I am trying to get drop down select value posted upon send.
Markup
<select name="taskOption">
<option value="1">First</option>
<option value="2">Second</option>
<option value="3">Third</option>
</select>
php
$selectOption = $_POST['taskOption'];
{
$message.=$value.'<br>';
}
$message.='
What am I doing wrong?
Is this what you're trying to do?
if(!empty($_POST['taskOption']))
{
$value = $_POST['taskOption'];
$message.=$value.'<br>';
}
you need to use post in your form, method="post"
if(isset($_POST['submit'])){
$dropdwon = $_POST['taskOption'];
//code here
}
Related
I have a problem which I'm not sure how to approach, I have a WP plugin which has a form, in the form I have a with result that could have varied amount of data (depending on what the user submits).
How can I add 'selected' to the selected item so when the user returns he can edit/view the item he selected?
<select name="supplier">
<option value="Supplier 1">Supplier 1</option>
<option value="Supplier 2">Supplier 2</option>
<option value="Supplier 3">Supplier 3</option>
<option value="Supplier 4">Supplier 4</option>
</select>
A loop comes to mind, shall I assign a number to each option? There may be 100 suppliers so it would need to count right?
I would use a loop like...
$theirChosenSupplier = $supplierVarFromDatabaseOrWherever;
?>
<select name="supplier">
foreach($allSuppliers as $individualSupplier) {
if($individualSupplier == $theirChosenSupplier) {
$selected = "selected";
} else {
$selected = "";
}
?><option value="<?php echo $individualSupplier; ?>" <?php echo $selected; ?>>
<?php echo $individualSupplier; ?>
</option>
<?php } ?>
</select>
This code isn't tested but should give you an idea. Apologies if I've misunderstood the question.
adding an id is always a good thing. Assuming this is wordpress generated and you don't have a loop already there is always the option of using javascript and maybe an onchange event.
I dont know if the form is submitted first or you do some other stuff but maybe:
<select id="foo">
<option id="provider_1">Provider 1</option>
<option id="provider_2">Provider 2</option>
</select>
<script type="text/javascript">
window.onload = function() {
var selectedValue = '<?php echo someEscapeFunction($_SESSION['id_provider']); // or _GET, _POST ?>';
document.getElementById('foo').selectedIndex = document.getElementById(selectedValue).index;
}
</script>
I have a drop down list where I select options
<form action="" method="POST" class="styled-select">
<select name="seasons" onchange='this.form.submit()'>
<option value="">Select a Season</option>
<option value="1">2002/2003</option>
<option value="2">2003/2004</option>
<option value="3">2004/2005</option>
<option value="4">2005/2006</option>
<option value="5">2006/2007</option>
<option value="6">2007/2008</option>
<option value="7">2008/2009</option>
<option value="8">2009/2010</option>
<option value="9">2010/2011</option>
<option value="10">2011/2012</option>
<option value="11">2012/2013</option>
<option value="12">2013/2014</option>
</select>
<noscript><input type="submit" value="Submit"></noscript>
</form>
You can see the list here footystat
I am using the following PHP
if(isset($_POST['seasons'])){ $seasonette = $_POST['seasons']; }
if(isset($_POST['year'])){ $yearette = $_POST['year']; }
if(isset($_POST['comp'])){ $competitionette = $_POST['comp']; }
if(isset($_POST['which'])){ $whichette = $_POST['which']; }
When I select something from the list, I want selected item in the list to continue showing. At the moment when I select (for example) 2013/2014, it will show the results but the drop down menu goes back to its original state instead of showing 2013/2014.
Get Option value selected when it gets posted value, like this,
<option value="1" <?php if(isset($_POST['seasons']) && $_POST['seasons'] == '1'){ ?> selected="selected" <?php } ?>>2002/2003</option>
Set value like this for each option
You can set the "selected" property to the option , just like you set a value !
<option value="8" selected>2009/2010</option>
Use a if statement in PHP to determine which one should be selected.
Thats because the page refreshes.
On page load check if there is post variable than match the value with each option's HTML and write selected attribute.
The shorter way is
<option value="1" <?php echo $_POST['seasons']==1?"selected":""; ?>2002/2003</option>
I have created the dropdown list in my index page i want to select the one value from that list and validate it if not selected any value the code for that is as follows:
<form action="" method="post">
<select value="state" name="state">
<option selected="">---please enter---</option>
<option value="1">andhra</option>
<option value="2">thamil</option>
<option value="3">kerela</option>
</select>
</form>
php code for the above file is as follows:
<?php
if(!empty($_POST['state'])) {
$state = $_POST['state'];
}
else {
echo "required";
}
?>
I dont want to be select the first option in selection list please enter to be selected but the code which I have used is taking that value also I want relevant code how to validate that list?
I prefer adding a disabled attribute to the placeholder option, that way, a user has to choose something if they click the drop-down:
<select value="state" name="state">
<option selected disabled>---please enter---</option>
<option value="1">andhra</option>
<option value="2">thamil</option>
<option value="3">kerela</option>
</select>
Quick demo: http://jsfiddle.net/hxxJZ/
something like this should to the trick:
<form action="" method="post">
<select name="state">
<option value="0" selected>---please enter---</option>
<option value="1">andhra</option>
<option value="2">thamil</option>
<option value="3">kerela</option>
</select>
</form>
php
<?php
if( 0 != $_POST['state'] ) {
$state = $_POST['state'];
}
else {
echo "required";
}
?>
your HTML is not right,
try edit this line
<option selected=""...
to this one
<option selected value="">---
select elements do not have an (HTML) value attribute, so remove value="state".
The selected attribute should be selected="selected" (or just SELECTED).
Im trying to implement the search feature in my website.
when the search keyword is entered in the textbox, and the category combo is selected, the form will be Posted and the result will be shown on the same page.
what i want is to keep the selected category of the combo by default in the form after posted
For eg., If i select the category 'Automobiles' in the combo and click search, after form submit, the combo should show the automobiles as default selected option. Please help me. Any help will be appreciated
I assume you get categories from database.
you should try:
<?php
$categories = $rows; //array from database
foreach($rows as $row){
if($row['name'] == $_POST['category']){
$isSelected = ' selected="selected"'; // if the option submited in form is as same as this row we add the selected tag
} else {
$isSelected = ''; // else we remove any tag
}
echo "<option value='".$row['id']."'".$isSelected.">".$row['name']."</option>";
}
?>
Assuming that by "combo" you mean "A regular select element rendering as a drop down menu or list box" and not "A combobox that is a combination of a drop down menu and free text input":
When outputting the <option> elements, check the value against the submitted data in $_POST / $_GET and output selected (in HTML) or selected="selected" (in XHTML) as an attribute of the option element.
Here is the JQuery way I am using.
<select name="name" id="name">
<option value="a">a</option>
<option value="b">b</option>
</select>
<script type="text/javascript">
$("#name").val("<?php echo $_POST['name'];?>");
</script>
But this is only if you have jquery included in your webpage.
Regards
<?php
$example = $_POST["friend"];
?>
<form method="POST">
<select name="friend">
<option value="tom" <?php if (isset($example) && $example=="tom") echo ' selected';?>>Thomas Finnegan</option>
<option value="anna" <?php if (isset($example) && $example=="anna") echo ' selected';?>>Anna Karenina</option>
</select>
<br><br>
<input type="submit">
</form>
This solved my problem.
This Solved my Problem. Thanks for all those answered
<select name="name" id="name">
<option value="a">a</option>
<option value="b">b</option>
</select>
<script type="text/javascript">
document.getElementById('name').value = "<?php echo $_GET['name'];?>";
</script>
$countries_uid = $_POST['countries_uid'];
while($row = mysql_fetch_array($result)){
$uid = $row['uid'];
$country = $row['country_name'];
$isSelected = null;
if(!empty($countries_uid)){
foreach($countries_uid as $country_uid){//cycle through country_uid
if($row['uid'] == $country_uid){
$isSelected = 'selected="selected"'; // if the option submited in form is as same as this row we add the selected
}
}
}else {
$isSelected = ''; // else we remove any tag
}
echo "<option value='".$uid."'".$isSelected.">".$country."</option>";
}
this is my solutions of multiple select dropdown box after modifying Mihai Iorga codes
After trying al this "solves" nothing work. Did some research on w3school before and remember there was explanation of keeping values about radio. But it also works for Select option. See here an example. Just try it out and play with it.
<?php
$example = $_POST["example"];
?>
<form method="post">
<select name="example">
<option <?php if (isset($example) && $example=="a") echo "selected";?>>a</option>
<option <?php if (isset($example) && $example=="b") echo "selected";?>>b</option>
<option <?php if (isset($example) && $example=="c") echo "selected";?>>c</option>
</select>
<input type="submit" name="submit" value="submit" />
</form>
Easy solution:
If select box values fetched from DB then to keep selected value after form submit OR form POST
<select name="country" id="country">
<?php $countries = $wpdb->get_results( 'SELECT * FROM countries' ); ?>
<option value="">
<?php if(isset($_POST['country'])){echo htmlentities($_POST['country']); } else { echo "Select Country *"; }?>
</option>
<?php foreach($countries as $country){ ?>
<option <?php echo ($_POST['country'] == $country->country_name ? 'selected="selected"':''); ?> value="<?php echo $country->country_name; ?>"><?php echo $country->country_name; ?>
</option>
<?php } ?>
</select>
i need some clarification on how to populate select(s) with data from mysql. Basically what I am trying to do is:
There will be a first select box with some data in it.
<select>
<option>1</option>
<option>2</option>
<option>3</option>
</select>
when the user selects a option in the first select,
there is a second select below that, which should reflect the values according to the selection made in the first select.
<select>
<option>1.1</option>
<option>1.2</option>
<option>1.3</option>
</select>
The data is commin from MySQL. I am not sure if need to post to same page, but if I do, how to retain the values alredy selected in the previous select boxes? do i need to use javascript?
any help?
Thanks.
You should use javascript so you don't need a page refresh. I just re-read your question and I'll have a solution involving an AJAX request in a second to pull dynamic data:
HTML
<select name="select1" id="select1">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
</select>
<select name="select2" id="select2">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
</select>
jQuery
<script type="text/javascript">
$(document).ready(function() {
$('#select1').change(getDropdownOptions);
});
function getDropdownOptions() {
var val = $(this).val();
// fire a POST request to populate.php
$.post('populate.php', { value : val }, populateDropdown, 'html');
}
function populateDropdown(data) {
if (data != 'error') {
$('#select2').html(data);
}
}
</script>
populate.php
<?php
if (!empty($_POST['value'])) {
// query for options based on value
$sql = 'SELECT * FROM table WHERE value = ' . mysql_real_escape_string($_POST['value']);
// iterate over your results and create HTML output here
....
// return HTML option output
$html = '<option value="1">1</option>';
$html .= '<option value="b">B</option';
die($html);
}
die('error');
?>