I'm at my first day with Yii
I've just learned how to filter columns
$dataProvider=new CActiveDataProvider('User',
array(
'criteria' => array(
'select' => 'id, username, realname, email, companyId, languageId, registrationDate',
'order' => 'username ASC',
)
)
);
$this->render('index',array(
'dataProvider'=>$dataProvider,
));
Now I've a problem: the 'registrationdDate' is a MySql datestamp.
So i need to show localized... how can I do this ?
the code of the view is a list. I'm listing using this:
<td><?php echo CHtml::encode($data->registrationDate); ?></td>
What can be done to detect user locale? And how to format a "2012-10-13 13:33:45", for example, into the correct localized string?
For italian it will be 13/10/2012 or 13-10-2012 with appended 13:33:45
The simplest way would be:
<td>
<?php echo Yii::app()->locale->dateFormatter
->formatDateTime($data->registrationDate); ?>
</td>
Related
I want to retrive custom post and want to sort it by title. However when i made a dump of what exact SQL request is sent, i found out the orderby is using menu_order instead of title
Here's the code:
$args=array(
'post_type' => 'custom_post',
'orderby' => 'title',
'order' => 'ASC',
'post_status' => 'publish',
'posts_per_page' => '-1',
);
Heres the dump
"SELECT wp_posts.* FROM wp_posts WHERE 1=1 AND wp_posts.post_type = 'custom_post' AND ((wp_posts.post_status = 'publish')) ORDER BY wp_posts.menu_order ASC "
Hence when i retrive the custom posts, its not in the order i want it to be.
Your help is appreciated
You can create new WP_Query instance to achieve the exact expected result.
<?php
$args = array(
'post_type' => 'custom_post_type',
'order' => 'ASC',
'orderby' => 'title',
);
$my_query = new WP_Query( $args );
if( $my_query->have_posts() ) :
while ( $my_query->have_posts() ) : $my_query->the_post(); ?>
<div class="smal-sec">
<h3><?php the_title();?></h3>
<?php the_content();?>
</div>
<?php
endwhile;
endif;
wp_reset_query(); // Restore global post data stomped by the_post().
?>
This query will help you to display all posts order by title (A -> Z). Please make sure no such plugins are there, which is actually overwriting the WP_Query instance.
I want to display the name of Shift instead of shift_id
I have a dropboxlist from other table that is like this
<div>
<?php echo $form->labelEx($model,'mon'); ?>
<?php echo $form->dropDownList($model, 'mon', CHtml::listData(
Shift::model()->findAll(), 'shft_id', 'name'),
array('prompt' => 'Select a Department')
); ?>
<?php echo $form->error($model,'mon'); ?>
I have two tables that is
Day: id_day,mon,tues,wed,etc
Shift: shft_id,start,end,name,status
Here is the relation in the day
'shift'=>array(self::HAS_MANY,'Shift','shft_id'),
For Shift:
'day'=>array(self::BELONGS_TO,'Day','id_day'),
It is already working. The choices in the dropbox was the name of the Shift, and puts the shft_id in the mon,tues,wed,etc. In the view of the form it looks like
id_user: 3
mon:5
tues:5
wed:6
what I wanted to be is that in the view.
id_user: 3
mon: 6am-5pm
tues: 7am-6pm
etc.etc.
I dont know what command it is. I have no idea. Help me please
Instead of User::getusername() method.
I think you can simply use this in one line.
// format models resulting using listData
<?php echo $form->dropdownlist($model,'user_id', CHtml::listData(User::model()->findAll(),'id', 'name')); ?>
below is an example which I think may solved your problem
in _form.php
<?php echo $form->dropdownlist($model,'user_id', User::getusername()); ?>
<?php echo $form->error($model,'user_id'); ?></th>
in User model
public function getusername()
{
$criteria2=new CDbCriteria;
$criteria2->select='*';
$quser=User::model()->findAll($criteria2);
foreach($quser as $r)
{
$user_id= $r->user_id;
$user_name=$r->user_name;
$user_array[$user_id]=$user_name;
}
return $user_array;
}
I'm trying to understand what you are really asking. I believe you were saying that you are already able to populate your records correctly from the dropdowns. But when you are displaying, the view is only showing the shift_id value? The relation for the day should be able to get your shift name. How are you displaying the list:
mon: 5
tue: 5
wed: 6
I'm trying to understand if you need help displaying the dropdown items properly, or how to display the shift name later after the values have been set in the database. Also, your 'Day' table has me confused. What are the field names? Are they 'day_id' and 'day'? Or do you actually name the fields after each day of the week?
