Question

我在PHP Like中有一个JSON对象

 $my_json = '[{
        "No": "1",
        "Id": "10",
        "msg": "Value is 10"

    },
    {
        "No": "2",
        "Id": "55",
        "msg": "value is 55"

    }, {
        "No": "3",
        "Id": "38",
        "msg": "value is 38 "
    }, {
        "No": "4",
        "Id": "95",
        "msg": "value is 95 "
    } ]';

我已使用$ obj_arr = json_decode（$ my_json）;将其转换为PHP对象数组；

现在，基于用户选择，我想过滤数组元素例如。如果用户选择“ ID”为38和55，则新数组应注意其ID为10和95，并且必须包含与该ID相关的所有其他数据，就像原始数据一样

EG

 $final_json = '[
{
    "No": "2",
    "Id": "55",
    "msg": "value is 55"

}, {
    "No": "3",
    "Id": "38",
    "msg": "value is 38 "
} ]';

Answer 1

尝试一下：

const inputScript = `
function test(callback, number, reason) {
    if (number != atob("MA==")) {
        console.log(atob("RmFpbGVkOiA=") + reason + atob("LiBQbGVhc2UgdHJ5IGFnYWluLg=="));
    } else {
        callback(number);
    }
}
`;
const outputScript = inputScript
  .replace(/atob\("([^\"]+)"\)/g, (_, p1) => `"${atob(p1)}"`)
  .trim();
console.log(outputScript);

输出：

<?php
$my_json = '[{
    "No": "1",
    "Id": "10",
    "msg": "Value is 10"

},
{
    "No": "2",
    "Id": "55",
    "msg": "value is 55"

}, {
    "No": "3",
    "Id": "38",
    "msg": "value is 38 "
}, {
    "No": "4",
    "Id": "95",
    "msg": "value is 95 "
} ]';

$selected = [38, 55];

$obj = json_decode($my_json);

$result = array_filter($obj, function($row) use ($selected) {
    return in_array($row->Id, $selected);
});

echo json_encode(array_values($result), JSON_PRETTY_PRINT);

如果要过滤掉上面选择的值而不是保留它们，请返回in_array的取反。

根据给定ID的键元素的值过滤数组元素

1 个答案: