how to create a for loop with a drop down box - javascript

Hi I am trying to create a set of drop down boxes that will use an array of data from the values that you pick and then runs through a loop to post them to the screen at the moment the data that i want to use will just be local but i want to edit this later so that it will loop through the data from my database and post that to the screen. i have looked at other questions on this subject and just wondering how i would change it for my code i have looked at this link questions on stack overflow that I have looked at i have just got a couple questions that im wondering if anybody has seen this before or if they have seen any examples i have also looked at for loops and i understand the concept
my questions to you are:
1) how would I post the values from my drop down boxes into a php array
2) how would I then check the values against and array of data and then choose which are correct and post them to the screen.
3)Would I need to use a second language like javascript or can it be done just in php
My drop down box code is
<div id="Content">
<?php include "select.class.php"; ?>
<form id="select_form">
Choose a category:<br />
<select id="category">
<?php echo $opt->ShowCategory(); ?>
</select>
<br /><br />
Choose a type:<br />
<select id="type">
<option value="%">any...</option>
</select>
<br /><br />
Choose a principle:<br />
<select id="principle">
<option value="%">any...</option>
</select>
<br /><br />
<input type="submit" value="confirm" />
</form>
<div id="result"></div>
<!-- end of the Options -->
below is the select.class.php
<?php
class SelectList
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database");
mysql_select_db($db,$this->conn) OR die("can not select the database $db");
return TRUE;
}
public function ShowCategory()
{
$sql = "SELECT * FROM subject";
$res = mysql_query($sql,$this->conn);
$category = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$category .= '<option value="' . $row['subject_id'] . '">' . $row['description'] . '</option>';
}
return $category;
}
public function ShowType()
{
$sql = "SELECT * FROM section WHERE subject_id=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$type = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$type .= '<option value="' . $row['section_id'] . '">' . $row['description'] . '</option>';
}
return $type;
}
public function ShowPrinciple()
{
$sql = "SELECT * FROM principle WHERE section_id=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$principle = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$principle .= '<option value="' . $row['principle_id'] . '">' . $row['description'] . '</option>';
}
return $principle;
}
}
$opt = new SelectList();
?>

1) how would I post the values from my drop down boxes into a php array
In the form tag add method="POST". Reference in PHP with $_POST array. Make sure to validate and escape the data before writing to your DB.
2) how would I then check the values against and array of data and then choose which are correct and post them to the screen.
If you don't have millions of categories, you are better off sending them all as a JSON array and using Javascript. Something like:
<script>
var categories = <?php echo json_encode($opt->ShowCategory()); ?>;
</script>
json_encode may require some options to be set, depening on your character set. More info here: http://php.net/manual/en/function.json-encode.php
Making a new request each time someone changes a dropdown box will drive them crazy, I know I hate that. If you have used jQuery before, this is very easy. This isn't that difficult without it.
3)Would I need to use a second language like javascript or can it be done just in php
For the sake of your users, use Javascript.

code for showCategory()
...
$categories = new array();
$category = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$categories[$row['subject_id']] = $row['description'];
}
$validCategories = $this->getValidCategories() // get the valid categories
foreach($categories as $index=>$cat){
// only choose the categories that are valid
if(array_search($cat,$validCategories) !== FALSE)
$category.= '<option value="'.$index.'">'.$cat.'</option>';
}
return $category;

Related

Generating SELECT OPTIONS with inline code?

