如何将有效的JSON发送回iOS应用程序？

Question

我有个问题。几个星期以来，我一直在尝试从数据库获取数据，将其编码为JSON，然后将其发送回我的iOS应用程序。问题是，每当JSON无效时，http://jsonviewer.stack.hu/会说，这是我现在拥有的代码：

//connection to the database
 $dbhandle = mysql_connect($hostname, $username, $password) 
 or die("Unable to connect to MySQL");
 //echo "Connected to MySQL<br>";


//select a database to work with
$selected = mysql_select_db("test",$dbhandle) 
or die("Could not select examples");

$result = mysql_query("SELECT * FROM test.debiteur WHERE SORT_NAAM LIKE '%eri%'");



while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

$deb_nr['deb_nr'] = $row['DEB_NR'];
$deb_naam['name'] = $row['DEB_NAAM'];
$deb_adres['adrs'] = $row['DEB_ADRES'];


$testje = array_merge($deb_nr, $deb_naam, $deb_adres);

$testjevervolg = array('klanten' => array($testje));

 sendResponse(200, json_encode($testjevervolg));
 }
 }

下面是它返回的内容：

{
"klanten": [
{
  "deb_nr": "10010",
  "name": "ERIKA Handelsonderneming",
  "adrs": "Aan de Heibloem 17"
}
]
}{
"klanten": [
{
  "deb_nr": "25071",
  "name": "Afdeling Heffing & Invordering",
  "adrs": "Postbus 1275"
}
]
}{
"klanten": [
{
  "deb_nr": "25247",
  "name": "v.d. Heerik b.v.",
  "adrs": "Flemingstraat 3-5"
}
]
}{
"klanten": [
{
  "deb_nr": "25454",
  "name": "Toering Automatisering",
  "adrs": "Appelhof 17a"
}
]
}{
"klanten": [
{
  "deb_nr": "25601",
  "name": "Ratering Bouw & Industrie",
  "adrs": "de Hogenkamp 1"
}
]
}

这就是我所得到的。问题是，应该有一个'klanten‘数组，并且应该有每个deb_nr、名称和adrs。现在每个事物都有了自己的“Klanten”，怎么解决这个问题呢？

谢谢。

Answer 1

创建了这个，但还没有测试，目前只是使用textwrangler，也许它可以帮助您继续前进。

你能试试这样的方法吗：

$completeJson = array();


while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

$deb_nr['deb_nr'] = $row['DEB_NR'];
$deb_nr['name'] = $row['DEB_NAAM'];
$deb_nr['adrs'] = $row['DEB_ADRES'];


 array_push($completeJson,$deb_nr);


 }
 $testjevervolg = array('klanten' => $completeJson);
 sendResponse(200, json_encode($testjevervolg));

Answer 2

我使用此函数将JSON返回给我的应用程序：

function sql2json($query) {
 $data_sql = mysql_query($query) or die("'';//" . mysql_error());
 $json_str = ""; 
 if($total = mysql_num_rows($data_sql)) { 
   $json_str .= "[\n";
    $row_count = 0;    
    while($data = mysql_fetch_assoc($data_sql)) {
        if(count($data) > 1) $json_str .= "{\n";
        $count = 0;
        foreach($data as $key => $value) {
         if(count($data) > 1) $json_str .= "\"$key\":\"$value\"";
         else $json_str .= "\"$value\"";
            $count++;
            if($count < count($data)) $json_str .= ",\n";
        }
        $row_count++;
        if(count($data) > 1) $json_str .= "}\n";
        if($row_count < $total) $json_str .= ",\n";
    }
   $json_str .= "]\n";
  }
  mysql_free_result($data_sql);
  return $json_str;
}

Answer 3

为了得到以下类型的结果：

[
    {
        "deb_nr": "25071",
        "name": "Afdeling Heffing & Invordering",
          "adrs": "Postbus 1275"
    },
    {
        "deb_nr": "25071",
        "name": "Afdeling Heffing & Invordering",
          "adrs": "Postbus 1275"
    },
    ...
]

=============================================

$result = array();
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$result[] = array(
  "deb_nr" => $row['DEB_NR'],
  "name" => $row['DEB_NAAM'],
  "adrs" => $row['DEB_ADRES'],
);
}

header('Content-type: application/json');
echo json_encode($result);

为了得到以下类型的结果：

[
    {
        "klanten" : {
            "deb_nr": "25071",
            "name": "Afdeling Heffing & Invordering",
              "adrs": "Postbus 1275"
        },
    }
    {
        "klanten" : {
            "deb_nr": "25071",
            "name": "Afdeling Heffing & Invordering",
              "adrs": "Postbus 1275"
        },
    },
    ...
]

=============================================

$result = array();
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$result[]["klanten"] = array(
  "deb_nr" => $row['DEB_NR'],
  "name" => $row['DEB_NAAM'],
  "adrs" => $row['DEB_ADRES'],
);
}

header('Content-type: application/json');
echo json_encode($result);

我不能假设你想要的是其他类型的响应，所以如果你想要其他类型的输出，请让我知道。