Codeigniter - How to create a form with undefined number of input? - php

So I've been trying to create an 'add' function for a contest table that is also linked to a joining table as the contest has many sport events associated to it (and a sport events will also be included in many contests). I have managed to create it as below, however it is limited to only adding one row into the joining table based on the chosen game in a dropdown.
So now, how do I go about creating the form if I don't know how many sport events there might be? How will the website create another dropdown if one dropdown is used so that more sport events can be selected? Similarly, how does the controller know that there are x amount of dropdowns selected, and hence x amount of rows to be added to the joining table?
I'm sorry if I am using incorrect or ambiguous terms to describe my problem. I have only started learning and creating websites recently.
Thank you.
UPDATE: I may have a solution per the updated codes below. However I am now stuck on the inserting into the database page. For some reason when I click submit, the next page loads but with no error and no rows inserted. Please help :)
View
<h1>Add New Contest</h1>
<?php
echo form_open('contests/add/');
?>
<?php
echo "<br />" "<br />"
echo "Contest Name";
$data = array( 'name' => 'contest_name',
'value' => set_value('contest_name'),
);
echo form_input($data);
echo "<br />" "<br />"
echo "Game";
?>
<div class="field_wrapper">
<div>
<select name="field_name[]">
<option value="" disabled selected>Select Game</option>
<?php
foreach($events_lists->result() as $row) {
$sports_events_id = $row->sports_events_id;
$sports_events_start_date = $row->sports_events_start_date;
$sports_events_start_time = $row->sports_events_start_time;
$home_team_shorthand = $row->home_team_shorthand;
$away_team_shorthand = $row->away_team_shorthand;
?>
<option value="<?php echo $sports_events_id; ?>"><?php echo $home_team_shorthand; ?> v <?php echo $away_team_shorthand; ?> - <?php echo $sports_events_start_date; ?> <?php echo $sports_events_start_time; ?></option>
<?php } ?>
</select>
<img src="add-icon.png"/>
</div>
</div>
<?php
echo "<br />" "<br />"
$data = array( 'value' => 'Add Contest',
'name' => 'submit',
'class' => 'submit-btn',
);
echo form_submit($data);
echo form_close();
?>
Controller
function add()
{
$league_id = $this->uri->segment(3);
$this->load->module('leagues');
$data['leagues_list'] = $this->leagues->get_where($league_id);
foreach ($data['leagues_list']->result() as $row) {
$league_insert_id = $row->id;
$this->load->module('sports_events');
$data['events_lists'] = $this->sports_events->get_events_list($league_insert_id);
}
if ($this->form_validation->run() == FALSE) {
$data['view_file'] = 'add_contest';
$this->load->module('template');
$this->template->cmslayout($data);
} else {
$data1 = array(
'contest_name' => $this->input->post('contest_name')
);
if(isset($_REQUEST['submit'])) {
$data2 = $_REQUEST['field_name'];
}
if ($this->_transactions_new_contest($data1, $data2)) {
return $query;
$this->session->set_flashdata('team_phase_created', 'The team phase has been set');
redirect('/contests/');
}
}
}
Model
function _transactions_new_contest($data1, $data2){
$this->db->trans_start();
$this->db->insert('contests', $data1);
$contest_id = $this->db->query('SELECT contests.id FROM contests ORDER BY contests.id DESC limit 1');
foreach ($contest_id->result() as $row) {
$contest_result_id = $row->id;
foreach($data2 as $value){
$this->db->query('INSERT INTO contests_has_sports_events (contests_id, sports_events_id) VALUES (' . $contest_result_id . ', ' . $value . ')');
} }
$this->db->trans_complete();
}

