It is possible to change dropdown to list in form?
can't embed an image because of the reputation
here's my form with dropdown :
then I've change the value inside dropdown to list like this:
but the problem, my first image can execute the button and save the value, but when I change the form interface like my 2nd image, the button can't execute (nothing effect showing after clicking the button)
my code on dropdown looks like :
<div class="form-group">
<label for="ik">Kriteria</label>
<select class="form-control" id="ik" name="ik">
<?php
$stmt2 = $pgn2->readAll();
while ($row2 = $stmt2->fetch(PDO::FETCH_ASSOC)){
extract($row2);
echo "<option value='{$id_kriteria}'>{$nama_kriteria}</option>";
}
?>
</select>
</div>
then I change the code of <select> to <form> like this to showing list :
<div class="form-group">
<form class="form-control" id="ik" name="ik">
<?php
$stmt2 = $pgn2->readAll();
while ($row2 = $stmt2->fetch(PDO::FETCH_ASSOC)){
extract($row2);
?>
<option value="<?php echo $id_kriteria; ?>"><?php echo $nama_kriteria; ?></option>
<input type="text" class="form-control" id="nn" name="nn">
<?php
}
?>
</form>
</div>
Problem:
the 2nd code is <form> inside <form>, because if I change the <form> tag to <label> or etc, the interface looks bad,
may someone help me how to do?
Related
how, after I choose the option and then it will appear #formid where the form is filled in by the $stock as selected in the option section
<div class="form-group">
<label for="exampleFormControlSelect1">Nama Item</label>
<select class="form-control" id="exampleFormControlSelect1" name="itemId">
<option selected>Choose</option>
<?php $no=1; while($row = $result->fetch_object() ) {
$stock = $row->stock;
$i_name = $row->i_name;
$id = $row->id;
?>
<option value="<?php echo $id; ?>"><?php echo $i_name; ?></option>
<?php } ?>
</select>
</div>
the form that will appear after selecting the option
<div class="form-group">
<label>Stock</label>
<input class="" type="number" name="" value="<?php echo $stock;?>">
</div>
I am new in php and javascript, can you write with a neat code & easy to understand, thank you :)
If I understand you correctly, you want to show the second div onSelect of the first one.
Because of the fact, that you use bootstrap I assume
that you have already included jQuery if not you have to do that for this example.
You have to add this JavaScript on your page.
<script>
$( document ).ready(function() {
$( "#exampleFormControlSelect1" ).change(function() {
$("#secondDiv").show();
});
});
</script>
And you need an id on your div.
<div id="secondDiv" class="form-group">
<label>Stock</label>
<input class="" type="number" name="" value="<?php echo $stock;?>">
</div>
You will need javascript for this if you want the data to appear without the page reloading.
First we make Javascript Function that takes your option id and passes it to your input value.
funcction chooseOption(clicked.id){
document.getElementById('myInput').value = clicked_id;
}
HTML and PHP we need to add a different id for each option. The best way it to pass the option name you want for each. We also added our onclick function to the options.
<div class="form-group">
<label for="exampleFormControlSelect1">Nama Item</label>
<select class="form-control" id="exampleFormControlSelect1" name="itemId">
<option selected>Choose</option>
<?php $no=1; while($row = $result->fetch_object() ) {
$stock = $row->stock;
$i_name = $row->i_name;
$id = $row->id;
echo "<option onclick=\"chooseOption(this.id);\" id=\"".$stock."\" value=\"".$id."\">".$i_name."</option>";
<?php } ?>
</select>
</div>
Added an id to your input so javascript can send the data to it.
<div class="form-group">
<label>Stock</label>
<input id="myInput" class="" type="number" name="" value="">
</div>
I have one dropdown box and one textbox. I want when I select the dropdown value then, it will fill the matching database textbox value automatically based on the database drop-down selection.
