使用key：value对

Question

我有JSON数组

[
{"id":"476","jobid":"324","remarks":"Job 500","status":"Open"},
{"id":"477","jobid":"324","remarks":"Job 501","status":"Closed"},
{"id":"481","jobid":"324","remarks":"Job 502","status":"Cancelled"}
]

如何使用PHP获得remarks的状态Closed？

可能的话最好不要循环

Answer 1

没有循环，这取决于。
没有任何循环都是不可能的，但是循环可以在后台而不是在php中完成。

这将获取状态列，并过滤掉所有未关闭的内容。
然后设置“备注”的键。

var_dump(array_intersect(array_column($arr, 'status', 'remarks'), ["Closed"]));

输出：

array(1) {
  ["Job 501"]=>
  string(6) "Closed"
}

https://3v4l.org/mNUr6

Answer 2

签出JQL：

$jql = new JQL('[
  {"id":"476","jobid":"324","remarks":"Job 500","status":"Open"},
  {"id":"477","jobid":"324","remarks":"Job 501","status":"Closed"},
  {"id":"481","jobid":"324","remarks":"Job 502","status":"Cancelled"}
]');

$result = $jql->select('remarks')->where('status = Closed')->fetch();