如何排序和显示数组以在 unity C# 中创建排行榜
How do I sort and display an array to create a leaderboards in unity C#
目前我遇到了一些问题,我必须与员工一起使用数据库。这些员工有不同的数据,这些是字符串和整数,数组中的示例是:Naam:Pino | Team:2 | VerkoopA:12 | VerkoopB:12 | NPS:40 | conversie:10。
这是一个只有一名员工的示例,数据库中还有更多员工,我已经在代码中拆分了他们。我想做的是用这些数据制作一个排行榜,但是我不知道如何分离数组中的不同项目,如果我已经分离了这个,我该如何对数组进行排序以查看谁拥有最多的 saleA 以及谁拥有saleB 最多,所以我可以将其设为带有过滤器的排行榜。
还有,有没有一种简单的方法可以制作一个可以显示一定数量员工的可视化排行榜?因为我现在显示它的方式它什么也没有显示。
我正在使用 Unity 和我自己的服务器 Xampp
谨致问候,
Dh
脚本:
using UnityEngine;
using UnityEngine.UI;
using System.Collections;
public class Leaderboard : MonoBehaviour
{
// Array of all the employees
public string[] leaderboard;
Text txt;
IEnumerator Start()
{
// Getting the leaderboard data from mySQL.
WWW leaderboardsData = new WWW("http://localhost/leaderboards.php");
yield return leaderboardsData;
// Split every employee in the Array so they are all apart from each other. |
// an example Message Naam:Pino | Team:2 | VerkoopA:12 | VerkoopB:12 | NPS:40 | Conversie:10
string leaderboardsDataString = leaderboardsData.text;
leaderboard = leaderboardsDataString.Split(';');
// for every employee in the array print every employee's data.
foreach (string employee in leaderboard)
{
string employeeApart = employee;
DisplayLeaderboards(employeeApart);
}
}
// Display the leaderboards 1 by 1.
void DisplayLeaderboards(string employeeApart)
{
print(employeeApart);
txt = gameObject.GetComponent<Text>();
txt.text = employeeApart;
}
string GetDataValue(string data, string index)
{
string value = data.Substring(data.IndexOf(index) + index.Length);
if (value.Contains("|")) value = value.Remove(value.IndexOf("|"));
return value;
}
}
PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "test3";
//Make Connection
$conn = new mysqli($servername, $username, $password, $dbName);
//Check Connection
if(!$conn){
die("Connection Failed. ". mysqli_connect_error());
}
$sql = "SELECT ID, Team, VerkoopA, VerkoopB, NPS, Conversie FROM Leaderboards";
$result = mysqli_query($conn ,$sql);
if(mysqli_num_rows($result) > 0){
//show data for each row
while($row = mysqli_fetch_assoc($result)){
echo "ID:".$row['ID'] . "|Team:".$row['Team']. "|VerkoopA:".$row['VerkoopA']. "|VerkoopB:".$row['VerkoopB'] . "|NPS:".$row['NPS']. "|Conversie:".$row['Conversie'] . ";";
}
}
?>
编辑:
经过程序员的帮助,我决定使用json
目前我遇到了一些问题,我必须与员工一起使用数据库。这些员工有不同的数据 我的 PHP 文件的输出是:
[{"ID":"1","Team":"2","VerkoopA":"35","VerkoopB":"12","NPS":"25","Conversie":"18"},{"ID":"2","Team":"1","VerkoopA":"55","VerkoopB":"2","NPS":"12","Conversie":"40"},{"ID":"3","Team":"2","VerkoopA":"12","VerkoopB":"12","NPS":"40","Conversie":"10"}]
我想做的是用这些数据制作一个排行榜,但是我不知道如何分隔数组中的不同项目通过使用 JsonHelper 我得到一个错误:ArgumentException: JSON must表示一个对象
脚本:
using UnityEngine;
using UnityEngine.UI;
using System.Collections;
public class Leaderboard: MonoBehaviour
{
// Array of all the employees
[System.Serializable]
public class LeaderBoard
{
public string ID;
public string Team;
public string VerkoopA;
public string VerkoopB;
public string NPS;
public string Conversie;
}
Text txt;
IEnumerator Start()
{
// Getting the leaderboard data from mySQL.
