如何修复，输入字符串格式不正确 unity 3d 和 serial

Question

我正在使用 unity 3d 来可视化 Arduino 的串行输出并不断收到错误

FormatException: Input string was not in a correct format. System.Number.StringToNumber (System.String str, System.Globalization.NumberStyles options, System.Number+NumberBuffer& number, System.Globalization.NumberFormatInfo info, System.Boolean parseDecimal) (at <599589bf4ce248909b8a14cbe4a2034e>:0) System.Number.ParseInt32 (System.String s, System.Globalization.NumberStyles style, System.Globalization.NumberFormatInfo info) (at <599589bf4ce248909b8a14cbe4a2034e>:0) System.Int32.Parse (System.String s, System.IFormatProvider provider) (at <599589bf4ce248909b8a14cbe4a2034e>:0) System.Convert.ToInt32 (System.String value) (at <599589bf4ce248909b8a14cbe4a2034e>:0) One_mputest.Update () (at Assets/One_mputest.cs:48)

我试过使用插件，但没有效果

C# 脚本：

using System.Collections;
using System.Collections.Generic;
using UnityEngine;
using System.IO.Ports;
using System;
using System.Linq;

public class One_mputest : MonoBehaviour {

private int x;
private int y;
private int z;

private int roll;
private int pitch;
private int yaw;

static SerialPort serial = new SerialPort("COM3", 9600);

// Use this for initialization
void Start () {

    serial.ReadTimeout = 5000;
    if (!serial.IsOpen) {
        try {
            serial.Open ();
        } catch (TimeoutException) {

        }

    } else
        Debug.LogError ("Port already open");
    }

// Update is called once per frame
void Update () {
    if (!serial.IsOpen)
        serial.Open ();

    string ypr = serial.ReadLine ();

    List<String> stringList = ypr.Split('\t').ToList();

    int[] intArray = new int[7];

    for (int s = 0; s < 7; s++)
    {
        intArray[s] = Convert.ToInt32(stringList[s]);
    }

    x = intArray[3];
    y = intArray[0];
    z = intArray[1];

    transform.localEulerAngles = new Vector3(z, y, x);
}

}

串行输入：

-104.65  -10     5   65536.00    0.00    0.00    0.00    
-102.74  -4  1   65536.00    0.00    0.00    0.00    
-102.19  -2  0   65536.00    0.00    0.00    0.00    
-102.09  0   0   65536.00    0.00    0.00    0.00    
-101.75  0   0   65536.00    0.00    0.00    0.00    
-101.61  0   -1  65536.00    0.00    0.00    0.00    
-101.62  0   -1  65536.00    0.00    0.00    0.00    
-101.68  0   -1  65536.00    0.00    0.00    0.00    
-101.76  0   -1  65536.00    0.00    0.00    0.00    
-102.02  0   -1  65536.00    0.00    0.00    0.00    
-102.20  0   -1  65536.00    0.00    0.00    0.00    
-102.33  0   -1  65536.00    0.00    0.00    0.00    
-102.43  0   -1  65536.00    0.00    0.00    0.00    
-102.57  0   -1  65536.00    0.00    0.00    0.00    
-102.77  0   -1  65536.00    0.00    0.00    0.00    
-102.90  0   -1  65536.00    0.00    0.00    0.00    
-102.68  0   -1  65536.00    0.00    0.00    0.00    
-102.48  0   -1  65536.00    0.00    0.00    0.00    
-102.26  0   -1  65536.00    0.00    0.00    0.00    
-102.01  0   -1  65536.00    0.00    0.00    0.00    
-101.83  0   -1  65536.00    0.00    0.00    0.00    
-102.15  0   -1  65536.00    0.00    0.00    0.00    
-102.34  0   -1  65536.00    0.00    0.00    0.00    
-102.17  0   -1  65536.00    0.00    0.00    0.00    
-101.89  0   -1  65536.00    0.00    0.00    0.00    
-101.74  0   -1  65536.00    0.00    0.00    0.00    
-101.88  0   -1  65536.00    0.00    0.00    0.00    
-101.95  0   -1  65536.00    0.00    0.00    0.00    
-102.01  0   -1  65536.00    0.00    0.00    0.00    
-102.05  0   -1  65536.00    0.00    0.00    0.00    
-102.06  0   -1  65536.00    0.00    0.00    0.00    
-102.00  0   -1  65536.00    0.00    0.00    0.00    
-101.96  0   -1  65536.00    0.00    0.00    0.00    
-102.06  0   -1  65536.00    0.00    0.00    0.00    
-102.14  0   -1  65536.00    0.00    0.00    0.00    
-102.37  0   -1  65536.00    0.00    0.00    0.00    
-102.81  0   -1  65536.00    0.00    0.00    0.00

我希望矢量变量能够更改并移动应用了脚本的对象

并且只收到关于以下内容的错误：输入字符串的格式不正确

Answer 1

Try & catch 是你的朋友。文件中可能有一个额外的 \t，而您正试图将 null 转换为某处的数字。另外一个好的提示是总是在 if 语句周围使用 {} 只是为了确保没有意外发生（例如：你认为某些东西在 if 语句内部但它在外部）

我要看的一点是，您正在尝试将代表 Float 的 String 转换为 Int32，这可能是问题所在。

尝试这样的事情：

void Update () {
    var whatdoing = "starting try";

    try {
        if (!serial.IsOpen) {
            serial.Open (); }

        whatdoing               = "reading line";
        string ypr              = serial.ReadLine ();

        List<String> stringList = ypr.Split('\t').ToList();
        int[] intArray          = new int[7];

         for (int s = 0; s < 7; s++) {
             whatdoing = "converting: " + stringList[s];

             if(!Int.TryParse(stringList[s], out intArray[s]) {
                 throw new Exception("Could not parse string to Int: "+whatdoing); } }
          whatdoing = "Assigning xyz";
          x         = intArray[3];
          y         = intArray[0];
          z         = intArray[1];

          transform.localEulerAngles = new Vector3(z, y, x);}
catch(Exception e) {
 BREAK HERE and check exception & whatdoing.
} }

如何修复，输入字符串格式不正确 unity 3d 和 serial

How to fix, Input string not in correct format unity 3d and serial

c#

serial-port

unity3d