Question 69800 in Ajax Question posted in
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List part of data in JSON file using php or javascript – Jquery ajax

anyone know how to list some/part of data from json file using php or javascript?

example if i using sql i must do code

select email from employees where name='Shyam'

but for json i don’t have any clue how to do it.

{"employees":[  
    {"name":"Shyam", "email":"[email protected]", "job":"police"},  
    {"name":"Bob", "email":"[email protected]", "job":"athelic"},  
    {"name":"Jai", "email":"[email protected]", "job":"king"}  
]}

by referring data above, i only want to select email where name=shyam by using php/javascript from json file.

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3

Answers


  1. 0

    With Javascript you could do something like that :

    const data = {"employees": [  
  {"name":"Shyam", "email":"[email protected]", "job":"police"},  
  {"name":"Bob", "email":"[email protected]", "job":"athelic"},  
  {"name":"Jai", "email":"[email protected]", "job":"king"}  
]};

const result = data.employees.filter(employe => employe.name === "Shyam");
console.log(result);

    And with PHP you can do that :

    $data = json_decode('{"employees": [  
  {"name":"Shyam", "email":"[email protected]", "job":"police"},  
  {"name":"Bob", "email":"[email protected]", "job":"athelic"},  
  {"name":"Jai", "email":"[email protected]", "job":"king"}  
]}', true);

$result = array_filter($data['employees'], function ($employee) {
    return $employee['name'] === 'Shyam';
});
var_dump($result);
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  2. 0

    Somethink like this (JS)?

          const getEmail = (json, name) => {
        const row = json.employees.find(e => e.name === name);
        return row ? row.email : null;
      }
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  3. 0

    In PHP you can do something like this.

    $json = '{"employees":[  
    {"name":"Shyam", "email":"[email protected]", "job":"police"},  
    {"name":"Bob", "email":"[email protected]", "job":"athelic"},  
    {"name":"Jai", "email":"[email protected]", "job":"king"}  
]} ';

$jsonArray = json_decode($json, true);
$emails = [];

foreach ($jsonArray['employees'] as $item){
    if ($item['name'] == "Shyam") {
        $emails[] = $item['email'];
    }
}
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