I want to know the structure of a table. How I can do it in CodeIgniter. Using database class I got 'Invalid SQL Statement' error when I ran $this->db->query('desc mytable');
Try:
$fields = $this->db->list_fields('table_name');
foreach ($fields as $field)
{
echo $field;
}
From manual
For more descriptive information, you should use
$fields = $this->db->field_data('table_name');
You're going to get something like this foreach field in fields as stdClass
name = "id"
type = "int"
max_length = 11
default = null
primary_key = 1
For Get Table Schema in CodeIgniter query:
$query = $this->db->query('SHOW CREATE TABLE yourTableName');
$queryResultArray = $query->result_array();
print_r( $queryResultArray[0]['Create Table'] );
Related
I am using laravel eager loading to load data on the jquery datatables. My code looks like:
$columns = array(
0 => 'company_name',
1 => 'property_name',
2 => 'amenity_review',
3 => 'pricing_review',
4 => 'sqft_offset_review',
5 => 'created_at',
6 => 'last_uploaded_at'
);
$totalData = Property::count();
$limit = $request->input('length');
$start = $request->input('start');
$order = $columns[$request->input('order.0.column')];
$dir = $request->input('order.0.dir');
$query = Property::with(['company','notices']);
$company_search = $request->columns[0]['search']['value'];
if(!empty($company_search)){
$query->whereHas('company', function ($query) use($company_search) {
$query->where('name','like',$company_search.'%');
});
}
$property_search = $request->columns[1]['search']['value'];
if(!empty($property_search)){
$query->where('properties.property_name','like',$property_search.'%');
}
if(!Auth::user()->hasRole('superAdmin')) {
$query->where('company_id',Auth::user()->company_id);
}
$query->orderBy($order,$dir);
if($limit != '-1'){
$records = $query->offset($start)->limit($limit);
}
$records = $query->get();
With this method I received error: Column not found: 1054 Unknown column 'company_name' in 'order clause' .
Next, I tried with following order condition:
if($order == 'company_name'){
$query->orderBy('company.name',$dir);
}else{
$query->orderBy($order,$dir);
}
However, it also returns similar error: Column not found: 1054 Unknown column 'company.name' in 'order clause'
Next, I tried with whereHas condition:
if($order == 'company_name'){
$order = 'name';
$query->whereHas('company', function ($query) use($order,$dir) {
$query->orderBy($order,$dir);
});
}else{
$query->orderBy($order,$dir);
}
But, in this case also, same issue.
For other table, I have handled this type of situation using DB query, however, in this particular case I need the notices as the nested results because I have looped it on the frontend. So, I need to go through eloquent.
Also, I have seen other's answer where people have suggested to order directly in model like:
public function company()
{
return $this->belongsTo('App\Models\Company')->orderBy('name');
}
But, I don't want to order direclty on model because I don't want it to be ordered by name everytime. I want to leave it to default.
Also, on some other scenario, I saw people using join combining with, but I am not really impressed with using both join and with to load the same model.
What is the best way to solve my problem?
I have table like: companies: id, name, properties: id, property_name, company_id, notices: title, slug, body, property_id
The issue here is that the Property::with(['company','notices']); will not join the companies or notices tables, but only fetch the data and attach it to the resulting Collection. Therefore, neither of the tables are part of the SQL query issued and so you cannot order it by any field in those tables.
What Property::with(['company', 'notices'])->get() does is basically issue three queries (depending on your relation setup and scopes, it might be different queries):
SELECT * FROM properties ...
SELECT * FROM companies WHERE properties.id in (...)
SELECT * FROM notices WHERE properties.id in (...)
What you tried in the sample code above is to add an ORDER BY company_name or later an ORDER BY companies.name to the first query. The query scope knows no company_name column within the properties table of course and no companies table to look for the name column. company.name will not work either because there is no company table, and even if there was one, it would not have been joined in the first query either.
The best solution for you from my point of view would be to sort the result Collection instead of ordering via SQL by replacing $records = $query->get(); with $records = $query->get()->sortBy($order, $dir);, which is the most flexible way for your task.
For that to work, you would have to replace 'company_name' with 'company.name' in your $columns array.