WWW leaderboardsData = new WWW("http://localhost/leaderboards.php");
yield return leaderboardsData;
string leaderboardsDataString = leaderboardsData.text;
print(leaderboardsDataString);
LeaderBoard[] leaderboard;
leaderboard = JsonHelper.FromJson<LeaderBoard>(leaderboardsDataString);
// for every employee in the array print every employee's data.
foreach (LeaderBoard employee in leaderboard)
{
DisplayLeaderboards(employee);
}
}
// Display the leaderboards 1 by 1.
void DisplayLeaderboards(LeaderBoard employeeApart)
{
Debug.Log("ID: " + employeeApart.ID);
Debug.Log("Team: " + employeeApart.Team);
Debug.Log("VerkoopA: " + employeeApart.VerkoopA);
Debug.Log("VerkoopB: " + employeeApart.VerkoopB);
Debug.Log("NPS: " + employeeApart.NPS);
Debug.Log("Conversie: " + employeeApart.Conversie);
}
}
PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "test3";
//Make Connection
$conn = new mysqli($servername, $username, $password, $dbName);
//Check Connection
if(!$conn){
die("Connection Failed. ". mysqli_connect_error());
}
$sql = "select * from Leaderboards";
$result = mysqli_query($conn, $sql) or die("Error in Selecting " . mysqli_error($conn));
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
?>
由于您可以访问服务器和服务器脚本(php),因此您应该使用json
或xml
来执行此操作,而不是使用[=18= 分隔数据].
从数据库发送数据为 json。在 Unity 端,使用 WWW
class 接收数据,然后使用 JsonUtility
将 json 转换回 class。
Unity 的 JsonUtility
不支持数组,因此您需要一个针对它的包装器。您可以获得 JsonHelper
class 允许 json 数组 .
我上面描述的应该是这样的:
[System.Serializable]
public class LeaderBoard
{
public string ID;
public string Team;
public string VerkoopA;
public string VerkoopB;
public string NPS;
public string Conversie;
}
Text txt;
IEnumerator _Start()
{
// Getting the leaderboard data from mySQL.
WWW leaderboardsData = new WWW("http://localhost/leaderboards.php");
yield return leaderboardsData;
string leaderboardsDataString = leaderboardsData.text;
LeaderBoard[] leaderboard = JsonHelper.FromJson<LeaderBoard>(leaderboardsDataString);
// for every employee in the array print every employee's data.
foreach (LeaderBoard employee in leaderboard)
{
DisplayLeaderboards(employee);
}
}
// Display the leaderboards 1 by 1.
void DisplayLeaderboards(LeaderBoard employeeApart)
{
Debug.Log("ID: " + employeeApart.ID);
Debug.Log("Team: " + employeeApart.Team);
Debug.Log("VerkoopA: " + employeeApart.VerkoopA);
Debug.Log("VerkoopB: " + employeeApart.VerkoopB);
Debug.Log("NPS: " + employeeApart.NPS);
Debug.Log("Conversie: " + employeeApart.Conversie);
}
在php端,您必须使用json_encode
函数将您的数据库信息编码为json数组,然后再将其发送到Unity。不是 php 人,但你可以了解更多 here and here。
正在对结果进行排序:
您可以在使用
从服务器接收数据后按ID
排序
leaderboard = leaderboard.OrderBy(c => c.ID).ToArray();
或NPS
leaderboard = leaderboard.OrderBy(c => c.NPS).ToArray();
这必须在 foreach
循环之前完成。
编辑:
php 生成的 json 无法直接在 Unity 中准备就绪。不擅长 php 并且无法修复 php 方面的 json 数据。虽然,我可以在 Unity 端修复它。
在接收到的字符串前添加{"Items":
,然后在最后添加}
。
我为此做了一个简单的函数:
string fixJson(string value)
{
value = "{\"Items\":" + value + "}";
return value;
}
将该函数包含到您的脚本中,然后只需替换
string leaderboardsDataString = leaderboardsData.text;
与
string leaderboardsDataString = fixJson(leaderboardsData.text);
现在一切正常。
目前我遇到了一些问题,我必须与员工一起使用数据库。这些员工有不同的数据,这些是字符串和整数,数组中的示例是:Naam:Pino | Team:2 | VerkoopA:12 | VerkoopB:12 | NPS:40 | conversie:10。
这是一个只有一名员工的示例,数据库中还有更多员工,我已经在代码中拆分了他们。我想做的是用这些数据制作一个排行榜,但是我不知道如何分离数组中的不同项目,如果我已经分离了这个,我该如何对数组进行排序以查看谁拥有最多的 saleA 以及谁拥有saleB 最多,所以我可以将其设为带有过滤器的排行榜。
还有,有没有一种简单的方法可以制作一个可以显示一定数量员工的可视化排行榜?因为我现在显示它的方式它什么也没有显示。
我正在使用 Unity 和我自己的服务器 Xampp
谨致问候,
Dh
脚本:
using UnityEngine;
using UnityEngine.UI;
using System.Collections;
public class Leaderboard : MonoBehaviour
{
// Array of all the employees
public string[] leaderboard;
Text txt;
IEnumerator Start()
{
// Getting the leaderboard data from mySQL.
