1 select:first-child'); //get first select element
var sel2 = jQuery('#dualselect1 select:last-child'); //get second select element
sel2.empty(); //empty it first from dom.
db.click(function(){
var t = (jQuery(this).hasClass('ds_prev'))? 0 : 1; // 0 if arrow prev otherwise arrow next
if(t) {
sel1.find('option').each(function(){
if(jQuery(this).is(':selected')) {
jQuery(this).attr('selected',false);
var op = sel2.find('option:first-child');
sel2.append(jQuery(this));
}
});
} else {
sel2.find('option').each(function(){
if(jQuery(this).is(':selected')) {
jQuery(this).attr('selected',false);
sel1.append(jQuery(this));
}
});
}
});
});
PHP:
if(isset($_POST['submit'])) {
var_dump($_POST['select2']);
}
Like I said, I have this sort of working. But, if I send a value to select2, I have to highlight it before I submit or else it wont POST. Any ideas?
I've come across this before and you have a couple of options. Using JS you can either push all of the values in the second box into a hidden field as well, or also using JS you can select all of the values in the second box as an onsubmit handler on the form.
I've actually done the latter before, and it works just fine.
Ultimately, a select box (multi or single select) only sends the values that are selected -- so that's why it only works if you select them first. It works a lot like checkboxes do, where the unchecked values just don't get posted.
This should "select" all of them:
$('#myform').submit(function() {
var sel2 = $('#dualselect1 select:last-child');
sel2.find('option').each(function(){
$(this).attr('selected',true);
});
});
OR this would put them into a series of hidden fields:
$('#myform').submit(function() {
var sel2 = $('#dualselect1 select:last-child');
sel2.find('option').each(function(){
var hidden = $('<input type="hidden" name="selectedOptions[]"/>');
hidden.val($(this).val());
sel2.after(hidden);
});
});
and then in PHP you'd get these values by using $_POST['selectedOptions'];
You can simply modify this line jQuery(this).attr('selected',false); in sel1.find....block
with jQuery(this).attr('selected',true); .
In this mode al selection moved from first to second box is automatically selected,
so when you submit form, you directly pass this value.
Try it.
this should work:
if(t) {
sel1.find('option').each(function(){
if(jQuery(this).is(':selected')) {
jQuery(this).attr('selected',true);
var op = sel2.find('option:first-child');
sel2.append(jQuery(this));
}
});
}
I have a drop down list (ddlAccount) which i can choose an item from it and do a db query in test.php to retrieve corresponding data, then output an input element with the returned data.
This is my javascript code:
function load1(){
$('#divAccountNum').load('test.php?accountInfo=' + document.getElementById('ddlAccount').value , '' ,function() {
alert('Load was performed.')});
}
load1 function called when i change the dropdown list items, and it takes the value of the selected option and sends it to test.php in a parameter called "accountInfo".
my html:
<select name="ddlAccount" id="ddlAccount" onchange="load1();">
<option value="1"> Account1</option>
<option value="2"> Account2</option>
<option value="3"> Account3</option>
</select>
<div id="divAccountNum" >
</div>
And test.php :
if($_GET['accountInfo'])
{
$account = $accountDAO->load($_GET['accountInfo']); //mysql query
//print an input holding account number as its value
echo "<input type='text' name='txtAccountNum' id='txtAccountNum' value='".$account->accountnumber."'/>";
}
The problem is that nothing happened when i choose an option (nothing appear in div (divAccountNum))
Any suggestions? Thanks in advance.
Edit:
I used #thecodeparadox 's bit of code and it works and i found a solution for the problem that i mentioned in the comments below which is that when choosing one item from the dropdown list it shows the value in input element and loads the form again. The solution is in:
jQuery Ajax returns the whole page
So my jquery code now looks like:
$(document).ready(function(){
$('#ddlAccount').on('change', function() {
$.get('testJquery.php?accountInfo=' + $(this).val(), function(accountNum) {
//console.log(accountNum);
$('input#txtAccountNum').val(accountNum);
});
});
And testJquery.php :
if($_GET['accountInfo'])
{
$account = $accountDAO->load($_GET['accountInfo']);
$accountNum = $account->accountnumber;
echo $accountNum;
}
And at last i added input element in divAccountNum which has id="txtAccountNum"
Though you don't give enough info about your problem, but you can try this:
function load1(){
$('#ddlAccount').on('change', function() {
$.get('test.php?accountInfo=' + $(this).val(), function(response) {
$('div#divAccountNum').html(response);
}, 'html');
});
}
NOTE:
$('#ddlAccount').on('change', fires when dropdown change
$.get('test.php?accountInfo=' + $(this).val().. send a get(ajax) request to test.php with value selected from drop down
parameter response with in $.get() second parameter callback function is the response from the server
'html' as third parameter of $.get() for data type you return, as you return a html so it is html.
for more info read:
change()
$.get()
To get selected option value from select input use:
$('#ddlAccount option:selected').val()