How to use JSON object to mysql query?

Question

From a recent query I am getting this as the result.

[{"BeneficiaryID":"2"},{"BeneficiaryID":"3"},{"BeneficiaryID":"4"},{"BeneficiaryID":"6"}]

I want to use these ID values for another mysql query using Laravel.
How can I just get only those numbers.
Final result should be like [2,3,4,6]

From mysql 5.7 you can use JSON data type dev.mysql.com/doc/refman/5.7/en/json.html — Abhik Chakraborty
– Abhik Chakraborty, Commented Jun 17, 2016 at 7:29
Just the ID value take for another query, so has to remove BeneficiaryID part. — Sachith Muhandiram
– Sachith Muhandiram, Commented Jun 17, 2016 at 7:47

Calin Blaga · Accepted Answer · 2016-06-17 09:52:22Z

1

 $data = json_decode('[{"BeneficiaryID":"2"},{"BeneficiaryID":"3"},{"BeneficiaryID":"4"},{"BeneficiaryID":"6"}]');

$id_array = [];

foreach($data as $data) {
    $id_array[] = $data->BeneficiaryID;
}

$query = Model::whereIn('column', $id_array)->get();

edited Jun 17, 2016 at 9:52

answered Jun 17, 2016 at 9:47

Calin Blaga

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5 Comments

Sachith Muhandiram Over a year ago

json_decode() expects parameter 1 to be string, array given

Calin Blaga Over a year ago

You're getting the result using toJson() method?

Calin Blaga Over a year ago

because in my answer a string is passed as parameter to json_decode function and the laravel toJson method is returning string as well? But if you don't need the json format you can use lists() method that will return an array

Sachith Muhandiram Over a year ago

If we supply '[{"BeneficiaryID":"2"},{"BeneficiaryID":"3"},{"BeneficiaryID":"4"},{"BeneficiaryID":"6"}]', we are getting the correct output. but we are getting this as a variable, when its used $data = json_decode($variable); we are getting errors.

Sachith Muhandiram Over a year ago

that variable in JSON format. this is the result [{"BeneficiaryID":"2"},{"BeneficiaryID":"3"},{"BeneficiaryID":"4"},{"Beneficiar‌yID":"6"}]

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