Laravel Select Query within helper file - mysql

I'm coming from codeigniter background. Unlike codeigniter helper directory, i just created helper directory within app directory of Laravel. Just want to know how to execute query within this common function. Here is my codeigniter function.
function show_menu($primary_key_col, $parent_id, $sort_order)
{
$output = "";
$ci =& get_instance();
$ci->db->select("*");
$ci->db->where('is_active', "Y");
$ci->db->where('is_delete', "N");
$ci->db->where('parent_id', $parent_id);
($sort_order!="")?$ci->db->order_by($sort_order, "ASC"):"";
$query = $ci->db->get('tbl_cms_menus');
foreach ($query->result() as $row){
$output .= '<option value="'.$row->$primary_key_col.'">'.$indent.$row->menu_name.'</option>';
}
return $output;
}
I tried something like this in laravel file. but this code did't give me any result. Please tell me where i'm doing wrong in this code. thanks
function databaseTable()
{
$table = DB::table('tbl_cms_menus');
$get_rows = $table->get();
$count_rows = $table->count();
if($count_rows > 0){
foreach ($get_rows as $tbl)
{
echo $tbl->menu_name;
}
}
}

This code will rot so hard that it shipped pre-rotten.
But, if you want to just.. ram it into the app all dry like that.. then add something like this to your base controller class...
$whatever = crazyChainingStuff;
foreach ($whatever ...) { $topMenu .= ... }
View::share('topMenu', $topMenu);
If you want to learn how to write code that will do less damage to your company and your clients then I recommend starting by watching Uncle Bob's "Fundamentals" videos. At least the first 5-6. http://cleancoders.com

It looks like you are trying to generate a drop-down/select with some data from your database, in this case, you should pass the data required for the drop-down/select from your controller to the view where you have written your HTML, for example, in your view, you may have a select like this:
echo Form::select('cms_menu', $cms_menu, Input::old('cms_menu'));
Or this (If you are using Blade):
{{ Form::select('cms_menu', $cms_menu, Input::old('cms_menu')) }}
From your controller you should pass the $cms_menu which should contain the menu-items as an arrtay and to populate that array you may try something like this:
$menuItems = DB::table('tbl_cms_menus')->lists('menu_name','id');
return View::make('your_view_name', array('cms_menu' => $menuItems));
Also, you may use something like this:
// Assumed you have a Page model
$menuItems = Page::lists('menu_name', 'id');
return View::make('your_view_name', array('cms_menu' => $menuItems));
You may also read this article which is about building a menu from database using view composer (More Laravelish way). Read more about Form::select on documentation.

It was too late to give an answer. I was also from CodeIgniter background and when I learnt Laravel then first I try to find how can I write a query in Helper. My Team leader helped me.
I have converted your code in a helper function.
function show_menu($primary_key_col, $parent_id, $sort_order)
{
$query = DB::table('tbl_cms_menus')
->select('*')
->where('is_active', '=', 'Y')
->where('is_delete', '=', 'N')
->where('parent_id', '=', $parent_id);
($sort_order != "")? $query->orderBy($sort_order, "ASC") : "";
$resultData = $query->get()->toArray();
}
Here $resultData will be array format. Now, you can create a foreach loop according to your requirement.