<body>
<H1>4a</H1>
<form action="hw4b.php" method="post">
<?php
$con = mysqli_connect("localhost","[credential]","","[credential]")
or die("Failed to connect to database " . mysqli_error());
?>
<select name="id" value="id">
<script>
for (x=1;x<=101;x++)
{
document.write("<option value="+x+">"+
<?php echo mysqli_query($con, "SELECT LASTNAME FROM CUSTOMERS WHERE CUSTOMERID == "+x+";")?>
+"</option>");
}
</script>
</select>
<input type="submit" value="SEND IT">
</form>
</body>
So this should put the corresponding LASTNAME into the select, but it just fills every row with "NaN". I'm sure this is some stupid minor error, but I've been staring at it too long.
you should query the results of mysqli_query
do something like this:
<select name="id" value="id">
<?php
$query = mysqli_query($con, "SELECT LASTNAME FROM CUSTOMERS WHERE WHERE CUSTOMERID >=1 and CUSTOMERID <= 101 ;");
while ($row = mysqli_fetch_array($query))
echo "<option id='".$row['LASTNAME']."'>".$row['LASTNAME']."</option>";
?>
</select>
notes:
no need for javascript usage
please escape the query parameter
id of the option is the value that will be sent to the server, makes more since to send LASTNAME
avoid using a query at a loop
Note that your for cycle is in javascript (between <script> tags), yet you try to fill in some data in php.
Everything in PHP happens on server side, i.e. is interpreted, packed into a http response and returned to the client, where it is unpacked and javascript is executed.
You need to either put both into javascript, or both into php.
<select>
<?php
for ($i = 0; $i < 100; i++){
///make some select here
echo "<option value="$i"> ...output the select </option>"
}
?>
</select>
This way, all options are generated on server side and transferred to client as text
<select>
<option value="0">...</option>
<option value="1">...</option>
...
Other option is to export the database data into javascript, and then access it in javascript.
<script>
//or perhaps better
var myOtherData = <?=json_encode($somePHPData)?>;
</script>
//now you can use for loop with document.write and one of the variables you exported...
You need to be very careful and sure which execution happens on server, and which on client side.
There are several issues I think. You are using a comparison operator in the SELECT statement, it should just be =, not ==. Also, mysqli_query returns a mysqli_result, not a value like "Johnson" for LASTNAME. And, maybe most importantly, it doesn't make sense to do this with javascript since you're writing the values to the document before sending it to the browser anyway.
The code should look something like this (not tested)
<select name="id" value="id">
<?php
$query = 'SELECT LASTNAME, CUSTOMERID FROM CUSTOMERS WHERE CUSTOMERID >= 1 AND CUSTOMERID <= 101 ORDER BY CUSTOMERID ASC';
$result = mysqli_query($con, $query);
if (!$result) {
echo 'some error handling';
} else {
while ($row = $result->fetch_assoc()) {
echo '<option value="' . $row['CUSTOMERID'] . '">' . $row['LASTNAME'] . '</option>';
}
}
?>
</select>

actual code for xferring javascript variable to post variable

I have looked at this site for three days. I admit I am new to using javascript. But I have used the many different solutions offered and none have worked. Please help.
I am trying to do something that should be simple: save a user choice of country from a dropdown box on an html5 page to a hidden post variable (using javascript onchange.) That is used in a post array on the same form for a php operation that sends the input to a mysql database. This is my code:
The hidden post variable doesn't update. From there I can't test the code logic. But my onchange code came from this site and is suppose to work.
References:
<script type="text/javascript" src="../../js/jquery-2.1.4.min_prod.js"> </script>
<script type="text/javascript" src="../../js/respond.min.js"> </script>
<script type="text/javascript" src="../../js/bootstrap.min.js"> </script>
form information:
<form name="form1" method="post" enctype="multipart/form-data" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>" />
</form>
form element
$query="SELECT * from country ";
$query=$query."ORDER BY country asc";
$data = mysqli_query($conn,$query);
//or die('could not connect to db:'. mysqli_connect_error() );
mysqli_error($conn);
If ($data) {
echo '<select id="country" onchange="send_name(this)">';
echo '<option value="204">United States</option>';
while ($row = mysqli_fetch_array($data)) {
echo '<option value="'.$row['country_id'].'">'.$row['country'].'</option>';
}//while
echo '</select>';
}//if
mysqli_close($conn);
?>
<input type="hidden" id="c_php_value" name="c_php_value" value="">
Javascript
function send_name(selectObjectI) {
var value = selectObjectI.value;
$.ajax({
type: "POST",
url: "http://localhost/php/protected/form_addnews.php",
data:{c_php_value: value}
});
Post Submit Code
$country_id1 = trim($_POST['c_php_value']);
Thanks to a coder on utube, speaking german I might add, I discovered I don't need javascript, ajax or anything complicated to accomplish what I am trying to do.
I simply needed to do the following:
(1) Add a name="" to the dynamically created SELECT flag(name="country_id2").
(2) user chooses input with drop down box created.
(3) gather the $_POST after the form submit is set.
($country_id = $_POST['country_id2'])
$data = mysqli_query($conn,$query);
//or die('could not connect to db:'. mysqli_connect_error() );
mysqli_error($conn);
If ($data) {
echo '<select name="country_id2" id="country_id2">';
echo '<option value="204">United States</option>';
while ($row = mysqli_fetch_array($data)) {
echo '<option value="'.$row['country_id'].'">'.$row['country'].'</option>';
}//while
echo '</select>';
}//if
mysqli_close($conn);

