如何检查2个charAt(i) [英] How to check 2 charAt(i)

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本文介绍了如何检查2个charAt(i)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有这行:

812.12 135.14 646.17 1
812.12 135.14 646.18 1
812.12 135.14 646.19 10
812.12 135.14 646.20 10
812.12 135.14 646.21 100
812.12 135.14 646.22 100

我只想删除最后一组,所以我做了如下代码:

I want to delete only the last group so I did code like this:

if(lines[i].charAt(lines[i].length())-1>= '0'&&lines[i].charAt(lines[i].length()-1)<= '9'){
    lines[i] = lines[i].substring(0, lines[i].length()-1);
   }

   else if(lines[i].charAt(lines[i].length())>='10'+c&&lines[i].charAt(lines[i].length()-1)<='99'){
   lines[i] = lines[i].substring(0, lines[i].length()-2);
        }

   else if(lines[i].charAt(lines[i].length())>='100'&&lines[i].charAt(lines[i].length()-1)<='999){
   lines[i] = lines[i].substring(0, lines[i].length()-3);
 }

现在它对我有用,我需要帮助

And it's now working for me I need help please

推荐答案

您有字符,并试图将它们比较为数字。

You have characters and trying to compare them as numbers. It doesn't work that way.

A char 在大多数高级语言(Java,.NET语言)中都是这样。 ..etc)编程语言默认采用 UTF-16 格式。由于每个 char 大小为 2字节,因此可以遵循您需要的任何编码。

A char in most higher-level (Java, .NET languages...etc) programming languages follows UTF-16 format by default. Since each char size is 2 bytes it can follow any encoding you need.

您无法将 char string 进行比较不能尝试将字符串放入像 '10'这样的char中。那是两个字符,而不是一个,它是一个 string

You cannot compare a char to string and you cannot attempt to put string in char like '10'. Those are two characters, not one, it's a string.

现在,根据所需的输出,您想删除最后一个空格之后的所有内容。为此,可以使用以下代码:

Now based on your desired output, you would like to remove everything after the last space. To do that, you can use the following code:

static void updateLines(String[] lines){
    for(int i=0;i<lines.length; i++){
        // get index of the last space
        int index = lines[i].lastIndexOf(" ");

        // remove everything after the last space
        lines[i] = lines[i].substring(0, index);
    }
}


public static void main(String[] args) {
    String[] lines = new String[]{
      "812.12 135.14 646.17 1",
      "812.12 135.14 646.18 1",
      "812.12 135.14 646.19 10",
      "812.12 135.14 646.20 10",
      "812.12 135.14 646.21 100", 
      "812.12 135.14 646.22 100",
      "812.12 135.14 646.23 1000",
      "812.12 135.14 646.24 1000"
    };

    updateLines(lines);

    for(int i=0;i<lines.length;i++)
        System.out.println(lines[i]);
}

输出:

812.12 135.14 646.17
812.12 135.14 646.18
812.12 135.14 646.19
812.12 135.14 646.20
812.12 135.14 646.21
812.12 135.14 646.22
812.12 135.14 646.23
812.12 135.14 646.24

这够了还是您需要比较数字?

Is this enough or do you need to compare numbers?

这篇关于如何检查2个charAt(i)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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