If you are displaying the data through a foreach and are getting each day (let's assume it is $day) object, then the relationship to 'shift' can give you the name like this: echo $day->shift->name; if the name displays as '#am-#pm'. Otherwise you can display it like this:
echo sprintf('%d a.m. - %d p.m.',$day->shift->start,$day->shift->end);
Thanks for the people who helped me. It did lead me to the answer anyone who has the same problem here is what i did.
In my model/Day
public function relations()
{
// NOTE: you may need to adjust the relation name and the related
// class name for the relations automatically generated below.
return array(
'monday'=>array(self::BELONGS_TO,'Shift','mon'),
'tuesday'=>array(self::BELONGS_TO,'Shift','tue'),
'wednesday'=>array(self::BELONGS_TO,'Shift','wed'),
'thursday'=>array(self::BELONGS_TO,'Shift','thurs'),
'friday'=>array(self::BELONGS_TO,'Shift','fri'),
'saturday'=>array(self::BELONGS_TO,'Shift','sat'),
'sunday'=>array(self::BELONGS_TO,'Shift','sun'),
);
}
and In my models/Shift
public function relations()
{
// NOTE: you may need to adjust the relation name and the related
// class name for the relations automatically generated below.
return array(
'day_mon' =>array(self::HAS_MANY,'Day','mon'),
'day_tue' =>array(self::HAS_MANY,'Day','tue'),
'day_wed' =>array(self::HAS_MANY,'Day','wed'),
'day_thurs'=> array(self::HAS_MANY,'Day','thurs'),
'day_fri'=>array(self::HAS_MANY,'Day','fri'),
'day_sat'=>array(self::HAS_MANY,'Day','sat'),
'day_sun'=>array(self::HAS_MANY,'Day','sun'),
);
}
The problem is with my relations. its not set properly, so you need to set it properly then go to Day/view
<?php $this->widget('bootstrap.widgets.TbDetailView',array(
'data'=>$model,
'attributes'=>array(
'id_day',
array(
'name'=>'mon',
'value'=>CHtml::encode($model->monday->name)
),
Here is the output
If you have a detail veiw then use the following code:
widget('zii.widgets.CDetailView', array(
'data'=>$model,
'attributes'=>array(
'id',
'name',
array(
'label' => $model->shift->getAttributeLabel('shift_name'),
'value' => $model->shift->shift_name
),)); ?>
In case anybody else will be in my boat, what you will need to do is edit view.php in view/modelname/view.php
widget('zii.widgets.CDetailView', array(
'data'=>$model,
'attributes'=>array(
'id',
'name',
'country_id')); ?>
into
widget('zii.widgets.CDetailView', array(
'data'=>$model,
'attributes'=>array(
'id',
'name',
array(
'label' => $model->country->getAttributeLabel('country'),
'value' => $model->country->name
),)); ?>
I want to display country name instead of country id in user profile view( country id is FK in user profile tbl) .
anyone can help me with this ?
This is my view controller/action in which I have Country model as well:
public function actionView($id)
{
$model = new TblUserProfile;
$model = TblUserProfile::model()->find('user_id=:user_id', array(':user_id' => $id));
$countrymodel = new Country;
$countrymodel = Country::model()->findAll();
// var_dump($countrymodel->name);
// die();
$this->render('view', array(
'model' => $model ,
'country' =>$countrymodel
));
}
This is my view
<div class="view">
<b><?php echo CHtml::encode($data->getAttributeLabel('id')); ?>:</b>
<?php echo CHtml::link(CHtml::encode($data->id), array('view', 'id'=>$data->id)); ?>
<br />
<b><?php echo CHtml::encode($data->getAttributeLabel('user_id')); ?>:</b>
<?php echo CHtml::encode($data->user_id); ?>
<br />
<b><?php echo CHtml::encode($data->getAttributeLabel('user_occuption')); ?>:</b>
<?php echo CHtml::encode($data->user_occuption); ?>
<br />
<b><?php
// $model = TblUserProfile::model()->find('country_id=:country_id', array(':user_id' => $id));
//echo CHtml::encode($model->getAttributeLabel('country')); ?>:</b>
<?php// echo CHtml::encode($model->name);
?>
<br />
</div>
Now I want to display country name in above view.
These are the relationships of country and user profile table
public function relations() {
// NOTE: you may need to adjust the relation name and the related
// class name for the relations automatically generated below.
return array(
'user' => array(self::BELONGS_TO, 'TblUser', 'user_id'),
'country' => array(self::BELONGS_TO, 'country', 'country_id'),
'state' => array(self::BELONGS_TO, 'state', 'state_id'),
'city' => array(self::BELONGS_TO, 'city', 'city_id')
);
}
Use the following code:
<b><?php echo CHtml::encode($data->country->getAttributeLabel('name')); ?>:</b>
<?php echo CHtml::encode($data->country->name); ?>
<br />
Provided that in your country model, the country name column is name.