Get all selected items of a Multi Select Dropdown on form Submit

I have a long form that initially populates the Multi Select Dropdown with values from the database. After this happens the user then has the option to add or remove the selected items from the dropdown. But every time the submit it and it goes to the next page, it always keeps the original array and nothing changes.
function getGrade($id, $grades_array)
{
$counter = 0;
$sql = "Select grade FROM grades";
$result = mysql_query($sql) or die (mysql_error());
echo '<select name="grade" multiple="multiple" id="grades_selected">';
while ($row = mysql_fetch_array($result)) {
if ($row['grade'] != $grades_array[$counter]) {
echo "<option>" . $row['grade'] . "</option>";
} else {
echo "<option selected=" . $row['grade'] . ">" . $row['grade'] . "</option>";
$counter = $counter + 1;
}
}
mysql_free_result($result);
echo '</select>';
}
All I need it to do is right when the submit button is pressed it needs to get all of the selected values from the dropdown and create a new array with all of those values.
You can use sessionStorage to implement this. This will enable you to save the array values as a string within the browser session.
Alternatively, you could use localStorage too, but this would save the values across sessions (in multiple tabs) which is probably not what you would want for such a scenario.

how would I post multiple values using $_Post and then use these values

Hi I am creating a website using html,css,php,mysql and javascript with little bits of jquery.
at the moment I have 3 drop down boxes that dynamically show you options for Example if I click english then the next drop down box shows all sections for english such as reading and then the next one will be related to reading this is all working perfectly.
What I am trying to do is use the values for each of these drop down boxes in a sql query to the return the specific videos that are related to these drop down boxes.
what type of array would I need to use for the $_Post and how would the sql query be structured so i can pull down all the information for the videos.
This is the code that I am currently using for the drop down menus
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script src="http://code.jquery.com/jquery-migrate-1.2.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("select#type").attr("disabled","disabled");
$("select#category").change(function(){
$("select#type").attr("disabled","disabled");
$("select#type").html("<option>wait...</option>");
var id = $("select#category option:selected").attr('value');
$.post("select_type.php", {id:id}, function(data){
$("select#type").removeAttr("disabled");
$("select#type").html(data);
});
});
$("select#principle").attr("disabled","disabled");
$("select#type").change(function(){
$("select#principle").attr("disabled","disabled");
$("select#principle").html("<option>wait...</option>");
var id = $("select#type option:selected").attr('value');
$.post("select_principle.php", {id:id}, function(data){
$("select#principle").removeAttr("disabled");
$("select#principle").html(data);
});
});
$("form#select_form").submit(function(){
var cat = $("select#category option:selected").attr('value');
var type = $("select#type option:selected").attr('value');
var princ = $("select#principle option:selected").attr('value');
if(cat>0 && type>0 && princ>0)
{
var result = $("select#principle option:selected").html();
$("#result").html('your choice: '+result);
}
else
{
$("#result").html("you must choose two options!");
}
return false;
});
});
</script>
</head>
<body>
<?php include "select.class.php"; ?>
<form id="select_form">
Choose a category:<br />
<select id="category">
<?php echo $opt->ShowCategory(); ?>
</select>
<br /><br />
Choose a type:<br />
<select id="type">
<option value="%">any...</option>
</select>
<br /><br />
Choose a principle:<br />
<select id="principle">
<option value="%">any...</option>
</select>
<br /><br />
<input type="submit" value="confirm" />
</form>
<div id="result"></div>
</body>
</html>
this is the section that runs the sql queries
<?php
class SelectList
{
protected $conn;
public function __construct()
{
$this->DbConnect();
}
protected function DbConnect()
{
include "db_config.php";