WWW leaderboardsData = new WWW("http://localhost/leaderboards.php");
yield return leaderboardsData;
// Split every employee in the Array so they are all apart from each other. |
// an example Message Naam:Pino | Team:2 | VerkoopA:12 | VerkoopB:12 | NPS:40 | Conversie:10
string leaderboardsDataString = leaderboardsData.text;
leaderboard = leaderboardsDataString.Split(';');
// for every employee in the array print every employee's data.
foreach (string employee in leaderboard)
{
string employeeApart = employee;
DisplayLeaderboards(employeeApart);
}
}
// Display the leaderboards 1 by 1.
void DisplayLeaderboards(string employeeApart)
{
print(employeeApart);
txt = gameObject.GetComponent<Text>();
txt.text = employeeApart;
}
string GetDataValue(string data, string index)
{
string value = data.Substring(data.IndexOf(index) + index.Length);
if (value.Contains("|")) value = value.Remove(value.IndexOf("|"));
return value;
}
}
PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "test3";
//Make Connection
$conn = new mysqli($servername, $username, $password, $dbName);
//Check Connection
if(!$conn){
die("Connection Failed. ". mysqli_connect_error());
}
$sql = "SELECT ID, Team, VerkoopA, VerkoopB, NPS, Conversie FROM Leaderboards";
$result = mysqli_query($conn ,$sql);
if(mysqli_num_rows($result) > 0){
//show data for each row
while($row = mysqli_fetch_assoc($result)){
echo "ID:".$row['ID'] . "|Team:".$row['Team']. "|VerkoopA:".$row['VerkoopA']. "|VerkoopB:".$row['VerkoopB'] . "|NPS:".$row['NPS']. "|Conversie:".$row['Conversie'] . ";";
}
}
?>
编辑:
经过程序员的帮助,我决定使用json
目前我遇到了一些问题,我必须与员工一起使用数据库。这些员工有不同的数据 我的 PHP 文件的输出是:
[{"ID":"1","Team":"2","VerkoopA":"35","VerkoopB":"12","NPS":"25","Conversie":"18"},{"ID":"2","Team":"1","VerkoopA":"55","VerkoopB":"2","NPS":"12","Conversie":"40"},{"ID":"3","Team":"2","VerkoopA":"12","VerkoopB":"12","NPS":"40","Conversie":"10"}]
我想做的是用这些数据制作一个排行榜,但是我不知道如何分隔数组中的不同项目通过使用 JsonHelper 我得到一个错误:ArgumentException: JSON must表示一个对象
脚本:
using UnityEngine;
using UnityEngine.UI;
using System.Collections;
public class Leaderboard: MonoBehaviour
{
// Array of all the employees
[System.Serializable]
public class LeaderBoard
{
public string ID;
public string Team;
public string VerkoopA;
public string VerkoopB;
public string NPS;
public string Conversie;
}
Text txt;
IEnumerator Start()
{
// Getting the leaderboard data from mySQL.