Related

Dependent dropdown for country and state in yii 1.13

I had created two database tables as,
Country
State
In Country table i had countryid & cname field. In State table i had state_id, state_name & country_cid.
Now i need to make a dependent dropdown i.e. when i select country dropdown box, then the particular state for that country should be displayed. I had done below coding. But it displays only country and not displays state.
This is in views/sample/register.php file.
<div class="row">
<?php echo $form->labelEx($model,'countryid');
$opts = CHtml::listData(Country::model()->findAll(),'countryid','cname');
echo $form->dropDownList($model,'country_id',$opts,
array(
'prompt'=>'Select Country',
'ajax' => array(
'type'=>'POST',
'url'=>CController::createUrl('SampleController/dynamicSubcategory'),
array('coountry_id'=>'js:this.value'),
'dataType' => 'JSON',
'success'=>'js:function(data)'
. '{'
. 'var html="<option value=>-----Select city-----</option>";'
. '$.each(data,function(i,obj)'
. '{'
. 'html+="<option value=\'"+obj.id+"\'>"+obj.name+"</option>";'
. '});'
. '$("#state_id").html(html);'
. '}'
)));
echo $form->error($model,'country_id'); ?>
</div>
<div class="row">
<?php echo $form->labelEx($model,'state_id');
echo CHtml::dropDownList('state_id','', array());
echo $form->error($model,'state_id'); ?>
</div>
This is in controllers/SampleController.php file
public function actiondynamicSubcategory()
{
$countryId=$_POST['coountry_id'];
$criteria=new CDbCriteria();
$criteria->select=array('state_id,state_name');
$criteria->condition='country_cid='.$countryId;
$criteria->order='state_name';
$cityAry= State::model()->findAll($criteria);
$ary=array();
foreach($cityAry as $i=>$obj)
{
$ary[$i]['state_id']=$obj->id;
$ary[$i]['state_name']=$obj->name;
}
echo json_encode($ary);
}
I had created Country model & State model. I had analyzed many sites. But i can't get right. Please anybody help.
You can try below code which I have used in one of my projects -
**Code in the view file :**
<?php $arrOptionsCountry = array('prompt'=>'Select Country','ajax'=> array('type'=>'POST','url' => ApplicationConfig::getURL('', 'site','SearchValueStates'),'update'=>'#Airports_state','beforeSend'=>'stateLoading'));
echo $form->dropDownList($model,'country',$arrCountryList,$arrOptionsCountry,array('class'=>'span11 customDropdown1_select'));
$arrOptionsState = array('prompt'=>'Select State','ajax'=> array('type'=>'POST','url' => ApplicationConfig::getURL('', 'site', 'SearchValueCities'),'update'=>'#station_name','beforeSend'=>'cityLoading'));
echo $form->dropDownList($model,'state',$arrStatesList,$arrOptionsState);
echo $form->dropDownList($model,'city',$arrCityList);
?>
**Site Controller :**
public function actionSearchValueStates()
{
$arrParam =array();
if(isset($_POST['Airports']['country']))
{
$Term = $_POST['Airports']['country'] ;
$case = "STATE-LIST";
$prompt = "Select State";
$arrParam['id'] = isset($Term)?$Term:null ;
$data = States::getList($case,$arrParam);
$data = CHtml::listData($data,'id','name');
echo CHtml::tag('option',array('value'=>''),$prompt,true);
foreach($data as $value=>$name)
{
echo CHtml::tag('option',array('value'=>$value),CHtml::encode($name));
}
}
Yii::app()->end();
}
I got the answer for my question.
This is in views/sample/register.php file
<div class="row">
<?php echo $form->labelEx($model,'country');
$opts = CHtml::listData(Country::model()->findAll(),'countryid','cname');
echo $form->dropDownList($model,'country_id',$opts,
array(
'prompt'=>'Select Country',
'ajax' => array(
'type'=>'POST',
'url'=>CController::createUrl('Sample/dynamicSubcategory'),
'update'=>'#state_name',
'data'=>array('country_id'=>'js:this.value'),
)));
echo $form->error($model,'country_id'); ?>
</div>
<div class="row">
<?php echo $form->labelEx($model,'state_id');
echo CHtml::dropDownList('state_name','', array(), array(
'prompt'=>'Select State'
));
echo $form->error($model,'state_id'); ?>
</div>
This is in controllers/SampleController.php file
public function actiondynamicSubcategory()
{
$data=State::model()->findAll('country_cid=:countryid',
array(':countryid'=>(int) $_POST['country_id']));
$data=CHtml::listData($data,'state_id','state_name');
echo "<option value=''>Select State</option>";
foreach($data as $value=>$state_name)
echo CHtml::tag('option', array('value'=>$value),CHtml::encode($state_name),true);
}
This code works fine for dependent dropdown in yii

Instead of submit get data on entering page php

I use this snippet to get vehicle data from a external database:
<form method="post" action="<?php echo $_SERVER['REQUEST_URI'] ?>" class="vwe-kenteken-widget">
<p><input type="text" name="open_data_rdw_kenteken" value="<?php echo $_POST['open_data_rdw_kenteken'] ?>" maxlength="8"></p>
<p><input name="submit" type="submit" id="submit" value="<?php _e('Kenteken opzoeken', 'open_data_rdw') ?>"></p>
</form>
<?php if($data): ?>
<h3><?php _e('Voertuiggegevens', 'open_data_rdw') ?></h3>
<table>
<?php
$categories = array();
foreach ($data as $d) {
if( !is_array($fields) || in_array($d['name'], $fields) ) {
if( !in_array($d['category'], $categories) ) {
$categories[] = $d['category'];
echo '<tr class="open-rdw-header">';
echo '<td colspan="2" style="font-weight: bold;">';
echo ''.$d['category'].'';
echo '</td>';
echo '</tr>';
}
echo '<tr style="display:none">';
echo '<td>'.$d['label'].'</td>';
echo '<td>'.$d['value'].'</td>';
echo '</tr>';
}
}
?>
</table>
<?php endif; ?>
What i want to accomplish is that the data is loaded without the user have to enter a value and hit the submit button. The input value will get loaded based on the product page the user is viewing.
EDIT:
The data is loaded based on:
public function get_json() {
if ( isset( $_POST['kenteken'] ) ) {
$data = $this->rdw->get_formatted($_POST['kenteken']);
foreach ($data as $row) {
$json['result'][$row['name']] = $row['value'];
}
if ($_POST['kenteken']) {
if ($data[0]['value'] == '') {
$json['errors'] = __( 'No license plates found', 'open_data_rdw' );
}
else {
$json['errors'] = false;
}
}
else {
$json['errors'] = __( 'No license plate entered', 'open_data_rdw' );
}
header('Content-type: application/json');
echo json_encode($json);
die();
}
}
So instead of a $_POST action just get the data based on a pre-declared value that is different on each page.
Hard to answer - but I'll try to use my crystal ball.
$data comes from a database query, right?
I assume further, that the query takes the value from the open_data_rdw_kenteken form field to gather $data.
To have the table rendered, you have to fill $data with the right data. That implies that you must have a kind of default value for open_data_rdw_kenteken to get the data out of the DB. The default can be "all" which should reflect on your SQL Query or as your wrote "defined by product page".
Pseudo Code
$data = getData('BT-VP-41');
function getData($open_data_rdw_kenteken="")
{
$where = "";
if(!empty($open_data_rdw_kenteken)) {
$where = 'WHERE rdw_kenteken = "'.mysqli_real_escape_string($open_data_rdw_kenteken)';
}
$data = myslqi->query("SELECT * FROM dbTbl ".$where)
return $data;
}
As I wrote - this is pseudo code and will not run out of the box. You'll have to adapt that to your environment.
TL;DR: The line
<?php if($data): ?>
keeps you from rendering the table. To render you need $data filled with the right data.
Hope that will get you in the right direction.

I want to display data when i click on customer name form dropdown list it will display only that customer's message and entry date

