Php json_encode Array output \ error

Question

Hello i have php mysql pdo array with output json encode and gives me \\\ chars in output and i want to delete them.

My php code

$stmt2 = $this->conn->prepare("SELECT ID,clientName FROM Clients WHERE userID='$userID' OR mainAccountID='$mainAccountID' ORDER BY ID DESC");
$stmt2->execute();
$result = $stmt2 -> fetchAll();

foreach( $result as $userRow2 ) {
  $private_list[] = '{"name":"'.$userRow2['clientName'].'","ID":"'.$userRow2['ID'].'"}';
}

echo json_encode($private_list);

And gives output

["{\"name\":\"zz\",\"ID\":\"312\"}","{\"name\":\"jv\",\"ID\":\"311\"}","{\"name\":\"fff2222\",\"ID\":\"309\"}","{\"name\":\"ffff\",\"ID\":\"308\"}","{\"name\":\"v\",\"ID\":\"288\"}","{\"name\":\"t\",\"ID\":\"286\"}","{\"name\":\"s\",\"ID\":\"285\"}","{\"name\":\"r\",\"ID\":\"284\"}","{\"name\":\"p\",\"ID\":\"283\"}","{\"name\":\"o\",\"ID\":\"282\"}","{\"name\":\"n\",\"ID\":\"281\"}","{\"name\":\"m\",\"ID\":\"280\"}","{\"name\":\"l\",\"ID\":\"279\"}","{\"name\":\"k\",\"ID\":\"278\"}","{\"name\":\"j\",\"ID\":\"277\"}","{\"name\":\"i\",\"ID\":\"276\"}","{\"name\":\"h\",\"ID\":\"275\"}","{\"name\":\"g\",\"ID\":\"274\"}","{\"name\":\"f\",\"ID\":\"273\"}","{\"name\":\"e\",\"ID\":\"272\"}","{\"name\":\"d\",\"ID\":\"271\"}","{\"name\":\"c\",\"ID\":\"270\"}","{\"name\":\"b\",\"ID\":\"269\"}","{\"name\":\"a\",\"ID\":\"268\"}"]

I want remove \ chars.

Thanks

That's actually a valid JavaScript string, because, ... Wait what? — Praveen Kumar Purushothaman
– Praveen Kumar Purushothaman, Commented Feb 26, 2016 at 10:57
@PraveenKumar i will use output for swift app and dont take values gives error when have \ — SwiftDeveloper
– SwiftDeveloper, Commented Feb 26, 2016 at 10:58
@PraveenKumar normally json output valid but have \ chars and gives on swift — SwiftDeveloper
– SwiftDeveloper, Commented Feb 26, 2016 at 10:59
You either cobble JSON together by hand or you use json_encode to do that for you; doing both results in garbage. — deceze
– deceze ♦, Commented Feb 26, 2016 at 11:08

Praveen Kumar Purushothaman · Accepted Answer · 2016-02-26 10:59:13Z

3

Change your code to include one forms. You are mixing JavaScript and PHP. So, do this:

$private_list = array();
$private_list[] = array(
  "name" => $userRow2['clientName'],
  "ID" => $userRow2['ID']
);

answered Feb 26, 2016 at 10:59

Praveen Kumar Purushothaman

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2 Comments

Praveen Kumar Purushothaman Over a year ago

@SwiftDeveloper Enjoy! :)

Praveen Kumar Purushothaman Over a year ago

You may remove the $private_list = array(); stupid declaration! LoL.

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