Access JSON array in PHP and send to MySQL SELECT

Question

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']));

exactly HOW is this not working for you? wrong data returned? no data returned? errors thrown? insults issued? dogs kicked? — Marc B
– Marc B, Commented Jul 8, 2015 at 18:57
@MarcB, in $where2 i'm not receiving the correct data, it was supossed to call a second database table — William Xavier
– William Xavier, Commented Jul 8, 2015 at 19:00
$ar2 = json_decode($processo2, true); maybe a typo to $ar2 = json_decode($processos, true); — vinayakj
– vinayakj, Commented Jul 8, 2015 at 19:01
@vinayakj, $processo2 is the input field that's sending the data to PHP. — William Xavier
– William Xavier, Commented Jul 8, 2015 at 19:06
then your root becomes ar2 not $processos, so try $ar2.processos['processo'] — vinayakj
– vinayakj, Commented Jul 8, 2015 at 19:09

waders group · Accepted Answer · 2015-07-08 19:11:27Z

2

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);

answered Jul 8, 2015 at 19:11

waders group

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10 Comments

William Xavier Over a year ago

your code worked like a charm, but with my variable is not working, maybe the way i'm encoding?

waders group Over a year ago

After encoding no character under the string should violet the JSON schema

William Xavier Over a year ago

I have updated my question with the json_encode code, can you check to see if there's something wrong please?

waders group Over a year ago

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

William Xavier Over a year ago

It returns "{"processos":Array}"

|

asiop · Accepted Answer · 2015-07-08 19:02:52Z

0

Your JSON string looks invalid, try instead:

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

answered Jul 8, 2015 at 19:02

asiop

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2 Comments

William Xavier Over a year ago

I've tried that, now it's returning "{"processos":[{"processo":"203"},{"processo":"1430"}]}" but still don't work.

vinayakj Over a year ago

so now try $ar2.processos['processo'] as $key=>$val

asiop · Accepted Answer · 2015-07-08 19:15:27Z

0

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

answered Jul 8, 2015 at 19:15

asiop

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