Php json_encode数组输出\错误

Php json_encode数组输出\错误

问题描述:

你好,我有php输出json编码的pdo数组,并给出了\\\字符输出,我想delete他们。Php json_encode数组输出错误

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); 

并给出了输出

["{\"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

+0

这实际上是一个有效的JavaScript字符串,因为......等等什么? –

+0

@PraveenKumar我将使用输出为快速的应用程序,并不采取值给出错误时有\ – SwiftDeveloper

+0

你不能JSON编码的JSON数组... ... LoL。 –

更改您的代码以包含一种形式。你在混合JavaScript和PHP。所以,这样做:

$private_list = array(); 
$private_list[] = array(
    "name" => $userRow2['clientName'], 
    "ID" => $userRow2['ID'] 
); 
+0

谢谢老兄工作!我会接受这个答案! – SwiftDeveloper

+0

@Swift开发者享受! ':'' –

+0

你可以删除'$ private_list = array();'stupid声明!大声笑。 –