$array = array($t, $c, $s);
foreach ($array as $a) {
receipt::create($a);
}
i want to pass the array content to database but i got this error
"Type error: Argument 1 passed to Illuminate\Database\Eloquent\Builder::create() must be of the type array, string given,
Actually while insert we should provide the array data even single field or more no of fields.
For example two fields : receipt::create($a). In this $a should contain two values in array format.
New example updated below as per your request. In this your $a variable should be like below format
$a = array(
'user_id' => $userId,
'receipt_description' => $desc
);
$receiptId = receipt::create($a)->id;
Related
The manual pagination I found while googling works fine but I was just wondering what does the 'query' => $request->query() in the option parameter does?
$total = count($form_list);
$per_page = 10;
$current_page = $request->input('page') ?? 1;
$starting_point = ($current_page * $per_page) - $per_page;
$form_list = array_slice($form_list, $starting_point, $per_page, true);
$form_list = new Paginator($form_list, $total, $per_page, $current_page, [
'path' => $request->url(),
'query' => $request->query(),
]);
Calling ->query() without any parameters returns all the values from the query string as an associative array.
Suppose you have a query string like this:
https://example.com/path/to/page?name=ferret&color=purple
You can retrieve the value of name by doing something like so:
$request->query('name')
which returns ferret. You can also pass a second parameter for a default value so if you call:
$request->query('size', 'Medium')
which doesn't exist on the query string, you'll get 'Medium' instead of null.
If you omit all parameters, you'll receive an associative array that looks something like this:
query = [
'name' => 'ferret',
'color' => 'purple',
]
The options parameter is not needed by the pagination itself but for your dataset query. If you do not pass the query parameter, when you click one of the pagination urls, you'll get something like this:
https://example.com/path/to/page?page=2&per_page=5
Sometimes, this works fine and will give us something that we want but sometimes, we need those additional query string to get the correct dataset. We pass in all values from our query to get something like this:
https://example.com/path/to/page?page=2&per_page=5&name=ferret&color=purple
Which will filter your dataset for all those purple ferrets. As for the question if you need it, it's up for you to decide if that is essential for your code or if you can get away with just pagination.
Good luck! I hope this helps.
My JSON looks something like this:
[
{"pet_type":"Dog","weight":"26","description":"Akita"},
{"pet_type":"Dog","weight":"6","description":"Pug"},
{"pet_type":"Cat","weight":"4","description":"Manx"},
{"pet_type":"Dog","weight":"12","description":"Beagle"},
{"pet_type":"Cat","weight":"5","description":"Siberian"}
]
How could I convert it to a string which would look like3 Dogs, 2 Cats?
The way I tried is filling an array with pet_type and than use array_count_values to count number of same records, and later I go through that array in a foreach and concat string like this:
foreach ($count_animals as $type => $number) {
$animals .= $number.' '.str_plural($type, $number).', ';
}
This works, but my question is, could I do it with less code, directly from JSON, without using one more foreach loop?
If it works, you can keep your code.
If you want less code, you can use this version :
$json = '[
{"pet_type":"Dog","weight":"26","description":"Akita"},
{"pet_type":"Dog","weight":"6","description":"Pug"},
{"pet_type":"Cat","weight":"4","description":"Manx"},
{"pet_type":"Dog","weight":"12","description":"Beagle"},
{"pet_type":"Cat","weight":"5","description":"Siberian"}
]';
print_r(array_count_values(array_map(function($item) {
return $item['pet_type'];
}, json_decode($json, true))));
Gonna display :
Array ( [Dog] => 3 [Cat] => 2 )
in your controller
$pet = Pet::get();
$petcount = Pet::where('pet_type','Dog')->get();
In your blade
<h1>{{count($petcount)}} Dog</h1>
it's a WHY-question, not How-to one:)
I have assigned a Query Bulder to a variable $query:
$query = table::where(['id'=>1, 'this_version'=> 1]);
$versions['slug1'] = $query->select('tourist_id', 'tourist_version')->get()->toArray();
print_r($versions);
outputs array with 2(!) sub-arrays:
Array
(
[slug1] => Array
(
[0] => Array
(
[tourist_id] => 1
[tourist_version] => 1
)
[1] => Array
(
[tourist_id] => 2
[tourist_version] => 1
)
)
)
But if I add another line using $query between my $query declaration and it's usage in getting $version[2] array, my $version[2] output is shortened to a 1-dimensional array:
$query = previoustour2_tourist::where(['tour2_id'=>$tour->id, 'this_version'=> 1]);
// Added line:
$versions['slug0'] = $query->select('version_created')->first()->version_created;
//
$versions['slug1'] = $query->select('tourist_id', 'tourist_version')->get()->toArray();
print_r($versions);
outputs (note slug1 now has only 1 nested array):
Array
(
[slug0] => 2017-08-08 08:25:26
[slug1] => Array
(
[0] => Array
(
[tourist_id] => 1
[tourist_version] => 1
)
)
)
it seems like the like this line:
$versions['slug0'] = $query->select('version_created')->first()->version_created;
has added "first()" method to the original $query . Am I right and, if yes, why does it happen?
