代码:
'我有一个PHP代码,首先计算从DB检索图像参数是否成功,然后插入响应3个参数(Id,名称,url)为每个图像。然后它返回json格式的响应。我收集响应在android的字符串的形式,然后我需要将其转换为jsonArray。当转换时,我得到异常。
PHP代码:
<?php
//Importing dbdetails file
require_once 'dbDetails.php';
//connection to database
$con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect...');
//sql query to fetch all images
$sql = "SELECT * FROM MyImages";
//getting images
$result = mysqli_query($con,$sql);
//response array
$response = array();
$response['error'] = false;
$response['images'] = array();
//traversing through all the rows
while($row = mysqli_fetch_array($result)){
$temp = array();
$temp['id']=$row['id'];
$temp['name']=$row['name'];
$temp['url']=$row['url'];
array_push($response['images'],$temp);
}
//displaying the response
echo json_encode($response);
?>
result是下面代码中PHP代码的响应
android code:
@Override protected void onPostExecute(String result) {
super.onPostExecute(result);
if (pd.isShowing()){
pd.dismiss();
}
txtJson.setText(result);
try {
JSONArray jsonArr = new JSONArray(result);
showGrid(jsonArr);
} catch (JSONException e) {
e.printStackTrace();
txtJson.setText("Exception in jsonArry");
}
}
那么如何正确地做呢?
响应字符串是->
{"error":false,"images":[{"id":"1","name":"1","url":"https://swulj.000webhostapp.com/AndroidUploadImage/uploads/1.jpg"},{"id":"2","name":"2","url":"https://swulj.000webhostapp.com/AndroidUploadImage/uploads/2.jpg"},{"id":"3","name":"3","url":"https://swulj.000webhostapp.com/AndroidUploadImage/uploads/3.jpg"},{"id":"4","name":"ygff","url":"https://swulj.000webhostapp.com/AndroidUploadImage/uploads/4.jpg"}]}
这是正确的json格式吗?我该怎么办?
你的android代码应该是这样的
protected void onPostExecute(String result) {
super.onPostExecute(result);
if (pd.isShowing()){
pd.dismiss();
}
txtJson.setText(result);
try {
JSONObject jsonObj = new JSONObject(result);
showGrid(jsonObj.optJsonArray("images"));
} catch (JSONException e) {
e.printStackTrace();
txtJson.setText("Exception in jsonArry");
}
}
由于字符串响应没有被"["]",我包围了它,没有例外发生。
代码:
protected void onPostExecute(String result) {
super.onPostExecute(result);
if (pd.isShowing()){
pd.dismiss();
}
txtJson.setText(result);
try {
String res = "[" + result + "]";
JSONArray jsonArr = new JSONArray(res);
showGrid(jsonArr);
/*JSONObject jsonObj = new JSONObject(result);
showGrid(jsonObj.optJsonArray("images"));*/
} catch (Exception e) {
e.printStackTrace();
txtJson.setText("Exception in jsonArry");
}
}