在预先存在的JSON对象中添加空json对象，并在给定的json对象索引之后添加 - Php

Question

so I have an Array of JSON Objects generated from mysql query.

JSON Object looks like:

[
  {"field1": "value1", "field2": "value2"},
  {"field1": "value3", "field2": "value4"},
  {"field1": "value5", "field2": "value6"},
  {"field1": "value7", "field2": "value8"}
]

now I want to add empty object like this: {} after 3rd JSON object in the array.

expected result:

[
  {"field1": "value1", "field2": "value2"},
  {"field1": "value3", "field2": "value4"},
  {"field1": "value5", "field2": "value6"},
  {},
  {"field1": "value7", "field2": "value8"}
]

I am able to loop through each json object and I am able to replace the current object's value but not sure how to add a whole new json empty object.

My code

for($j = 0; $j < $length; $j++) {

            if($j == 1 ){
                // HERE I dont want to replace Jth JSON Obj but add new empty Obj, something like $data.push('{}')
                $data[$j]['field1'] = '';
                $data[$j]['field2'] = '';
            }else{
                $data[$j]['question'] = $ques_arr[$j];
                $data[$j]['response'] = $allresponses[$j]['response'];
            }
        }

Answer 1

Try to add it via new \StdClass()OR [(object)[]] as example

$test = '[
{"field1": "value1", "field2": "value2"},
{"field1": "value3", "field2": "value4"},
{"field1": "value5", "field2": "value6"},
{"field1": "value7", "field2": "value8"}
]';
 $temp = json_decode($test);
 array_splice($temp, 3, 0, new \StdClass());
$result = json_encode($temp);
var_dump($result);

return [{"field1":"value1","field2":"value2"},{"field1":"value3","field2":"value4"},{},{"field1":"value7","field2":"value8"}]

update Use array_splice() and try :

    $test = '[
{"field1": "value1", "field2": "value2"},
{"field1": "value3", "field2": "value4"},
{"field1": "value5", "field2": "value6"},
{"field1": "value7", "field2": "value8"}
]';
    $temp = json_decode($test);
    $value = (new \StdClass());
    array_splice($temp, 3, 0, [(object)[]]);
    $result = json_encode($temp,true);
    var_dump($result);

as you wanted.