CODEWITHSUNDEEP
phpjquerymysqlajaxjson
William Xavier
William Xavier

Reputation: 478

Access JSON array in PHP and send to MySQL SELECT

I'm trying to receive the data from JSON with PHP and use it in my SELECT query. I have searched everywhere and every solution didn't worked for me.

My Returned JSON data: 

"{processos:[{"processo":"203"},{"processo":"1430"}]}"

My PHP:

$ar2 = json_decode($processo2, true);

    $where2 = array();

    foreach($ar2['processo'] as $key=>$val){ 
        $where2[] = $val;
    }
    $where2 = implode(',', $where2);


$pegaSonda = $pdo->prepare("SELECT * FROM sonda WHERE n_processo IN ($where2)");
$pegaSonda->execute();

What's wrong with my code?

EDIT

The code from @wadersgroup worked correctly, but when i change to my variable it stops. This is the way i'm encoding it:

$jsonData = array();
$jsonData[] = array("processo" => $automovel->n_processo);
$data['processo2'] .= '{"processos":'.json_encode($jsonData).'}';

$data['processo2'] is sending to my AJAX to populate the input and then it receive data back with:

$processo2 = strip_tags(stripslashes($_POST['n_processo2']));

Upvotes: 2

Views: 152

Answers (3)

asiop
asiop

Reputation: 727

$processo2 = '{"processos": [{ "processo": "203"},{ "processo": "1430"}]}'
$ar2 = json_decode($processo2, true);    
$arr = array_map(function($i) {return $i['processo']; }, $ar2["processos"]);
$where = implode(", ", $arr);

Upvotes: 0

waders group
waders group

Reputation: 538

There are many errors in this code. Try this

    $ar2 = json_decode('{"processos":[{"processo":"203"},{"processo":"1430"}]}', true);

    $where2 = array();

    foreach($ar2['processos'] as $key=>$val){ 
        $where2[] = $val['processo'];
    }
    $where = implode(',', $where2);
    print_r($where);

Upvotes: 2

asiop
asiop

Reputation: 727

Your JSON string looks invalid, try instead:

'{"processos":[{"processo":"203"},{"processo":"1430"}]}'

Upvotes: 0

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