Related

Best way to query a model with search and pass its id into another model

Can I get some help? I am trying to query my model database as a search. When the results are fetched, I want to pass the id of the collection through a form into another model and I don't want to use a package.
So, far, I have been getting errors.
Basically, I am querying the database with
$services = Service::where('name' 'LIKE' Request::('input')):
But if I try to pass to pass $servces to view with
return view('search.result')->with($services);
It returns **Illegal Offset Type. **
So, instead I have to chain the $services in details
Like so:
return view('search.results')->withDetails($services);
On my results.blade.php, I have:
#foreach($details as $d)
{{$d->name}}
<form action="{{route('book', ['id' =>$d->id} method="POST">
<button type="submit">Book</button>
</form>
#endforeach
This displays the services collections
But, if I try to pass to pass the Id of $d collection into the Book model like this:
public function book($id) {
$service = Service::find($id);
//or
$service = Service::where('id', $id) ;
App\Models\Book::insert([
'service_I'd => $service->id]) ;
It tells me that id does not exist on this
Collection instance.
Can someone help me with a better way to achieve this?
in your line:
$services = Service::where('name' 'LIKE' Request::('input')):
it should be:
$services = Service::where('name', 'LIKE', Request::('input'))->get();
or if you want to get similar names:
$services = Service::where('name', 'LIKE','%'. Request::('input').'%')->get();
and in this line:
$service = Service::where('id', $id) ;
you can use:
$service = Service::firstWhere('id', $id) ;
So, to pass the id of the returned collection to the new model, I had to use
///Then,
Book::insert([
'service_Id => $service->id]);
Thanks a lot everyone.

mySQL query for building dynamically populated model

I have some code below which demonstrates a hard-coded example of what I would like to accomplish dynamically.
At a high level, I wish to do something like select * from view_data_$app_state and then get all of the data from that views table into my mustache templates dynamically.
The code I currently must use to group multiple rows of data for a specific column along with the views data is:
<?php
error_reporting(E_ALL);
class Example {
function __construct(){
try {
$this->db = new PDO('mysql:host=localhost;dbname=Example', 'root','drowssap');
}
catch (PDOException $e) {
print($e->getMessage());
die();
}
}
function __destruct(){
$this->db = null;
}
function string_to_array($links_string){
return explode(",", $links_string);
}
function get_view_data(){
$q = $this->db->prepare('select *, GROUP_CONCAT(`links`) as "links" from `view_data_global` ');
$q->execute();
$result = $q->fetchAll(PDO::FETCH_ASSOC);
return $result;
}
}
$Example = new Example();
$result = $Example->get_view_data();
$result[0]["links"] = $Example->string_to_array($result[0]["links"]);
echo json_encode($result);
This gives me the perfect object while
GROUP_CONCAT seems to be doing the trick this way, however I MUST know the column name that will contain multiple rows before writing the query. I am trying to figure out an approach for this and wish to make a custom query + code example that will transform cols with multiple rows of null null and not empty data into an array like above - but return the data.. again like the code above.
Below is an output of the actual data:
[{"id":"1","title":"This is the title test","links":["main","about","store"]}];
How can I replicate this process dynamically on each view table?
Thank you so much SO!
You can use PDOStatement::fetch to retrieve your results, with fetch_style set to PDO::FETCH_ASSOC (some other values will also provide the same information). In this case, the result set will be array indexed by column name. You can access this information with foreach (array_expression as $key => $value)
See the documentation for additional information.

Retrieve every model and return three values

I am looking for a way to retrieve all models in a database. Then loop through all of the models and read out the values for name, firstname and phonenumber.
So far I've gotten this and failed to go past it:
$searchModel = new EmployeeSearch();
$dataProvider = $searchModel->search(Yii::$app->request->queryParams);
I am then looking to implement those three values in a simple HTML table:
<tr><td>$firstname</td><td>$name</td><td>$phone</td></tr>
The table should be part of a PDF output, so ideally I would save it to a variable:
$html_table = '<tr><td>$firstname</td><td>$name</td><td>$phone</td></tr>';
I would need to get this for every model that fulfills the criteria of status = 'active' in the database.
So far I've only been able to get tables via gridView and not in a HTML template either.
You don't really need a data provider to achieve this, you could simply try :
$models = Employee::find()->where(['status'=>'active'])->orderBy('name ASC')->all();
foreach ($models as $model) {
echo "<tr><td>{$model->firstname}</td><td>{$model->name}</td><td>{$model->phone}</td></tr>";
}
Read more : http://www.yiiframework.com/doc-2.0/guide-db-active-record.html#querying-data
You can get all models like this:
$employees = Employee::find()
->select('firstname, name, phone')
->asArray()
->where(['status'=>'active'])
->all();
This way you will get an array of arrays containing the 3 selected fields, so now you only need to use a foreach to loop through them and create the table:
$html = '<table>';
foreach($employees as $employee) {
$html .= '<tr><td>'.$employee['firstname'].'</td><td>'.$employee['name'].'</td><td>'.$employee['phone'].'</td></tr>';
}
$html .= '</table>'