Well, this is because by default an object (in your case is the Query builder object) in PHP is passed by reference. You can read more about this here: PHP OOP References.
I quote from the above reference:
A PHP reference is an alias, which allows two different variables to
write to the same value.
When you pass the query builder object to the $query variable, you actually just pass the reference to this object and not the copy of it.
$query = previoustour2_tourist::where(['tour2_id'=>$tour->id, 'this_version'=> 1]);
So when you call the first() method on the second line, it actually modifies the query builder object.
$versions['slug0'] = $query->select('version_created')->first()->version_created;
Thus causing the upcoming query result to be limited to 1. In order to work around this issue, you can clone the query object like this:
$query = previoustour2_tourist::where(['tour2_id'=>$tour->id, 'this_version'=> 1]);
$versions['slug0'] = (clone $query)->select('version_created')->first()->version_created;
$versions['slug1'] = (clone $query)->select('tourist_id', 'tourist_version')->get()->toArray();
print_r($versions);
Hope this help!
I have a record "id_zone" in database that store cites id like this $abilitato->id_zone = '1, 2, 3, 9, 50'
From laravel i need to check if a value passed from url is present in this field. The following work only if i have a single value
$abilitato = \App\Models\Servizi::where('id', $servizio)->('id_zone', $verifica->id_citta)->first();
if(!empty($abilitato) && empty($verifica->multicap)) {
$json = ['CAP' => '1' , 'DEB' => $cap ];
}else{
$json = ['CAP' => '0' , 'DEB' => $cap];
}
i need to check if
If you want to know if a value is in your array you can use this functión to know it.
here is a example:
// First convert your string to array
$myString = $abilitato->id_zone; // Here your string separate with comas
$array = explode(',', $myString); // Here convert it to array
$value = 7; // Here you put the value that you want to check if is in array
// Check if exist a value in array
if (in_array($value, $array)){
echo "Exist in array";
}else{
echo "No exist in array";
}
This is the documentation of these functions: explode , in_array
Regards!
I tried to get a umat_id with this SQL query :
SELECT umat_id FROM msumat WHERE nama = $nama
I converted this SQL query into CI's Active Record :
$this->db->select('umat_id');
$terdaftar = $this->db->get_where('msumat', array('nama' => $nama));
So this query should return a string (example : "John").
But I got an error when I tried to echo it :
Object of class CI_DB_mysql_result could not be converted to string
I have tried something like this : echo (string)$terdaftar;, but it's not working.
All I want is to echo "John"
EDIT
Just said I want to insert "John" into a variable. How to do that?
$john = ????
As some of the users already pointed the solution, I'm only explaining why you did get this error so you can understand better the querying results that codeigniter gives.
This error:
But I got an error when I tried to echo it : Object of class
CI_DB_mysql_result could not be converted to string
Happens because you were trying to echo an object.
This piece of code
$terdaftar = $this->db->get_where('msumat', array('nama' => $nama));
Will return an object, this object will have information about the query you've done.
With this object you can get the result(rows) as objects doing this:
$results = $terdaftar->result();
Or you if you feel more comfortable with arrays you can return the results(rows) as an array doing this:
$results = $terdaftar->result_array();
You can also get the number of results doing this:
$number_results = $terdaftar->num_rows()
And this is just an example you can read more about the results here
http://ellislab.com/codeigniter/user-guide/database/results.html
EDIT
A better explanation: imagine that we use the result_array() function to get the result in a pure array format:
$results = $terdaftar->result_array();
Now your variable $results is an array, to iterate through it and get the data you want you'll do something like this:
foreach ($results as $key => $row) {
//the row variable will have each row of your database returned by your query
//so if you want to access a field from that row,
//let's say for example the name field. You would do something like this
if($row['name']=='John')
echo $row['name'];
}
Try:
$this->db->select('umat_id');
$terdaftar = $this->db->get_where('msumat', array('nama' => $nama));
foreach ($terdaftar->result() as $row)
{
echo $row->umat_id;
}
Read the documentation for more information.
Try this:
$this->db->select('umat_id');
$terdaftar = $this->db->get_where('msumat', array('nama' => $nama));
$row = $terdaftar->row_array();
$your_variable = $row['umat_id']; /*Here comes your john*/