我發現很多關於這個主題的問題只有很少的投票,但我似乎無法得到任何迄今爲止的工作答案。我在Web服務器上創建了一個PHP頁面,將其轉換爲application/json,將mysqli查詢轉換爲數組,並使用json_encode對其進行編碼,使其成爲JSON對象。我現在正在嘗試使用javascript解碼JSON對象,但唯一能找到的解決方案是處理數組而不是對象。最終,我想解碼JSON並遍歷它,以便我可以將數據插入到客戶端上的Sqlite數據庫中。我很難在客戶端上得到任何結果,除非能夠對JSon進行字符串化,以便我能夠看到它已被檢索。我的代碼如下:使用Javascript解碼PHP編碼的JSON對象
Web服務器retrieveJSON.php頁面
<?php header('Content-Type: application/json');
$mysqli= new mysqli("host","user","password","database");
mysqli_select_db($mysqli,"database");
$query = "SELECT * FROM AppCustomers";
$result = mysqli_query($mysqli,$query) or die('Errant query: '.$query);
$customers = array();
if(mysqli_num_rows($result)) {
while($customer = mysqli_fetch_assoc($result)) {
$customers[] = array('customer'=>$customer);
}
}
$json_array = array('customers'=>$customers,);
echo json_encode($json_array);
mysqli_close($mysqli);
?>
客戶端的JavaScript
<script>
$.ajax({
url : 'http://webserver/retrieveJSON.php',
dataType : 'json',
type : 'get',
success : function(Result){
//ResultAlert = JSON.stringify(Result);
//alert(ResultAlert);
}
});
</script>
當我字符串化的結果我得到下面的JSON對象的長版:
{"customers":[{"customer":{"id":"1","customerName":"Customer Alpha","customerID":" custA","customerAddress":" Alpha Way","customerCity":" Alpha","customerState":" AL","customerZip":"91605"}},{"customer":{"id":"2","customerName":"Customer Beta","customerID":" CustB","customerAddress":" Beta Street","customerCity":" Beta","customerState":" BE","customerZip":"91605"}}]}
我有一個數據庫和下面的插入功能已經設置。
function insertCustomer(customerName, customerID, customerAddress, customerCity, customerState, customerZip) {
db.transaction(function (tx) {
tx.executeSql('INSERT INTO Customers (customerName, customerID, customerAddress, customerCity, customerState, customerZip) VALUES (?, ?, ?, ?, ?, ?)', [customerName, customerID, customerAddress, customerCity, customerState, customerZip],CountReturns);
});
};
如何將對象轉回到數組中,以便我可以遍歷它並將每個字段插入到Sqlite數據庫中?換句話說,我用什麼替換// stringify部分?我希望這個問題不是太本地化,但我試圖闡明整個過程,以防其他人試圖做同樣的事情。該流程中的任何其他提示或建議也受到歡迎。謝謝。
感謝Quentin我能夠訪問數組並使用下面的代碼來代替stringify輸入字段到數據庫中;然而,由於某種原因,所有的輸入都是不確定的。
for (var i = 0, len = Result.customers.length; i < len; ++i) {
var customer = Result.customers[i];
insertCustomer(customer.customerName, customer.customerID, customer.customerAddress, customer.customerCity, customer.customerState, customer.customerZip);
}
您正在尋找['JSON.parse'](https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/JSON/parse)。 – 2013-05-03 16:32:13