$this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database");
mysql_select_db($db,$this->conn) OR die("can not select the database $db");
return TRUE;
}
public function ShowCategory()
{
$sql = "SELECT * FROM subject";
$res = mysql_query($sql,$this->conn);
$category = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$category .= '<option value="' . $row['subject_id'] . '">' . $row['description'] . '</option>';
}
return $category;
}
public function ShowType()
{
$sql = "SELECT * FROM section WHERE subject_id=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$type = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$type .= '<option value="' . $row['section_id'] . '">' . $row['description'] . '</option>';
}
return $type;
}
public function ShowPrinciple()
{
$sql = "SELECT * FROM principle WHERE section_id=$_POST[id]";
$res = mysql_query($sql,$this->conn);
$principle = '<option value="0">choose...</option>';
while($row = mysql_fetch_array($res))
{
$principle .= '<option value="' . $row['principle_id'] . '">' . $row['description'] . '</option>';
}
return $principle;
}
}
$opt = new SelectList();
?>
Here is something basic:
HTML - create the interface to gather your data
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.js"></script> (put in head)
<div class='nameinput'>Name: <input type=text /></div>
<div class='ageinput'>Age: <input type=text /></div>
<button id = 'clickme'>CLICK</button>
JS - gather the data and present to php with ajax
$('#clickme').click(function(){
var username = $('.nameinput').val();
var userage = $('.ageinput').val();
$.ajax({
url: 'nameofphpfile.php',
type: POST,
data: { name: username,
age: userage },
dataType: 'json'
})
.done(function(){ alert('Great job!') })
.fail(function(){ alert('The server does not like you!) });
});
(NB: The more difficult part for me was understanding the callbacks - I've switched to the "newer" forms of .done, .fail, .always).
PHP - grab the data passed from client with ajax, and put in the db.
<!php
$name = $_POST['username'];
$age = $_POST['userage'];
$host = "xxxxx";
$user = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
$cxn = mysqli_connect($host,$user,$password,$dbname);
if (mysqli_connect_errno()) {echo "No connection" . mysqli_connect_error();}
$query = " INSERT INTO nameofyourtable
( name, age)
VALUES
('$name', '$age') " ;
$result = mysqli_query($cxn, $query) or die ("could not query database 1");
?>
The approach is different if you want to read data, but that's a different question.
Best of luck!
I went through this about a year ago, so I know what you're going through.

Two Selectboxes - load by selection

I want, if I select a value in selectbox 1, that selectbox2 is loaded with values that are dependet on selectbox1.
Here´s how I tried:
<head>
<script src="jquery/jquery-1.8.2.min.js" type="text/javascript"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#selectbox1").change(function(){
var value = $(this).children('option.selected').val();
$("#selectbox2").load("php/loadteams.php", {value: value} );
});
});
</script>
</head>
<body>
<?php
include 'php/db.php';
$res = mysql_query("select * from liga", $connection);
echo '<select id="selectbox1">';
while($val = mysql_fetch_array($res)) {
echo '<option value="'.$val['id'] .'">' . $val['name'] . '</option>';
}
echo '</select>';
mysql_close($connection);
?>
<select id="selectbox2"> </select>
Like you see, I tried with jquery. The file loadteams.php looks like that:
<?php
include 'php/db.php';
$ligaid = mysql_real_escape_string($_POST['value']);
$result = mysql_query("select * from mannschaft where liga = '" . $ligaid . "'", $connection);
while($te = mysql_fetch_array($result)){
echo '<option> ' . $te . '</option>';
}
?>
I don´t know where my mistake is - can you help me?
You should try using the :selected selector, instead of .selected which would match a class with the name selected.
Also, I believe the .load() method posts data through HTTP GET, so I believe you would have to use $_GET["value"] instead of $_POST['value'].
Correction:
As you are providing the data as an object, it IS posted through HTTP POST, so strike my above statement. From the jQuery documentation:
The POST method is used if data is provided as an object; otherwise, GET is assumed.

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