Question

嗨，大家需要一些帮助。

所以我试图像这样回应一个json数组格式：

{"ttype": [{ "id":"1", "type":"Order No.","code":"ORDNO" },
            {"id":"2", "type":"Orderline", "code":"ORDLINE"} ]

,"input":[{"id":"1","order":"Order No.", "limit":"5"},
{"id":"2","order":"Order No.", "limit":"5"},
{"id":"2", "order":"Order Line No.","limit":"10"}]}

所以这是我的代码：

$query = "SELECT * FROM `order` WHERE is_deleted = 0 ORDER BY id ASC";
$que = mysql_query($query);
while($row = mysql_fetch_array($que)){
    $id = $row['id'];
    $keyword = $row['keyword'];
    $parameters = $row['parameters'];

    $json = array(
            'ttype'=>array(array(
                'id'=>$id,
                'type'=>'',
                'code'=>$keyword
                )),'input'=>array(array(
                        'id' =>$id,
                        'order'=>'',
                        'limit'=>20
                 ))
        );

    echo json_encode($json);
}

我遇到的问题是：

 {"ttype":[{"id":"1","type":"","code":"ORDNO"}],"input":[{"id":"1","order":"","limit":20}]}{"ttype":    [{"id":"2","type":"","code":"ORDLINE"}],"input":    [{"id":"2","order":"","limit":20}]}

知道如何正确地给出正确的格式吗？感谢

Answer 1

你必须在while循环中有两个变量，然后相应地使用你的结构变量json_encode。这是您完成工作的方式

Answer 2

我已经解决了我的问题。

所以这是我的代码：

$query = "SELECT * FROM `order` WHERE is_deleted = 0 ORDER BY id ASC";
$que = mysql_query($query);
while($row = mysql_fetch_array($que)){
    $id = $row['id'];
    $keyword = $row['keyword'];
    $parameters = $row['parameters'];


    $array1[] = array(
        'id'=>$id,
        'type'=>'',
        'code'=>$keyword
        );

    $array2[] = array(
            'id'=>$id,
            'order'=>'',
            'limit'=>20
        );
}

 $globalarray = array(
                    'ttype'=>$array1,
                    'input'=>$array2
    );

echo json_encode($globalarray);

我现在得到了这个结果：

{"ttype":[{"id":"1","type":"","code":"ORDNO"},    {"id":"2","type":"","code":"ORDLINE"}],"input":[{"id":"1","order":"","limit":20},{"id":"2","order":"","limit":20}]}

感谢您的帮助！特别是对Waleed Ahmed先生来说。我欠你的，你的答案给了我一个让这成为可能的想法。

非常感谢！

Answer 3

尝试以下代码。

$query = "SELECT * FROM `order` WHERE is_deleted = 0 ORDER BY id ASC";
        $que = mysql_query($query);
        while($row = mysql_fetch_array($que)){
            $id = $row['id'];
            $keyword = $row['keyword'];
            $parameters = $row['parameters'];

            $json[] = array(
                    'ttype'=>array(array(
                        'id'=>$id,
                        'type'=>'',
                        'code'=>$keyword
                        )),'input'=>array(array(
                                'id' =>$id,
                                'order'=>'',
                                'limit'=>20
                         ))
                );

        }
        echo json_encode($json);

PHP echo json只有两个类别和多个记录

3 个答案: