JSON對象內它

我有一些PHP從MySQL數據庫生成從數據的JSON對象JSON對象內它

$addressData = mysql_query("SELECT * FROM address WHERE ContactID = $contactID")or die("<br/><br/>".mysql_error()); 

while($r = mysql_fetch_assoc($addressData)){ 
    $rows[] = array('data' => $r); 
} 

// now all the rows have been fetched, it can be encoded 
echo json_encode($rows);

這將生成以下JSON對象：

[ 
    {"address": 
    {"AddressID":"10011","AddressType":"Delivery","AddressLine1":"4 Caerleon Drive","AddressLine2":"Bittern","AddressLine3":"","CityTown":"Southampton","County":"Hampshire","PostCode":"SO19 5LF","Country":"United Kingdom","ContactID":"10011"}}, 
    {"address": 
    {"AddressID":"10012","AddressType":"Home","AddressLine1":"526 Butts Road","AddressLine2":"Sholing","AddressLine3":"","CityTown":"Southampton","County":"Hampshire","PostCode":"SO19 1DJ","Country":"England","ContactID":"10011"}} 
]

當接收它返回到Ajax並通過以下方式運行它：

$.each(data, function(key, val) { 
    string =string + "Key: " + key + " Value:" + val + "<br />"; 
});

它打印以下內容：

鍵：0值：[對象的對象]

鍵：1的值：[對象的對象]

有關如何我可以在數據內的關鍵0和1訪問的對象的任何想法？

來源

2012-09-11 user1662233

這是因爲對象的默認toString實現在連接到String對象時導致「[object Object]」。您可以訪問VAL對象的字段像往常一樣，例如：

val.data.AddressID

這樣的：

$.each(data, function(key, val) { 
    string += "Key: " + key + " Value:" + val.data.AddressID + "<br />"; 
});

請注意在環路VAL上面是{「數據」：... } JSON代碼的一部分，這就是爲什麼你需要指定val.data。可以訪問裏面的部分地址數據。

此外，你可以簡單地刪除數據嵌套級別，導致JSON佈局類似：

[ 
    {"AddressID":"10011","AddressType":"Delivery","AddressLine1":"4 Caerleon Drive","AddressLine2":"Bittern","AddressLine3":"","CityTown":"Southampton","County":"Hampshire","PostCode":"SO19 5LF","Country":"United Kingdom","ContactID":"10011"}, 
    {"AddressID":"10012","AddressType":"Home","AddressLine1":"526 Butts Road","AddressLine2":"Sholing","AddressLine3":"","CityTown":"Southampton","County":"Hampshire","PostCode":"SO19 1DJ","Country":"England","ContactID":"10011"} 
]

使用這個PHP代碼：

while($r = mysql_fetch_assoc($addressData)){ 
    $rows[] = $r; 
}

然後，你可以訪問迴路中的地址數據字段如下：

$.each(data, function(key, address) { 
    string += "Key: " + key + " Value:" + address.AddressID + "<br />"; 
});

來源

2012-09-11 08:31:54

否，因爲'val'是{data：{}}對象。他應該改變他的json的格式並去除數據：嵌套 –

同意，數據字段的嵌套級別是無用的，我會將其添加到答案中。 –

我會改變你的json的佈局，你不需要嵌套。嘗試：

[ 
    {"AddressID":"10011","AddressType":"Delivery",...}, 
]

，那麼你可以遍歷返回值與主要=> VAL數組，其中val將成爲對象，因爲現在，val是抱着你要的對象的對象...

來源

2012-09-11 08:36:44

很好的建議。只要改變'$ rows [] = array（'data'=> $ r）;'到'$ rows [] = $ r;'和你的舊代碼就可以正常工作。 –

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