I have a drop down menu that I want to change to a multiple select box. The code below is working if you only select 1 option (the way I had it before), but of you select 2 it will only show 1 of the two, how can I make it show both options selected, here is the code:
<?php $makes = array("volvo","Saab","Opel","Audi","BMW") ?>
<form method="post" name="store" action="<?php $_SERVER['PHP_SELF'] ?>" >
<select multiple="multiple" name="cars">
<?php foreach ($makes as $make){echo "<option value=\"$make\">". $make ."</option>"; $vehicles = $_POST['cars'];} ?>
<input name="submit" type="submit">
</select>
</form>
<?php
if($_POST['submit']){
echo $vehicles;
}
?>
</body>
</html>
I hope I read this correctly that you would like to retrieve an array of results from the HTML multi select box.
By the way your code snippit is not correct; the HTML <form> closing tag should be after your closing <select> tag and I'm not sure why you have the following in your PHP for() loop:
$vehicles = $_POST['cars'];
You will want to make the HTML tags' name attribute an array as follows (Note I did not test this code):
<select multiple="multiple" name="cars[]">
<?php
foreach ($makes as $make) {
echo "<option value=\"$make\">". $make ."</option>";
}
?>
</select>
<?php
if($_POST['submit']) {
print_r($_POST['cars']);
}
?>
PHP.net - How do I get all the results from a select multiple HTML tag?
It isn't completely clear from your question but I think you mean the following bit of code only echoes one value:
if($_POST['submit']){
echo $vehicles;
}
To turn your selected cars into an array you need to add [] onto the end of the name:
<select multiple="multiple" name="cars[]">
Then to echo each of the selections you can use a foreach loop:
foreach ($_POST['cars'] as $car)
echo $car.'<br />';
you can try with if($_POST['submit']){
print_r($vehicles);
}
please let me know if any issue then..
Related
I have a dropdown box where the data is coming from my database. I want to send the selected drop-down item to the controller action method. And then I will send these values to the model to do the further works. Here, the problem is I have two database values in a single item of drop-down box. And, I am not figuring out how to send those two the method. Here is my code given below,
<select name='select'>
<option selected disabled>Choose Stations</option>
<?php foreach ($get_stations as $get_stations_item): ?>
<option>Station <?php echo $get_stations_item['sourcestationid']; ?> - Station <?php echo $get_stations_item['destinationstationid']; ?></option>
<?php endforeach; ?>
</select>
From this dropdown items, I want to send the sourcestationid and destinationstationid separately as 2 parameters to my controller action method. Here is my controller code though this is not correct way I think,
function getdata(){
$iotdata['test'] = $this->input->post('select2');
//rest of the code according to the source and destination id item
}
Thanks in advance.
You haven't kept the variable inside option's value tag.
<select name='select'>
<option selected disabled>Choose Stations</option>
<?php foreach ($get_stations as $get_stations_item): ?>
<option value="<?php echo $get_stations_item['sourcestationid'].','.$get_stations_item['destinationstationid']; ?>">Station <?php echo $get_stations_item['sourcestationid']; ?> - Station <?php echo $get_stations_item['destinationstationid']; ?></option>
<?php endforeach; ?>
</select>
Also here at PHP end, you can explode the string into an array and store it further into db.
<?php
$select = explode(',', $select)
print_r($select); // this will have your two values
?>
Thanks #BitsPlease for our suggestion. Your idea is working. But I need to do a small change to get it instantly. My select tag needs to be under form tag to get them without reloading the page again. Here is my view,
<form method="post" accept-charset="utf-8" action="<?php echo site_url("controller/action"); ?>">
<select name='select' onchange="this.form.submit()">
<option selected disabled>Choose Stations</option>
<?php foreach ($get_stations as $get_stations_item): ?>
<option value="<?php echo $get_stations_item['sourcestationid'].','.$get_stations_item['destinationstationidr']; ?>">Station <?php echo $get_stations_item['sourcestationid']; ?> - Station <?php echo $get_stations_item['destinationstationidr']; ?></option>
<?php endforeach; ?>
</select>
</form>
Here is the php code,
$select = explode(',', $this->input->post('select'));
print_r($select)
I have a HTML DataList input box, and I would like to have the options in the DataList to be things from my SQL database. I have been trying to make this work, but for some reason I cant. Does anyone know how this could be achieved?
Here is my code:
<?php
include_once 'connect.php';
$sql="select * FROM table";
?>
<form>
<input list="list" name="name">
<datalist id="datalist">
<?php
foreach ($dbo->query($sql) as $row) {
echo "<option value="$row[name]"/>";
}
?>
</datalist>
<input type="submit">
</form>
The option I get with this is literally $row[name]
Could someone tell me what I'm doing wrong?
Datalist id has to match the input list attribute and change the loop to "<option value=" . $row['name'] . "/>"
Could someone kindly help out with this rather basic issue: In a simple HTML form (functionally all good) two select elements are included.
