Php json_编码数组输出\错误

您好，我有一个带有输出json encode的php mysql pdo数组，在输出中给我

\\\\

字符，我想

删除它们

我的php代码

$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);

并给出输出

["{\"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\"}"]

我想删除
\
字符

谢谢
将您的代码更改为包含一个表单。您正在混合JavaScript和PHP。因此，请这样做：

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

这实际上是一个有效的JavaScript字符串，因为。。。等等什么？@PraveenKumar我将使用swift应用程序的输出，并且在出现错误时不接受值\n您无法对JSON数组进行JSON编码。。。LoL.@PraveenKumar通常json输出有效，但在swiftYou上有\chars和gives手动拼凑json或使用
json\u encode
为您完成这项工作；这两种方法都会产生垃圾。@SwiftDeveloper享受
：）
您可以删除
$private\u list=array（）愚蠢的声明！英雄联盟