My code works properly, it is display all messages and entry dates. I want to display message and entry date when I select customer name from drop down list.
Here is my controller CustomerlifecycleController
public function actionCustomerlifecycleanalytic() {
$customername = Customer::model()->findall("store_id='" . Yii::app()->session['store_id'] . "'");
$customerlifecycle = CustomerLifecycle::model()->findAll();
$this->renderPartial('customerlifecycleanalytic', array( 'customername' => $customername, 'customerlifecycle' => $customerlifecycle), false, true);
}
Here is my view file customerlifecycleanalytic.php
Dropdown list
Customer name fetch from customer model
and there is column name id for customer
<select class="form-control selectpicker customerfilter">
<option value=''>Select Customer</option>
<?php
if (isset($customername)) {
foreach ($customername as $customernames) {
echo '<option value="' . $customernames['id'] . '" >' . $customernames['firstname'].'&nbsp'. $customernames['lastname']. '</option>';
}
}
?>
</select>
Message and entry date fetch from Customerlifecycle model there is column name is customer_id
$msg = '<li {classstr}>
<div class="tl-circ">
</div>
<div class="timeline-panel">
<div class="tl-body">
<p>
{msg} on {date}</p>
</div>
</div>
</li>';
$cnt = 0;
$htmlstring = '';
foreach ($customerlifecycle as $key => $row) {
$htmlstring .= $msg;
$classString ='';
if ($cnt % 2 == 0) {
$classString = " class='timeline-inverted' ";
}
$htmlstring = str_replace("{classstr}", "$classString", $htmlstring);
$htmlstring = str_replace("{msg}", "$row->message", $htmlstring);
$htmlstring = str_replace("{date}", "$row->entrydate", $htmlstring);
$cnt++;
}
echo $htmlstring;
?>
My code is already running completely but I want when I select customer name form drop down list at that time it will display only that customer's message and entry date.
To display selected customer's message and entry date, you can call javascript function on onChange event of dropdownlist as follows:
<script type="text/javascript" language="Javascript">
function submitForm() {
document.getElementById("form-id").submit();
}
</script>
<select class="form-control selectpicker customerfilter" name="Customer[id]" onChange="js: return submitForm();">
Using this javascript function, submit the form and make changes in your controller as follows:
public function actionCustomerlifecycleanalytic() {
$customername = Customer::model()->findall("store_id='" . Yii::app()->session['store_id'] . "'");
$strCondition = "";
if(isset($_POST['Customer'])) {
if (!empty($_POST['Customer']['id'])) {
$strCondition .= "store_id = '" . Yii::app()->session['store_id'] . "' AND customer_id = '" . $_POST['Customer']['id'] . "'";
}
}
$customerlifecycle = CustomerLifecycle::model()->findAll($strCondition);
$this->renderPartial('customerlifecycleanalytic', array( 'customername' => $customername, 'customerlifecycle' => $customerlifecycle), false, true);
}
Hope this helps!

How to transfer values from controller to view codeigniter into dropdownlist

if (is_array($jsonhome)) {
foreach ($jsonhome as $query) {
foreach ($query['results']['place'] as $places) {
if (is_array($places['country'])) {
echo "Country:\n";
echo "Content: " . $places['country']['content'] . "\n\n";
}
if (is_array($places['admin1'])) {
echo "State:\n";
echo "Content: " . $places['admin1']['content'] . "\n\n";
}
if (is_array($places['admin2'])) {
echo "District/City:\n";
echo "Content: " . $places['admin2']['content'] . "\n\n";
}
}
}
}
$location_data['user_current_country'] = $places['country']['content'];
$this->load->view('header_register', $location_data);
$this->load->view('body_complete_register', $location_data);
$this->load->view('footer_register');
I want to transfer the country to input type select in the view codeigniter:
Say i get two countries via the above queries say India and USA i want to pass it to the view
Do i do it via ajax?
Please help i am new to codeigniter
So in your controller you want to build an array of options for your <select>. Something like
//Build array of country options
$aData['countryOptions'] = array();
foreach ($jsonhome as $query) {
foreach ($query['results']['place'] as $places) {
$aData['countryOptions'][] = $places['country']['content'];
}
}
$this->load->view('body_complete_register', $aData);
Then in your view you can use the form_dropdown function providing you have the form helper loaded
$this->load->helper('form');
echo form_dropdown('countries', $countryOptions);
Of if you don't want to use the helper you could do
<select name="countries" id="countries">
<?php foreach($countryOptions as $key => $countryName) { ?>
<option value="<?php echo $key ?>"><?php echo $countryName ?></option>
<?php } ?>
</select>
Please note that you shouldn't be echoing anything out in the controller. The controller is just for passing data between models, libraries and views

