Question

我有这段代码从文本文件中读取数据并将其分配给2d数组

Scanner read_blockage = null;
General_Inputs.Blockage_Number=new double[Input.General_Inputs.Num_Of_Analysis_Years*Input.General_Inputs.Num_Of_States][Input.General_Inputs.Num_Of_Ppes];
    try{
    read_blockage=new Scanner(new File("Blockage Output1"));
    int row = -1; // since we're incrementing row at the start of the loop
    while(read_blockage.hasNext()) {
        row++;
        String[] line = read_blockage.nextLine().split("\t");
        for(int j=0;j<Input.General_Inputs.Num_Of_Ppes;j++){
            try {
                General_Inputs.Blockage_Number[row][j] = Double.parseDouble(line[j]);
            } catch (NumberFormatException e) {
                e.printStackTrace();
            }
        }}
    read_blockage.close();}catch (FileNotFoundException e) {
        e.printStackTrace();
    }

但是我收到了这个错误：

java.io.FileNotFoundException: Blockage Output1 (The system cannot find the file specified)
2
2
2
3
    at java.io.FileInputStream.open0(Native Method)
    at java.io.FileInputStream.open(FileInputStream.java:195)
    at java.io.FileInputStream.<init>(FileInputStream.java:138)
    at java.util.Scanner.<init>(Scanner.java:611)
    at Input.Get_Inputs(Input.java:270)
    at Input.main(Input.java:288)

我不确定为什么会出现此错误？

编辑：我摆脱了上述错误，但现在我有一个新错误：

java.lang.NumberFormatException: For input string: "0.2810496821150867                           0.3455471819235053                           0.1247760656600859                           0.1925735036025203                           0.16475561749067555                           0.3267969645821732                           0.5325079154577266                           0.7311354592633524                           0.29828747755582985                           0.3983939064000447                           1.6432540332118697                           2.242416989842468                           0.8199042126197025                           1.1448149650482649                           0.6569387483611318                           0.35521248909704994                           0.8311372587904973                           1.2599707232227086                           1.4153816162469934                           1.091443886313361                           0.7492391207620115                           1.4029328027711394                           1.3060173850919903                           3.0212129386585675                           1.185220575726193                           3.2093022651230037                           2.2304670167490195                           4.028061408800144                           1.1957020911741867                           2.3250822033050813                           6.144104904071859                           9.634733755857885                           3.3148373093880736                           9.740483573762857                           3.857137427951027                           4.527035922001198                           7.248709304936811                           10.112180962036412                           12.688211002013142                           3.5445943135631026                           5.87022858087266                           11.490999298946353                           13.75534054772614"
    at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:2043)
    at sun.misc.FloatingDecimal.parseDouble(FloatingDecimal.java:110)
    at java.lang.Double.parseDouble(Double.java:538)
    at Input.Get_Inputs(Input.java:277)
    at Input.main(Input.java:288)
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
    at Input.Get_Inputs(Input.java:277)
    at Input.main(Input.java:288)

任何建议？

Answer 1

该错误消息非常有用，并且它说该程序无法通过您的文件声明找到该文件。

尝试使用文件的绝对路径＆＃34; Blockage Output1＆＃34;，并记住包含文件扩展名（.txt，.conf，.bla）。

//correcting the escape sequence usage
new File("C:\\workspace\\project\\Blockage Output1.txt")

Eclipse将在项目的根目录中查找相对定义的文件，因此如果您将文件放在src，bin，res等文件夹中，那么您需要像这样声明文件。 / p>

//correcting the escape sequence usage
new File("src\\Blockage Output1.txt")

希望其中一种解决方案适合您！

在java eclipse中读取文本文件为2d数组错误

1 个答案: