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java - 无法用 PHP 更新 MySQL

转载 作者:行者123 更新时间:2023-11-29 03:29:47 26 4
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我正在制作一个使用 MySQL 和 PHP 进行注册的 Android 应用程序,但我的 MySQL 数据库没有更新。我正在使用 php 方法,但没有成功更新表格。这是 java 代码的重要 fragment :

public void updateGroupName() {
submitButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
EditText groupName = (EditText) dialog.findViewById(R.id.group);
final String name = groupName.getText().toString();
System.out.println("Name of group is: " + name);
updateUser(name);
System.out.println("Trying to update user group");
if (!name.isEmpty()) {

//add to sqlite database
updateUser(name);
} else {
Toast.makeText(getApplicationContext(),
"Please enter your details!", Toast.LENGTH_LONG)
.show();
}
String tag_string_req = "req_update";

//showDialog();

StringRequest strReq = new StringRequest(Request.Method.POST,
AppConfig.URL_REGISTER, new Response.Listener<String>() {

@Override
public void onResponse(String response) {
Log.d(TAG, "Update Response: " + response.toString());
//hideDialog();

try {
JSONObject jObj = new JSONObject(response);
boolean error = jObj.getBoolean("error");
if (!error) {
System.out.println("User group successfully updated in mysql");
// User successfully stored in MySQL
} else {
System.out.println("The use group failed to update");
// Error occurred in registration. Get the error
// message
String errorMsg = jObj.getString("error_msg");
Toast.makeText(getApplicationContext(),
errorMsg, Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
e.printStackTrace();
}

}
}, new Response.ErrorListener() {

@Override
public void onErrorResponse(VolleyError error) {
Log.e(TAG, "Update Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_LONG).show();
// hideDialog();
}
}) {
String email = db.getUserDetails().get("email");
@Override
protected Map<String, String> getParams() {
System.out.println("trying to return params");
// Posting params to register url
Map<String, String> params = new HashMap<>();
params.put("tag", "update");
params.put("user_group", name);
params.put("email", email);
return params;
}
};
System.out.println("updating group to mysql database");
// Adding request to request queue
AppController.getInstance().addToRequestQueue(strReq, tag_string_req);
// System.out.println("here1");
// System.out.println("user group: " + db.getUserDetails().get("user_group"));
dialog.dismiss();
}
});
//System.out.println("here2");
}

以下是我的 PHP 代码的重要部分:

else if ($tag == "update") {
$user_group = $_POST['user_group'];
$uid = $_POST['email'];
//update user
$user = $db->updateUser($user_group, $email);
if ($user) {
$response["error"] = FALSE;
$response["uid"] = $user["unique_id"];
$response["user"]["name"] = $user["name"];
$response["user"]["email"] = $user["email"];
$response["user"]["user_group"] = ["user"]["user_group"];
$response["user"]["created_at"] = $user["created_at"];
$response["user"]["updated_at"] = $user["updated_at"];
echo "The update User function was called and a user was returned";
echo json_encode($response);
} else {
// user failed to store
$response["error"] = TRUE;
$response["error_msg"] = "Error occured in Updating user group";
echo json_encode($response);
}

public function updateUser($user_group, $email) {
$result = mysql_query("UPDATE users SET user_group = '$user_group' WHERE email = '$email'");
if ($result) {
// get user details
$uid = mysql_insert_id(); // last inserted id
$result = mysql_query("SELECT * FROM users WHERE uid = $uid");
// return user details
return mysql_fetch_array($result);
} else {
return false;
}
}

最佳答案

您正在将 $email 传递给您的方法 updateUser,但电子邮件地址存储在 $uid 中。

请更改

$user = $db->updateUser($user_group, $email);

$user = $db->updateUser($user_group, $uid);

关于java - 无法用 PHP 更新 MySQL,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31933314/

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