The only other option I see is to ->join('companies', 'companies.id', 'properties.company_id'), which will join the companies table to the first query.
Putting it all together
So, given that the rest of your code works as it should, this should do it:
$columns = [
'company.name',
'property_name',
'amenity_review',
'pricing_review',
'sqft_offset_review',
'created_at',
'last_uploaded_at',
];
$totalData = Property::count();
$limit = $request->input('length');
$start = $request->input('start');
$order = $columns[$request->input('order.0.column')];
$dir = $request->input('order.0.dir');
$query = Property::with(['company', 'notices']);
$company_search = $request->columns[0]['search']['value'];
$property_search = $request->columns[1]['search']['value'];
if (!empty($company_search)) {
$query->whereHas(
'company', function ($query) use ($company_search) {
$query->where('name', 'like', $company_search . '%');
});
}
if (!empty($property_search)) {
$query->where('properties.property_name', 'like', $property_search . '%');
}
if (!Auth::user()->hasRole('superAdmin')) {
$query->where('company_id', Auth::user()->company_id);
}
if ($limit != '-1') {
$records = $query->offset($start)->limit($limit);
}
$records = $query->get()->sortBy($order, $dir);
I don't want laravel to format my query result to an array or object ..etc. All I want, is to run the result set from database and then I will manually do the fetch myself in my custom code.
At the moment, I ran my select query and get my result in an array. The reasons for that, because the result is huge and I want to stream it directly to API.
$result = self::$db->select('select * from customer');
How can I tell laravel, to return my query result set without any format at all?
You can use DB:Raw like:
$results = DB::table('users')->select(DB::raw("*"))->get()
Or
$results = DB::select('select * from users where id = ?', [1]);
These two will return a neat object without any casts or relations etc. You can also make any object or array your API need by simple eloquent models by the way. Please explain more about data type you wanna extract from model query.
You must be use ->toSql() or ->dd()
Exapmle
Customer::toSql(); // select * from `customer`
if you want some condition
$query = Customer::where(`some conditions`);
$sql = $query->toSql();
$bindings = $query->getBindings();
$sql = str_replace('?', '%s', $sql);
$sql = sprintf($sql, ...$bindings);
Thanks everyone, I end up writing a raw function to query the data I want from database.
public static function dataStreamJSON($stmt, $headers)
{
return Response::stream(function() use ($stmt){
$conn = self::getConnection();
$result = sqlsrv_query($conn, "exec $stmt");
echo '
{
"Customers": {
"Customer": [';
$counter = 0;
while($customer = sqlsrv_fetch_object($result)) {
if($counter !== 0){
echo ",";
}
$counter++;
$row = [
'Firstname' => $customer->Firstname,
'Lastname' => $customer->Lastname,
...
];
echo json_encode($row);
unset($row);
unset($customer);
}
echo ']
}
}';
#sqlsrv_free_stmt($result);
#sqlsrv_close($conn);
}, 200, $headers);
}
The purpose of this code is to stream the data out to JSON format on browser without store the data in any variable, which will caused “out of memory” error.
I managed to stream 700MB of JSON data to the browser without any error. With this code, you will never run into “out of memory” error.
Best way to test this, is to use CURL to access your API and download the data to a JSON file. If you open on browser, it will freeze your screen because browser can't handle large data.
You can use toArray() or toJson() methods like below:
$array = Customer::all()->toArray();
$json = Customer::all()->toJson();
echo '<pre>';
print_r($array);
print_r($json);
If you want to run raw SQL Queries, you can do as below
$users = DB::select('select * from users where 1');
echo '<pre>';
print_r($users);
You can use
1) query builder way:-
DB::table('your_table_name)->select('your_col_names')->get();
eg:- DB::table('shop')->select('product_id','product_name')->get();
2) use laravel Raw
$orders = DB::table('orders')->selectRaw('price * ? as price_with_tax', [1.0825])->get();
3) for select raw
$product_count = DB::table('product')->select(DB::raw('count(*) as total_product_count'))->where('status',1)->get();
I have the following query.