Reduce number of MySQL Queries and Output Data in Correct Order

My main issue is the number of times I query the database (see below). Also, I would like to check that the current product (optionsToProducts.productID) has options for the current optionName before outputting the select statement! See the final image below to see the blank select box...
I have 8 tables in total, but the 3 that matter are:
optionNames
http://www.grabb.co.uk/stack/001.png
productOptions
http://www.grabb.co.uk/stack/002.png
optionsToProducts
http://www.grabb.co.uk/stack/003.png
<?php
$i=0;
$optionsquery = "SELECT * FROM optionNames WHERE categoryID = ".$categoryID."";
$optionsresult= mysql_query($optionsquery) or die(mysql_error());
while ($optionnames = mysql_fetch_array($optionsresult)) {
$i++;
$optionname = $optionnames["optionName"];
$optionID = $optionnames["optionNamesID"];
//echo $optionname."<br />";
?>
<label for="option<?php echo $i; ?>"><?php echo $optionname; ?></label>
<select name="option<?php echo $i; ?>" id="<?php echo $i; ?>">
<?php
//$optionvalues = "SELECT * FROM (optionsToProducts,productOptions) WHERE optionsToProducts.productID = ".$productID." AND productOptions.optionNamesID = ".$optionID."";
//echo $optionvalues."<br /><br />";
$optionvalues = "SELECT * FROM optionsToProducts WHERE productID = ".$productID."";
$valuesresult= mysql_query($optionvalues) or die(mysql_error());
while ($optionvals = mysql_fetch_array($valuesresult)) {
$valueName = $optionvals["optionValue"];
$valueID = $optionvals["productOptionsID"];
//echo $valueName."<br />";
$optionfinal = "SELECT * FROM productOptions WHERE productOptionsID = ".$valueID." AND optionNamesID = ".$optionID."";
$finalresult= mysql_query($optionfinal) or die(mysql_error());
while ($optionlast = mysql_fetch_array($finalresult)) {
$optionValueName = $optionlast["optionValue"];
$optionValueID = $optionlast["productOptionsID"];
$num_rows = mysql_num_rows($finalresult);
?>
<option value="<?php echo $optionValueID; ?>"><?php echo $optionValueName; ?></option>
<?php
}
}
echo "</select>";
}
?>
final Output:
http://www.grabb.co.uk/stack/004.png
As always, your help is appreciated. Thank you.
Since you tagged this question with the join tag, you probably know you need to write a join query to get what you need.
<?php
$i=0;
$query = "SELECT options.optionName, options.optionNamesID, po.optionValue, po.productOptionsID
FROM optionNames AS options
INNER JOIN productOptions AS po ON po.optionNamesID=options.optionNamesID
INNER JOIN optionsToProducts AS otp ON otp.productOptionsID=po.productOptionsID
WHERE otp.productID=" . (int) $productID
. " AND options.categoryID=" . (int) $categoryID;
$result = mysql_query($query);
if($result) {
$rows = array();
while($row = mysql_fetch_assoc($result) ) {
$rows[] = $row;
}
$i = 0;
$optionId = null;
foreach($rows as $row) {
if($optionId != $row['optionNamesID']) {
$optionId = $row['optionNamesID'];
?>
<label for="option<?php echo $optionId; ?>"><?php echo $row['optionName']; ?></label>
<select name="option<?php echo $optionId; ?>" id="<?php echo $optionId; ?>">
<?php } ?>
<option value="<?php echo $row['productOptionsID']; ?>"><?php echo $row['optionValue']; ?></option>
<?php
//Close select element when the optionNamesID changes or on the last row
if( (isset($rows[$i + 1]) && $rows[$i + 1]['optionNamesID'] != $optionId) ||
!isset($rows[$i + 1]) ) { ?>
</select>
<?php }
$i++;
}
} else {
//Debug query, remove in production
echo mysql_error();
}
?>
I also made some small changes - I use the optionNamesID in the select and label tag names - I don't know how you knew previously which select belonged to which option.
I also assumed that categoryID and productID came from somewhere, since it's not specified in the code.
Pushing all the rows to an array at the beginning is optional, but it makes the code a bit more organized (since you can check ahead in the array to see where to close the select tags).
NOTICE - this code is untested so there could some minor typos. Please make the needed corrections if necessary.

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