我有问题。我现在几周试图从数据库中获取数据,将其编码为JSON,然后将其发送回我的iOS应用程序。问题是,每次JSON无效时都说http://jsonviewer.stack.hu/ 这是我现在的代码:
//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($testjevervolg));
}
}
这就是它的回报:
{
"klanten": [
{
"deb_nr": "10010",
"name": "ERIKA Handelsonderneming",
"adrs": "Aan de Heibloem 17"
}
]
}{
"klanten": [
{
"deb_nr": "25071",
"name": "Afdeling Heffing & Invordering",
"adrs": "Postbus 1275"
}
]
}{
"klanten": [
{
"deb_nr": "25247",
"name": "v.d. Heerik b.v.",
"adrs": "Flemingstraat 3-5"
}
]
}{
"klanten": [
{
"deb_nr": "25454",
"name": "Toering Automatisering",
"adrs": "Appelhof 17a"
}
]
}{
"klanten": [
{
"deb_nr": "25601",
"name": "Ratering Bouw & Industrie",
"adrs": "de Hogenkamp 1"
}
]
}
这就是我得到的。问题是,应该有一个'klanten'数组 应该每个deb_nr,名称和adrs都是。现在每件事都有自己的'Klanten' 怎么能解决这个问题?
感谢。
答案 0 :(得分:1)
我使用此函数将JSON返回到我的应用程序:
function sql2json($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;
}
答案 1 :(得分:0)
创建了这个但尚未测试过,目前只使用textwrangler它可以帮助你前进。
你可以尝试这样的事情:
$completeJson = array();
while($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));
答案 2 :(得分:0)
为了获得以下类型的研究:
[
{
"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);
为了获得以下类型的研究:
[
{
"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其他类型的反复,所以如果你想要其他类型的输出,请告诉我。