我正在：
W / System.err：org.json.JSONException：值
这是我的JAVA代码：
public void onResponse(String response) {
try {
JSONObject jsonResponse = new JSONObject(response);
boolean success = jsonResponse.getBoolean("success");
if (success){
String first_name = jsonResponse.getString("first_name");
String last_name = jsonResponse.getString("last_name");
Intent intent = new Intent(LoginActivity.this, HomePageActivity.class);
intent.putExtra("first_name", first_name);
intent.putExtra("last_name", last_name);
}else {
AlertDialog.Builder builder= new AlertDialog.Builder(LoginActivity.this);
builder.setMessage("Login Failed")
.setNegativeButton("Retry", null)
.create()
.show();
}
} catch (JSONException e) {
e.printStackTrace();
}
}
我的PHP代码：
<?php
$con = mysqli_connect("*****", "********", "********", "*********");
$email = $_POST["email"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "SELECT * FROM user WHERE email = ? AND password = ?");
mysqli_stmt_bind_param($statement, "ss", $email, $password);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $userid, $first_name, $last_name, $email, $password);
$response = array();
$response["success"] = false;
while(mysqli_stmt_fetch($statement)){
$response["success"] = true;
$response["first_name"] = $first_name;
$response["last_name"] = $last_name;
$response["email"] = $email;
$response["password"] = $password;
}
echo json_encode($response);
?>
上面问题中的调试错误 这是Debug：
在com.thecreds.credsapp.LoginActivity $ 2 $ 1.onResponse（LoginActivity.java:47） 在com.thecreds.credsapp.LoginActivity $ 2 $ 1.onResponse（LoginActivity.java:43）
而且，仍然不知道我哪里出了错，我按照这些步骤操作了，但是它不起作用。