joomla select list from api

How do I generate a dropdown list with custom item in joomla 3.1. I took a look on couple of examples but I don not get the thing to work. I have been trying the following but the list is not genarated the html works.
public function getInput() {
$jinput = JFactory::getApplication()->input;
$sub_id = $jinput->get('sub_id');
$db = JFactory::getDbo();
$query = $db->getQuery(true)
->select('*')
->from('#__unis_faculties')
->order('faculty_name');
$db->setQuery($query);
$rows = $db->loadObjectList();
//array_unshift($rows, JHtml::_('select.option', '', JText::_('COM_UNIS_FACULTIES'), 'value', 'text'));
return JHTML::_('select.genericlist',$rows,'faculties',array('class'=>'nourritures','option.attr'=>'data'));
}
Your code actually does not look to have problems.
As long as the query returns something you are on the right track.
Change select('*') to select('COL_A as value, COL_B as text').
Make sure you echo the result of the method getInput (not a great name btw, how about getFacultiesDropdown()

How to see MySQL statements and error (if any) in CakePHP shell

I am using CakePHP 1.3 and writing custom shells to run mundane tasks in cronjobs. I am seeing failed Model->save() from time to time but I don't know anyway to find out what the exact problem is.
Is there a way to display the actual SQL statements executed and warning/error returned by MySQL in a CakePHP shell?
Thanks.
You can use the following SQL dump task for shells.
http://bakery.cakephp.org/articles/carcus88/2011/04/08/sql_dump_task_for_shells
One way to do this would be to watch the MySQL log file in a separate terminal.
A couple ways of doing this are listed here:
MySQL Query Logging in CakePHP
I found a way to do it. In your shell, add:
function initialize()
{
Configure::write('debug', 2);
$this->_loadDbConfig();
$this->_loadModels();
}
Then whenever you like to see the log, call this function:
function dump_sql()
{
$sql_dump = '';
if (!class_exists('ConnectionManager') || Configure::read('debug') < 2)
return false;
$noLogs = !isset($logs);
if ($noLogs)
{
$sources = ConnectionManager::sourceList();
$logs = array();
foreach ($sources as $source):
$db =& ConnectionManager::getDataSource($source);
if (!$db->isInterfaceSupported('getLog')):
continue;
endif;
$logs[$source] = $db->getLog();
endforeach;
}
if ($noLogs || isset($_forced_from_dbo_))
{
foreach ($logs as $source => $logInfo)
{
$text = $logInfo['count'] > 1 ? 'queries' : 'query';
$sql_dump .= "cakeSqlLog_" . preg_replace('/[^A-Za-z0-9_]/', '_', uniqid(time(), true));
$sql_dump .= '('.$source.') '. $logInfo['count'] .' '.$text. ' took '.$logInfo['time'].' ms';
$sql_dump .= 'Nr Query Error Affected Num. rows Took (ms)';
foreach ($logInfo['log'] as $k => $i)
{
$sql_dump .= $i['query'];
}
}
}
else
{
$sql_dump .= 'Encountered unexpected $logs cannot generate SQL log';
}
}
One other approach would be to have all your custom queries in the models/behaviors, and just calling the data/updates from shells. This would give you an extra benefit of being able to reuse those custom SQL in other parts of the project. For example, in unit tests.
In CakePHP 1.2, I was able to get the SQL queries to show up in my console output by adding a Configure::write('debug', 2); call to the bottom of the __bootstrap method in the cake/console/cake.php file.
No need to mess around with specifically calling a dump_sql function like some of these answers, I just automatically get the normal queries like at the bottom of a web page.