Here is my code:-
Dropdown and Textbox:-
<select name="product" required>
<option value=""></option>
<?php
while($productRow = mysqli_fetch_assoc($productResult)){?>
<option value="<?php echo $productRow['ID'];?>">
<?php echo $productRow['Category']; ?></option><?php } ?>
</select>
<input placeholder="Phone Number" name="phoneNumber" id="pnum" type="text">
Script :-
<script>
$(document).ready(function(){
$('select[name="product"]').change(function()
{
$('#pnum').val($('select[name="product"] option:selected').data('Mobile'));
}
);
});
</script>
Now textbox fills the empty value in every time when dropdown value change.
pls help a lot.
You need to change the html to
<select name="product[]" size=20 multiple required>
<?php
while($productRow = mysqli_fetch_assoc($productResult)){?>
<option value="<?php echo $productRow['ID'];?>">
<?php echo $productRow['Category']; ?></option><?php } ?>
</select>
Then remove the script because it is no longer required. All the selection of items is done in the <select> element.
On the server-side you will pickup the values in the product array.
I'm currently using Selectize.js with a 'select' item and multiple options. My html is something like:
<div class="form-group">
<label for="record-genre">Genre(s):</label>
<select name="record_genre" id="record-genre" class="form-control multi-selectize" multiple="multiple" placeholder="Enter genre(s)">
<?php foreach ($record_genre_opts as $key => $val): ?>
<option value="<?php echo $key; ?>" <?php echo (in_array($key, $record_genre)) ? "selected" : ''; ?>
<?php echo set_select('record_genre', $val); ?>
>
<?php echo $val; ?>
</option>
<?php endforeach;?>
</select>
</div>
So far so good. The items are being loaded correctly and until this point everything works great. The issue is that when the form is submitted only the last value selected is the one that is available, so if I selected multiple items I'm getting just one:
["record_genre"]=> string(1) "1"
Here's my JS setup:
$('.multi-selectize').each(function(){
//console.log($(this))
$(this).selectize({})
})
Any suggestions?
Thanks.
My mistake. The name of the select item should be record_genre[] since it is declared as a multiple select instead of just record_genre.
I use multiselect JS with php to get more than one selection of my form select.
Impossible for me not to show the selection' results on the select tags.
At least, if It's possible to show just the number of sélection(s) , it would be nice.
Thanks.
Sorry by impossible for to enter my code for the moment ! (seem not working)
<label for="showroom">Type/products* :</label>
<select class="form-control" id="materials" name="materials" multiple="multiple">
<?php while($prod=mysqli_fetch_assoc($prodQ)): ?>
<option value="<?php echo $prod['choc_id']; ?>"><?php echo $prod['choc_nom']; ?></option>
<?php endwhile; ?>
</select>
<script type="text/javascript">
$(document).ready(function() {
$('#materials').multiselect();
});
</script>
I have two dropdown boxes for car makes and models, when a make is selected i need all its models to show in the next box. As of now using the following code the models show up when the dropdown is clicked.
window.onmousedown = function(e){
this.id = e.target.id;
if(this.id == "vehicle_makes"){
var make = document.getElementById("vehicle_makes").value;
ajaxCall(make);
}
}
However i need the javascript to trigger when an option from the dropdown is selected rather than when the drop down is opened.
here also is the html - with some php
<div class="vehicle-search-wrap">
<form method="get" id="searchform" action="<?php bloginfo('url'); ?>">
<div>
<select id="vehicle_makes" name="s">
<?php
foreach($makes as $make){ //put all makes in dropdown
echo "<option value='". $make ."'>". $make ."</option>";
}
?>
</select>
<select id="model_drop" name="vmodels">
<?php
//nothing to start
?>
</select>
<input type="submit" value="Search Vehicle" />
</div>
</form>
</div>
use the onchange event of selectbox.
here is a demo
Try this:
<select name="Pizza" size="5"
onchange="myFunctionHere()">
<option value="P101">Pizza Napoli</option>
<option value="P102">Pizza Roma</option>
</select>