$projects = Project::orderBy('created_at', 'desc');
$data['sorted'] = $projects->groupBy(function ($project) {
return Carbon::parse($project->created_at)->format('Y-m-d');
})->simplePaginate(5);
When I try to paginate with the simplePaginate() method I get this error.
stripos() expects parameter 1 to be string, object given
How can I paginate grouped data in this case?
The created_at attribute is already casted as a Carbon Object (by default in laravel models). that's why you are getting that error. Try this:
$projects = Project::orderBy('created_at', 'desc')->get();
$data['sorted'] = $projects->groupBy(function ($project) {
return $project->created_at->format('Y-m-d');
})->simplePaginate(5);
this answer is just for the error you're getting. now if you want help with the QueryBuilder, can you provide an example of the results you're expecting to have and an example of the database structure ?
The pagination methods should be called on queries instead of collection.
You could try:
$projects = Project::orderBy('created_at', 'desc');
$data['sorted'] = $projects->groupBy('created_at');
The problem was solved. I was create custom paginator via this example:
https://stackoverflow.com/a/30014621/6405083
$page = $request->has('page') ? $request->input('page') : 1; // Use ?page=x if given, otherwise start at 1
$numPerPage = 15; // Number of results per page
$count = Project::count(); // Get the total number of entries you'll be paging through
// Get the actual items
$projects = Project::orderBy('created_at', 'desc')
->take($numPerPage)->offset(($page-1)*$numPerPage)->get()->groupBy(function($project) {
return $project->created_at->format('Y-m-d');
});
$data['sorted'] = new Paginator($projects, $count, $numPerPage, $page, ['path' => $request->url(), 'query' => $request->query()]);
simplePaginate Method is exist in the path below:
Illuminate\Database\Eloquent\Builder.php::simplePaginate()
I'm facing this problem when use database in Laravel. How can i convert that to Array the most simpletest?
$data = DB::table('users')->get();
Please try this. this will return array of objects.
$result = json_decode(json_encode($data, true));
*Updated
if you want to convert all nested properties to the array, try this.
$result = json_decode(json_encode($data, true), true);
get() will return a collection. If you want to get an array of objects, use the toArray() method:
$data->toArray();
If you want to convert every object to an array too, do this:
$data->map(function($i) {
return (array)$i;
})->toArray();
I usually run into this problem when I use DB::select and manually write my sql:
$sql = 'SELECT *
FROM ba_pics ba
INNER JOIN pages pgs
ON ba.service_page_id = pgs.id';
$baPics = DB::select($sql);
$baPics = json_decode(json_encode($baPics, true), true);
return view('beforeAndAfter',['baPics'=>$baPics, 'lodata'=>'no lodata yet']);
Here is another way using PHP's array_map function:
$sql = 'SELECT *
FROM ba_pics ba
INNER JOIN pages pgs
ON ba.service_page_id = pgs.id';
$baPics = DB::select($sql);
$baPics = array_map(function($i) {
return (array)$i;
}, $baPics);
return view('beforeAndAfter',['baPics'=>$baPics, 'lodata'=>'no lodata yet']);
I'm using Eloquent Query Builder to produce raw sql for me. My goal is to return a sql query like this
select * from accounts where id = '120'
now i tried
$query = \App\Account::query();
$query = $query->where('id', 120);
dd($query->toSql());
Output:
select * from `accounts` where `id` = ?
is there a way to display the value instead of (?) because this is not a valid query on the third party API that i'm using.
I'm expecting to produce a query on my first example that i give.
Note: I'm not querying on my database, i will use that raw query to pass on the 3rd party API. All i want is to produce only a raw sql. Any suggestion would be appreciated. Thanks!
Yes, you can print the parameters value by using getBindings()
print_r( $query->getBindings() );
Array
(
[0] => instructor
[1] => 1
)
Found a solution for this, maybe it can help to someone that has same problem with me.
You can check this link: https://gist.github.com/JesseObrien/7418983
public function getSql()
{
$builder = $this->getBuilder();
$sql = $builder->toSql();
foreach($builder->getBindings() as $binding)
{
$value = is_numeric($binding) ? $binding : "'".$binding."'";
$sql = preg_replace('/\?/', $value, $sql, 1);
}
return $sql;
}