我在使用 php-mysql 和 JSON 时收到此错误，说真的，我花了 2 天，所以这是我的代码
require "includes/connexion.php";
$requete="SELECT * FROM contacts;";
$rep=$pdo->query($requete);
while($data=$rep->fetch(PDO::FETCH_ASSOC)){
$sortie[]=$data;
}
print(json_encode($sortie));
$rep->closeCursor();
下面是 Java 代码的一部分，结果参数中的 jsonArray 包含该值（来自控制台）：
[{"id":"1","nom":"Fedson Dorvilme","email":"fedsondorvilme@gmail.com","phone":"34978238","img":"ic_launcher"},{"id":"2","nom":"Darlene Aurelien","email":"darleneaurelien@yahoo.fr","phone":"34171191","img":"ic_launcher"}]
JSONArray array = new JSONArray(result);
for (int i = 0; i < array.length(); i++) {
JSONObject json_data = array.getJSONObject(i);
Log.i("MyLogFlag", "Nom Contact: " + json_data.getString("nom"));
returnString = "'n't" + array.getJSONArray(i);
System.out.println(" ********** RS [ "+returnString+" ]****************");
}
} catch (Exception e) {
Log.e("log_Exception_tag", "Error parsing data " + e.toString());
}
提前感谢您的帮助
这是你的 json 结构：
[
{
"id": "1",
"nom": "Fedson Dorvilme",
"email": "fedsondorvilme@gmail.com",
"phone": "34978238",
"img": "ic_launcher"
},
{
"id": "2",
"nom": "Darlene Aurelien",
"email": "darleneaurelien@yahoo.fr",
"phone": "34171191",
"img": "ic_launcher"
}
]
从您的 Java 代码：
JSONArray array = new JSONArray(result); // OK
for (int i = 0; i < array.length(); i++) {
JSONObject json_data = array.getJSONObject(i); // OK
returnString = "'n't" + array.getJSONArray(i); // ??? wrong!!
}
您尝试从数组中获取数组，但它JSONObject。
正确的方式：
JSONArray gb = new JSONArray(str);
for (int j = 0; j < gb.length(); j++) {
JSONObject element = gb.getJSONObject(j);
int id = element.getInt("id");
int phone = element.getInt("phone");
String nom = element.getString("nom");
String email = element.getString("email");
String img = element.getString("img");
}