Since the relation is in place we can use it to access the related country for each user.
In detailview it'll change to:
// instead of 'country_id'
array(
'name'=>'country_id',
'value'=>$model->country->name
)
Read the CDetailView doc to see how you can change even more things.
I would use:
$model = TblUserProfile::model()->with('country')->find('user_id=:user_id', array(':user_id' => $id));
with('country') explained: the string country comes from the array relations key.
Then to display in the view you would use:
$model->country->CountryName;
$model contains
country contains a "model" of country joined by the relation foreign key
CountryName is the column name in the Country table
This is the recommended way to do it in the MVC way. Both my example and the one above work, but the one above has more process/logic decisions in the view, which breaks the concept of MVC where the Model should be the fat one and contain all logic/processing possible, then controller will pull this data and provide it to the view.
You can also follow this below answer:
Just put the below code in your view.php file
[
"attribute"=>"User_Country_Id",
'value' =>$model->country->conName ,
],
[
"attribute"=>"User_State_Id",
'value' =>$model->states->stsName ,
],
[
"attribute"=>"User_City_Id",
'value' =>$model->cities->ctName ,
],
I have a multiple select-list-box for Staff in Create-Service-Form, used to select multiple staff when creating a new service. for this i can assign multiple staff on a single service.
I saved staff_id field as:
$model->staff_id = serialize($model->staff_id);
Here the update-view code for multiple-select-list-box:
<div class="row">
<?php echo $form->labelEx($model,'staff_id'); ?>
<?php
$data = array('1' => 'Sam', '2' => 'john', '3' => 'addy');
$htmlOptions = array('size' => '5', 'prompt'=>'Use CTRL to Select Multiple Staff', 'multiple' => 'multiple');
echo $form->ListBox($model,'staff_id', $data, $htmlOptions);
?>
<?php echo $form->error($model,'staff_id'); ?>
</div>
Problem is, when i load form for updating a service. how do i select those staff, which are previously saved in database?
I tried this dropDownList-attributes, but it not working.
$select | string | the selected value
if someone has solution, then suggest me. Thanks All Mates...
Here's a quick code I wrote for you, its an example that will help you understand how it works.
<div class="row">
<?php echo $form->labelEx($model,'staff_id'); ?>
<?php
$data = array('101' => 'Faraz Khan', '102' => 'Depesh Saini', '103' => 'Nalin Gehlot', '104' => 'Hari Maliya');
$selected = array(
'102' => array('selected' => 'selected'),
'103' => array('selected' => 'selected'),
);
$htmlOptions = array('size' => '5', 'prompt'=>'Use CTRL to Select Multiple Staff', 'multiple' => 'true', 'options' => $selected);
echo $form->listBox($model,'staff_id', $data, $htmlOptions);
?>
<?php echo $form->error($model,'staff_id'); ?>
</div>
Have Fun Ya !!!
So I have the User Id of someone in Buddypress.
What is the function to print out their Avatar?
What is the function to link to their profile?
This will display the user avatar by user ID and also make the avatar a clickable link to their profile.
<?php $member_id = bp_core_get_userid( '1' ) ?>
<a href="<?php echo bp_core_get_user_domain( $member_id ) ?>"
title="<?php echo bp_core_get_user_displayname( $member ) ?>">
<?php echo bp_core_fetch_avatar ( array( 'item_id' => $member_id, 'type' => 'full' ) ) ?></a>
Obviously replace the number in the first line with the userID that you want.
Simply you can use bp_activity_avatar() function:
<?php bp_activity_avatar(array('user_id' => $user_id)); ?>
Output the avatar of the user for specified User ID
bp_activity_avatar( 'user_id=' . $user_id );
Returns a HTML formatted link for a user with the user's full name as the link text for specified User ID
echo bp_core_get_userlink( $user_id );
I've managed do it with code like this (without php warnings):
$imgTagAva = apply_filters( 'bp_group_request_user_avatar_thumb', bp_core_fetch_avatar( array( 'item_id' => $user_id, 'type' => 'thumb', 'alt' => sprintf( __( 'Profile picture of %s', 'buddypress' ), bp_core_get_user_displayname( $user_id ) ) ) ) );
found it. The best solution is below:
get_avatar(get_the_author_meta('ID'), 40);
// OR
get_avatar($author_id_or_email, $size);