[英]List part of data in JSON file using php or javascript
有人知道如何使用 php 或 javascript 從 json 文件中列出部分/部分數據嗎?
例如如果我使用 sql 我必須做代碼
select email from employees where name='Shyam'
但是對於 json 我不知道該怎么做。
{"employees":[
{"name":"Shyam", "email":"shyamjaiswal@gmail.com", "job":"police"},
{"name":"Bob", "email":"bob32@gmail.com", "job":"athelic"},
{"name":"Jai", "email":"jai87@gmail.com", "job":"king"}
]}
通過參考上面的數據,我只想使用 json 文件中的 php/javascript 來 select email where name=shyam。
使用 Javascript 您可以執行以下操作:
const data = {"employees": [ {"name":"Shyam", "email":"shyamjaiswal@gmail.com", "job":"police"}, {"name":"Bob", "email":"bob32@gmail.com", "job":"athelic"}, {"name":"Jai", "email":"jai87@gmail.com", "job":"king"} ]}; const result = data.employees.filter(employe => employe.name === "Shyam"); console.log(result);
使用 PHP 您可以做到這一點:
$data = json_decode('{"employees": [
{"name":"Shyam", "email":"shyamjaiswal@gmail.com", "job":"police"},
{"name":"Bob", "email":"bob32@gmail.com", "job":"athelic"},
{"name":"Jai", "email":"jai87@gmail.com", "job":"king"}
]}', true);
$result = array_filter($data['employees'], function ($employee) {
return $employee['name'] === 'Shyam';
});
var_dump($result);
在 PHP 你可以做這樣的事情。
$json = '{"employees":[
{"name":"Shyam", "email":"shyamjaiswal@gmail.com", "job":"police"},
{"name":"Bob", "email":"bob32@gmail.com", "job":"athelic"},
{"name":"Jai", "email":"jai87@gmail.com", "job":"king"}
]} ';
$jsonArray = json_decode($json, true);
$emails = [];
foreach ($jsonArray['employees'] as $item){
if ($item['name'] == "Shyam") {
$emails[] = $item['email'];
}
}
像這樣(JS)?
const getEmail = (json, name) => {
const row = json.employees.find(e => e.name === name);
return row ? row.email : null;
}
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