Android 随机字符串生成器

Android random string generator

提问by Defus

I have a problem. I want to draw a random String something like this aXcFg3s2. What i doing bad ? How change my random()

我有个问题。我想绘制一个像 aXcFg3s2 这样的随机字符串。我在做什么不好？如何改变我的random()

private String random;
private String charsEntered;
private EditText et;
private Button ok;
CaptchaInterface.OnCorrectListener mCorrectListener;

public void setOnCorrectListener(CaptchaInterface.OnCorrectListener listener) {
    mCorrectListener = listener;
}

public TextCaptcha(Context context) {
    super(context);
    getWindow().requestFeature(Window.FEATURE_NO_TITLE);
}

public static String random() {
    Random generator = new Random();
    String x = (String) (generator.nextInt(96) + 32);
    return x;
}

public void onCreate(Bundle icicle) {
    setContentView(R.layout.main);
    random = random();
    TextView display = (TextView) findViewById(R.id.textView1);
    display.setText("Random Number: " + random); // Show the random number
    et = (EditText) findViewById(R.id.etNumbers);
    ok = (Button) findViewById(R.id.button1);
    ok.setOnClickListener(this);

}

public void onClick(View arg0) {
    // TODO Auto-generated method stub
    try {
        charsEntered = et.getText().toString();
    } catch (NumberFormatException nfe) {
        Toast.makeText(et.getContext(), "Bla bla bla",
                Toast.LENGTH_LONG).show();
    }

    if (random == charsEntered) {
        Toast.makeText(et.getContext(), "Good!", Toast.LENGTH_LONG).show();
    } else {
        Toast.makeText(et.getContext(), "Bad!", Toast.LENGTH_LONG).show();
    }
}

回答by android developer

the problem is that you've handled only a single character instead of using a loop.

问题是您只处理了一个字符而不是使用循环。

you can create an array of characters which has all of the characters that you wish to allow to be in the random string , then in a loop take a random position from the array and add append it to a stringBuilder . in the end , convert the stringBuilder to a string.

您可以创建一个字符数组，其中包含您希望允许在随机字符串中的所有字符，然后在循环中从数组中获取随机位置并将其添加到 stringBuilder 中。最后，将 stringBuilder 转换为字符串。

---

---

EDIT: here's the simple algorithm i've suggested:

编辑：这是我建议的简单算法：

private static final String ALLOWED_CHARACTERS ="0123456789qwertyuiopasdfghjklzxcvbnm";

private static String getRandomString(final int sizeOfRandomString)
  {
  final Random random=new Random();
  final StringBuilder sb=new StringBuilder(sizeOfRandomString);
  for(int i=0;i<sizeOfRandomString;++i)
    sb.append(ALLOWED_CHARACTERS.charAt(random.nextInt(ALLOWED_CHARACTERS.length())));
  return sb.toString();
  }

and on Kotlin:

在 Kotlin 上：

companion object {
    private val ALLOWED_CHARACTERS = "0123456789qwertyuiopasdfghjklzxcvbnm"
}

private fun getRandomString(sizeOfRandomString: Int): String {
    val random = Random()
    val sb = StringBuilder(sizeOfRandomString)
    for (i in 0 until sizeOfRandomString)
        sb.append(ALLOWED_CHARACTERS[random.nextInt(ALLOWED_CHARACTERS.length)])
    return sb.toString()
}

回答by bennettaur

There are a few things wrong with your code.

您的代码有一些问题。

You cannot cast from an int to a string. Cast it to a char instead. This however will only give you a single char so instead you could generate a random number for the length of your string. Then run a for loop to generate random chars. You can define a StringBuilder as well and add the chars to that, then get your random string using the toString()method

您不能从 int 转换为字符串。改为将其转换为字符。但是，这只会给您一个字符，因此您可以为字符串的长度生成一个随机数。然后运行 ​​for 循环以生成随机字符。您也可以定义一个 StringBuilder 并将字符添加到其中，然后使用该toString()方法获取随机字符串

example:

例子：

public static String random() {
    Random generator = new Random();
    StringBuilder randomStringBuilder = new StringBuilder();
    int randomLength = generator.nextInt(MAX_LENGTH);
    char tempChar;
    for (int i = 0; i < randomLength; i++){
        tempChar = (char) (generator.nextInt(96) + 32);
        randomStringBuilder.append(tempChar);
    }
    return randomStringBuilder.toString();
}

Also, you should use random.compareTo()rather than ==

此外，您应该使用random.compareTo()而不是 ==

回答by ??-Bham

You need to import UUID. Here is the code

您需要导入 UUID。这是代码

import java.util.UUID;

id = UUID.randomUUID().toString();

回答by Defus

this is how i generate my random strings with desired characters and desired length

这就是我生成具有所需字符和所需长度的随机字符串的方式

          char[] chars1 = "ABCDEF012GHIJKL345MNOPQR678STUVWXYZ9".toCharArray();
                                StringBuilder sb1 = new StringBuilder();
                                Random random1 = new Random();
                                for (int i = 0; i < 6; i++)
                                {
                                    char c1 = chars1[random1.nextInt(chars1.length)];
                                    sb1.append(c1);
                                }
                                String random_string = sb1.toString();  

回答by Android Dev

You can simply use the following method to generate random String with 5 character and it will return arrayList of random String

您可以简单地使用以下方法生成具有 5 个字符的随机字符串，它将返回随机字符串的 arrayList

public ArrayList<String> generateRandomString() {

    ArrayList<String> list = new ArrayList<>();
    Random rnd = new Random();
    String str = "";

    String randomLetters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    String randomLetterSmall = "abcdefghijklmnopqrstuvwxyz";

    for (int n = 0; n < 50; n++) {
        str = String.valueOf(randomLetters.charAt(rnd.nextInt(randomLetters.length())));

        str += String.valueOf(randomLetterSmall.charAt(rnd.nextInt(randomLetters.length())));
        str += String.valueOf(randomLetterSmall.charAt(rnd.nextInt(randomLetters.length())));
        str += String.valueOf(randomLetterSmall.charAt(rnd.nextInt(randomLetters.length())));
        str += String.valueOf(randomLetterSmall.charAt(rnd.nextInt(randomLetters.length())));

        //Copy above line to increase character of the String
        list.add(str);
    }
    Collections.sort(list);
    return list;
}

回答by Aryan Moradi

This function run in kotlin ->

此函数在 kotlin 中运行 ->

fun randomString(stringLength: Int): String {
    val list = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz".toCharArray()
    var randomS = ""
    for (i in 1..stringLength) {
        randomS += list[getRandomNumber(0, list.size - 1)]
    }
    return randomS
}

fun getRandomNumber(min: Int, max: Int): Int {
   return Random().nextInt((max - min) + 1) + min
}

Or you can use my library https://github.com/Aryan-mor/Utils-Library

或者你可以使用我的图书馆 https://github.com/Aryan-mor/Utils-Library

回答by BitByteDog

Quick one liner using the org.apache.commons.lang3.RandomStringUtilspackage.

使用org.apache.commons.lang3.RandomStringUtils包装快速的一个衬垫。

String randonString = RandomStringUtils.randomAlphanumeric(16);

Requires the library dependency in the gradle build file:

需要 gradle 构建文件中的库依赖项：

implementation 'org.apache.commons:commons-text:1.6'

回答by ORY

final  String[] Textlist = { "Text1", "Text2", "Text3"};

TextView yourTextView = (TextView)findViewById(R.id.yourTextView);

Random random = new Random();

Random random = new Random();

String randomText = TextList[random.nextInt(TextList.length)];

String randomText = TextList[random.nextInt(TextList.length)];

yourTextView.setText(randomText);

回答by LuxuryMode

You cannot cast an int to a String. Try:

您不能将 int 转换为 String。尝试：

 Random generator = new Random();
 String x = String.valueOf (generator.nextInt(96) + 32);