Code:
<select id="dept" name="dept" required>
<?php
foreach($stddept as $item0) {
echo "<option value='$item0'";
if ($_POST['dept'] == $item0) echo 'selected="selected"';
echo ">$item0</option>";
}
?>
</select>
<br><br>
<select id="lev" name="lev" required>
<?php
foreach($stdlev as $item1) {
echo "<option value='$item1'";
if ($_POST['lev'] == $item1) echo selected="selected"';
echo ">$item1</option>";
}
?>
</select>
<input type='submit' id='submituser' name='submituser' value='Submit'>`
These two drop-down boxes behave differently in that only the second one honours the 'required' attribute. The first one can be submitted blank, which is obviously not right.
They also appear differently (see image) in that the first displays the first option in the list (from MySQL DB) but the second displays a blank. I have checked everything I could to see what else could cause the difference - without any joy.
Does anyone know what I am missing here? Thank you.
You can do this using jquery/javascript by getting the id of the first selected value and then select the result for the second box based on that id.
I think the problem with the code ( as it is shown ) is that you are incorrectly echoing the option in the second dropdown - it is missing quotes. You can streamline the code somewhat like this:
<select id="dept" name="dept" required='required'>
<?php
foreach($stddept as $item0){
$selected=$_POST['dept'] == $item0 ? " selected='selected'" : "";
echo "<option value='$item0'{$selected}>";
}
?>
</select>
<br /><br />
<select id="lev" name="lev" required='required'>
<?php
foreach($stdlev as $item1){
$selected=$_POST['lev'] == $item1 ? " selected='selected'" : "";
echo "<option value='$item1'{$selected}>";
}
?>
</select>
<input type='submit' id='submituser' name='submituser' value='Submit'>
on calculatePC.php, I have this code to display the finish_product
Select product:
<select class="itemTypes">
<?php while ($row1 = mysqli_fetch_array($result)) { ?>
<option value="<?php echo $row1['finish_product']; ?>">
<?php echo $row1['finish_product']; ?>
</option>
<?php } ?></select>
<br/><br/>
Let's say I have chosen Table as the finish_product.
On docalculate.php, I would like to display what I've chosen based on the dropdown list I've selected.
I tried this but there is error.
<?php echo $_POST['finish_product'] ?>
May I know how to display the result?
This doesn't exist:
$_POST['finish_product']
because you don't have a form element named "finish_product" in your markup. Add that name to the form element:
<select name="finish_product" class="itemTypes">
You need to do two things:-
Create a form before select and give it an action
give name attribute to your select box, then only you can get data in $_POST
So do like below:-
<form method="POST" action = "docalculate.php"> // give the php file paht in action
<select class="itemTypes" name="finish_product"> // give name attribute to your select box
<?php while ($row1 = mysqli_fetch_array($result)) { ?>
<option value="<?php echo $row1['finish_product']; ?>">
<?php echo $row1['finish_product']; ?>
</option>
<?php } ?></select>
<br/><br/>
<input type ="submit" name="submit" value ="Submit">
</form>
Now on docalculate.php:-
<?php
echo "<pre/>";print_r($_POST['finish_product']); // to see value comes or not
Note:- through Jquery its also possible. Thanks
I have the below code. It's a function creating a from containing a dropdown menu and a selectionbox, and is being called from a different page).
<?php
require_once 'forbindtilDB.php';
function populerDropdownsOpretProgram()
{
global $mysqliOOP;
$stmt = $mysqliOOP->prepare("SELECT DISTINCT primaer FROM oevelser ORDER BY primaer ASC");
?>
<form action="temp.php" method="post">
<select multiple>
<option value="0">Vælg ønskede øvelse(r)</option>
<?php
$stmt->execute();
$stmt->bind_result($primaereMuskelgruppe);
while ($stmt->fetch())
{
?>
<option value = "<?php echo $primaereMuskelgruppe;?>" >
<?php echo $primaereMuskelgruppe;
?>
</option>
<?php
}
$stmt->close();
?>
</select>
<?php
$stmt = $mysqliOOP->prepare("SELECT antalreps FROM antalreps ORDER BY antalreps ASC");
?>
<select>
<option value="0">Vælg ønskede antal gentagelser</option>
<?php
$stmt->execute();
$stmt->bind_result($antalreps);
while ($stmt->fetch())
{
?>
<option value = "<?php echo $antalreps;?>" >
<?php echo $antalreps;
?>
</option>
<?php
}
$stmt->close();
?>
</select>
<input type="submit">
</form>
<?php
}
I want to post the user input on a different page (currently temp.php), but I don't know how to handle the fact that the form contents is variables fetched by a mysqli call.
So far I've tried different versions of the below on the temp.php-page
<?php
echo $_POST[$antalreps];
?>
<?php
echo $_POST[$primaereMuskelgruppe];
?>
But I'm getting errors stating that there is undefined variables (antalreps and primaeremuskelgruppe) and undefined indexes...
Also, there's the added complexity that the selection box may return more than one result.
Pretty sure echo $_POST[$antalreps]; etc. is the wrong way to go about this, but I haven't been able to figure out alternatives...
Any hints?
You should add 'name' to your select and then use it.
For example instead of:
<select>
you should have:
<select name="yourname">
and then in php to display its value use:
echo $_POST['yourname'];
and for select multiple you can use:
<select name="othername[]">
and then in PHP use it as array:
foreach ($_POST['othername'] as $item) {
echo $item;
}
of course where I put yourname and othername you should put descriptive name such as colours for color box and similar