語法JSON陣列

爲了在網格中輸出它，我必須有像如下所示的JSON語法：語法JSON陣列

{ 
"data": [ 
    { 
     "SupplierID": 1, 
     "CompanyName": "Exotic Liquids", 
     "ContactName": "Charlotte Cooper", 
     "ContactTitle": "Purchasing Manager", 
     "Address": "49 Gilbert St.", 
     "City": "London", 
     "Region": null, 
     "PostalCode": "EC1 4SD", 
     "Country": "UK", 
     "Phone": "(171) 555-2222", 
     "Fax": null, 
     "HomePage": null 
    }, 
    { 
     "SupplierID": 2, 
     "CompanyName": "New Orleans Cajun Delights", 
     "ContactName": "Shelley Burke", 
     "ContactTitle": "Order Administrator", 
     "Address": "P.O. Box 78934", 
     "City": "New Orleans", 
     "Region": "LA", 
     "PostalCode": "70117", 
     "Country": "USA", 
     "Phone": "(100) 555-4822", 
     "Fax": null, 
     "HomePage": "#CAJUN.HTM#" 
    },....more data... 
     ] 
}

這是我使用的代碼：

mysql_connect('localhost','root','')or die(mysql_error()); 
mysql_select_db('testdb')or die(mysql_error()); 
$result = mysql_query("select * from city"); 
$data[]=array(); 
while($rows=mysql_fetch_array($result,MYSQL_ASSOC)){ 
$jsondata=json_encode($rows); 
echo $data[$jsondata]; 
}

但我有一定的錯誤。 Notice: Undefined index:

我怎樣才能解決這個問題？，感謝

來源

2012-02-24 tomexsans

您編碼數據過早。通常你應該構建整個多維陣列，然後去編碼它

mysql_select_db('testdb') or die(mysql_error()); 
$result = mysql_query("select * from city"); 

$data = array('data' => array()); 
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { 
    $data['data'][] = $row; 
} 

echo json_encode($data);

來源

2012-02-24 14:32:39 gintas

是啊！多數民衆贊成它，我忘記了一個多陣列多麼愚蠢的我。謝謝.. – tomexsans 2012-02-24 14:41:07

你需要的編碼數據添加到陣列第一

$data =array(); 
while($rows=mysql_fetch_array($result,MYSQL_ASSOC)){ 
$jsondata=json_encode($rows); 
$data[] = $jsondata; 
}

現在你可以重複整個事情（有可能是一個更簡單的方法）

echo explode($data, ',');

另外，我想你可能能夠一次編碼整個數組。

來源

2012-02-24 14:32:13 mindandmedia

語法JSON陣列

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