在预先存在的json对象和给定json对象索引之后添加空json对象-Php

所以我有一个从mysql查询生成的JSON对象数组

JSON对象看起来像：

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

现在我想在数组中的第三个JSON对象之后添加如下空对象：

{}

预期结果：

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

我可以循环遍历每个json对象，可以替换当前对象的值，但不知道如何添加一个全新的json空对象

我的代码 （$j=0；$j<$length；$j++）的

{
如果（$j==1）{
//这里我不想替换Jth JSON Obj，而是添加新的空Obj，比如$data.push（“{}”）
$data[$j]['field1']=''；
$data[$j]['field2']=''；
}否则{
$data[$j]['question']=$ques_arr[$j]；
$data[$j]['response']=$allresponses[$j]['response']；
}
}

例如，尝试通过

new\StdClass（）

或

[（对象）[]]]

添加它

$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);

返回

[{“field1”：“value1”，“field2”：“value2”}，{“field1”：“value3”，“field2”：“value4”}，{}，{“field1”：“value7”，“field2”：“value8”}]

更新 使用并尝试：

    $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);

---

根据您的需要。

尝试通过

new\StdClass（）

或

[（对象）[]]

添加它

$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);

    $json = '[
    {"field1": "value1", "field2": "value2"},
    {"field1": "value3", "field2": "value4"},
    {"field1": "value5", "field2": "value6"},
    {"field1": "value7", "field2": "value8"}
    ]';
    $temp = json_decode($json);
    $inserted = array( (object)[]);
    array_splice( $temp, 3, 0, $inserted );
    $result = json_encode($temp);
    var_dump($result);

返回

[{“field1”：“value1”，“field2”：“value2”}，{“field1”：“value3”，“field2”：“value4”}，{}，{“field1”：“value7”，“field2”：“value8”}]

更新 使用并尝试：

    $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);

---

你可以演示一个例子说明你所说的

stdClass（）

？

array\u-splice（$data，$j，0，new-stdClass（））

这将在$jtry

array\u-splice

和

new-stdClass（）插入一个空对象

@Fei你能举例说明一下你所说的

stdClass（）

？

array\u拼接（$data，$j，0，new stdClass（））

这将在$j0处插入一个空对象。你已经将第三个Json对象替换为空{}。我不想替换它，但要添加一个新的object@MurlidharFichadia我解决了它，请再次检查我的答案。您已将第三个Json对象替换为空{}。我不想替换它，但要添加一个新的object@MurlidharFichadia我解决了，请再次检查我的答案。

    $json = '[
    {"field1": "value1", "field2": "value2"},
    {"field1": "value3", "field2": "value4"},
    {"field1": "value5", "field2": "value6"},
    {"field1": "value7", "field2": "value8"}
    ]';
    $temp = json_decode($json);
    $inserted = array( (object)[]);
    array_splice( $temp, 3, 0, $inserted );
    $result = json_encode($temp);
    var_dump($result);