CODEWITHSUNDEEP
laravellaravel-5
Mohammed
Mohammed

Reputation: 658

How to save dynamic data in database in laravel

My project is on angularjs+Laravel.

From frontend I am sending dynamically generated input to laravel controller.

Scenario is I am taking family details where child field is dynamic.So if you click of Add more It will generate another input field.

Initially Person X have 1 child namely ROohan . Now he Have 2 more, Now when he is sending data of 3 new children, I want that 2 children data get added without over writing present 2 children.

My request which I am receiving in controller

family_data : 
      {"famdet":[{"id":"choice1","cname":"Rohan","doa":"2017-01-12"},  
       {"id":"choice2","cname":"sohan","cdob":"2017-01-13"},           
       {"id":"choice3","cname":"nitesh","cdob":"2017-01-07"}
          ]}

my problem is How to make this data save in separate row and if data is present then row should not be get created.

Upvotes: 1

Views: 872

Answers (1)

Alexey Mezenin
Alexey Mezenin

Reputation: 163978

First you need to use json_decode() to convert data to an array, for example:

$familyData = json_decode($data, true)['famdet'];

Then you should iterate over this array and use updateOrCreate() method to update existing rows or create new ones. For example, if you're using and ID to identify a row:

foreach ($familyData as $row) {
    Model::updateOrCreate(['id' => $row['id']], $row)
}

Upvotes: 1

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