如何在PHP中构建一个json返回？

function startChatSession($items) {


    foreach ($test as $chatbox => $void) {
            $items .= $this->chatBoxSession($chatbox);
        }

    if ($items != '') {
        $items = substr($items,-1);
    }

    header('Content-type: application/json');
    ?>
    {
        "username": "<?php echo $_SESSION['username'];?>","items": [
            <?php echo $items;?>
        ]
    }

    <?php
    exit(0);
}

我对这个功能的第二部分很感兴趣.我想把它翻译成像：

echo json_encode(array(
            'username'=>$_SESSION['username'],'items'=>$items
        )); 

    exit;

这种作品,但并不完全.

响应应如下所示：

{
    "username": "johndoe","items": [{
        "s": "1","f": "janedoe","m": "dqwdqwd"
    },{
        "s": "1","m": "sdfwsdfgdfwfwe"
    },"m": "werwefgwefwefwefweg"
    }]
}

在我的情况下它看起来像：

{
    "username": "johndoe","items": "ttttt   {rnttt"s": "1",rnttt"f": "babydoe",rnttt"m": "test"rnt   },ttttt   {rnttt"s": "1",rnttt"f": "",rnttt"m": ""rnt   }"
}

有任何想法吗？

谢谢

编辑：

如果我转储$items我喜欢的东西：

{
    s: 1,f: babydoe,m:  test
},{
    s: 1,{
    s:1,m:  test
}