WWW leaderboardsData = new WWW("http://localhost/leaderboards.php");
yield return leaderboardsData;
string leaderboardsDataString = leaderboardsData.text;
print(leaderboardsDataString);
LeaderBoard[] leaderboard;
leaderboard = JsonHelper.FromJson<LeaderBoard>(leaderboardsDataString);
// for every employee in the array print every employee's data.
foreach (LeaderBoard employee in leaderboard)
{
DisplayLeaderboards(employee);
}
}
// Display the leaderboards 1 by 1.
void DisplayLeaderboards(LeaderBoard employeeApart)
{
Debug.Log("ID: " + employeeApart.ID);
Debug.Log("Team: " + employeeApart.Team);
Debug.Log("VerkoopA: " + employeeApart.VerkoopA);
Debug.Log("VerkoopB: " + employeeApart.VerkoopB);
Debug.Log("NPS: " + employeeApart.NPS);
Debug.Log("Conversie: " + employeeApart.Conversie);
}
}
PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbName = "test3";
//Make Connection
$conn = new mysqli($servername, $username, $password, $dbName);
//Check Connection
if(!$conn){
die("Connection Failed. ". mysqli_connect_error());
}
$sql = "select * from Leaderboards";
$result = mysqli_query($conn, $sql) or die("Error in Selecting " . mysqli_error($conn));
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
?>
由于您可以访问服务器和服务器脚本(php),因此您应该使用json
或xml
来执行此操作,而不是使用[=18= 分隔数据].
从数据库发送数据为 json。在 Unity 端,使用 WWW
class 接收数据,然后使用 JsonUtility
将 json 转换回 class。
Unity 的 JsonUtility
不支持数组,因此您需要一个针对它的包装器。您可以获得 JsonHelper
class 允许 json 数组
我上面描述的应该是这样的:
[System.Serializable]
public class LeaderBoard
{
public string ID;
public string Team;
public string VerkoopA;
public string VerkoopB;
public string NPS;
public string Conversie;
}
Text txt;
IEnumerator _Start()
{
// Getting the leaderboard data from mySQL.
WWW leaderboardsData = new WWW("http://localhost/leaderboards.php");
yield return leaderboardsData;
string leaderboardsDataString = leaderboardsData.text;
LeaderBoard[] leaderboard = JsonHelper.FromJson<LeaderBoard>(leaderboardsDataString);
// for every employee in the array print every employee's data.
foreach (LeaderBoard employee in leaderboard)
{
DisplayLeaderboards(employee);
}
}
// Display the leaderboards 1 by 1.
void DisplayLeaderboards(LeaderBoard employeeApart)
{
Debug.Log("ID: " + employeeApart.ID);
Debug.Log("Team: " + employeeApart.Team);
Debug.Log("VerkoopA: " + employeeApart.VerkoopA);
Debug.Log("VerkoopB: " + employeeApart.VerkoopB);
Debug.Log("NPS: " + employeeApart.NPS);
Debug.Log("Conversie: " + employeeApart.Conversie);
}
在php端,您必须使用json_encode
函数将您的数据库信息编码为json数组,然后再将其发送到Unity。不是 php 人,但你可以了解更多 here and here。
正在对结果进行排序:
您可以在使用
从服务器接收数据后按ID
排序
leaderboard = leaderboard.OrderBy(c => c.ID).ToArray();
或NPS
leaderboard = leaderboard.OrderBy(c => c.NPS).ToArray();
这必须在 foreach
循环之前完成。
编辑:
php 生成的 json 无法直接在 Unity 中准备就绪。不擅长 php 并且无法修复 php 方面的 json 数据。虽然,我可以在 Unity 端修复它。
在接收到的字符串前添加{"Items":
,然后在最后添加}
。
我为此做了一个简单的函数:
string fixJson(string value)
{
value = "{\"Items\":" + value + "}";
return value;
}
将该函数包含到您的脚本中,然后只需替换
string leaderboardsDataString = leaderboardsData.text;
与
string leaderboardsDataString = fixJson(leaderboardsData.text);
现在一切正常。