参考链接：
源码：
/**
* Parses the string argument as a signed decimal integer. The
* characters in the string must all be decimal digits, except
* that the first character may be an ASCII minus sign {@code '-'}
* ({@code '\u005Cu002D'}) to indicate a negative value or an
* ASCII plus sign {@code '+'} ({@code '\u005Cu002B'}) to
* indicate a positive value. The resulting integer value is
* returned, exactly as if the argument and the radix 10 were
* given as arguments to the {@link #parseInt(java.lang.String,
* int)} method.
*
* @param s a {@code String} containing the {@code int}
* representation to be parsed
* @return the integer value represented by the argument in decimal.
* @exception NumberFormatException if the string does not contain a
* parsable integer.
*/
public static int parseInt(String s) throws NumberFormatException {
return parseInt(s,10);
}
分析：底层是通过parseInt(s,10)进行转换的，当输入的字符串中的内容不是数字时，会抛出NumberFormatException异常。
举例：
（1）正常时
String number = "12345";
// Integer.parseInt()：返回原始整数
int i = Integer.parseInt(number);
System.out.println("i = " + i);// i = 12345
（2）不正常时
String number1 = "12345A";
int i1 = Integer.parseInt(number1);
System.out.println("i1 = " + i1);
输出结果：
Exception in thread "main" java.lang.NumberFormatException: For input string: "12345A" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:580) at java.lang.Integer.parseInt(Integer.java:615) at string.StringToNumberTest.testStringToIntOrInteger(StringToNumberTest.java:29) at string.StringToNumberTest.main(StringToNumberTest.java:12)
为此需要加try...catch...：
String number1 = "12345A";
try{
int i1 = Integer.parseInt(number1);
System.out.println("i1 = " + i1);
}catch (NumberFormatException e){
System.out.println("Unable to convert input string (" + number1 + ") to int");
}
输出结果：
Unable to convert input string (12345A) to int
源码：
/**
* Returns an {@code Integer} object holding the
* value of the specified {@code String}. The argument is
* interpreted as representing a signed decimal integer, exactly
* as if the argument were given to the {@link
* #parseInt(java.lang.String)} method. The result is an
* {@code Integer} object that represents the integer value
* specified by the string.
*
* In other words, this method returns an {@code Integer}
* object equal to the value of:
*
*
* {@code new Integer(Integer.parseInt(s))}
*
*
* @param s the string to be parsed.
* @return an {@code Integer} object holding the value
* represented by the string argument.
* @exception NumberFormatException if the string cannot be parsed
* as an integer.
*/
public static Integer valueOf(String s) throws NumberFormatException {
return Integer.valueOf(parseInt(s, 10));
}
分析：底层也是通过parseInt(s,10)进行转换的，当输入的字符串中的内容不是数字时，会抛出NumberFormatException异常。
举例：
（1）正常时
String number = "12345";
// Integer.valueOf()：返回一个Integer对象
Integer integer = Integer.valueOf(number);
System.out.println("integer = " + integer);// integer = 12345
（2）不正常时
String number1 = "12345A";
Integer integer1 = Integer.valueOf(number1);
System.out.println("integer1 = " + integer1);
输出结果：
Exception in thread "main" java.lang.NumberFormatException: For input string: "12345A" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:580) at java.lang.Integer.valueOf(Integer.java:766) at string.StringToNumberTest.testStringToIntOrInteger(StringToNumberTest.java:40) at string.StringToNumberTest.main(StringToNumberTest.java:12)
为此需要加try...catch...：
String number1 = "12345A";
try {
Integer integer1 = Integer.valueOf(number1);
System.out.println("integer1 = " + integer1);
}catch (NumberFormatException e){
System.out.println("Unable to convert input string (" + number1 + ") to int");
}
输出结果：
Unable to convert input string (12345A) to int
测试代码：
package string;
/**
* @ClassName StringToNumberTest
* @Description TODO
* @Author Jiangnan Cui
* @Date 2022/11/15 20:45
* @Version 1.0
*/
public class StringToNumberTest {
public static void main(String[] args) {
// String number = "12345";// 输出12345
// String number = "12345A";// 抛异常
// String number = "-12345";// 输出12345
String number = "012345";// 输出12345
if (isDigit(number)) {
System.out.println(Integer.parseInt(number));
} else {
System.out.println("Please provide a valid digit [0-9]");
}
}
/**
* @MethodName isDigit
* @Description 判断输入字符串内容是否是数字，检查输入，但是代价很高
* @param: input
* @return: boolean
* @Author Jiangnan Cui
* @Date 23:09 2022/11/16
*/
public static boolean isDigit(String input) {
// null or length < 0, return false.
if (input == null || input.length() < 0)
return false;
// empty, return false
input = input.trim();// 从当前字符串中删除所有前导和尾随空白字符
if ("".equals(input))
return false;
// negative number in string
if (input.startsWith("-")) {
// cut the first char
return input.substring(1).matches("[0-9]*");
} else {
// positive number, good, just check
return input.matches("[0-9]*");
}
}
}
使用前要先导入依赖：
com.google.guava guava 15.0
测试代码：
package test;
import com.google.common.primitives.Ints;
import java.util.Optional;
public class TestInts {
public static void main(String[] args) {
Integer integer = Optional.ofNullable("123").map(Ints::tryParse).orElse(0);
System.out.println("integer = " + integer);// integer = 123
}
}
测试代码：
package string;
import java.util.Optional;
/**
* @ClassName StringToNumberTest2
* @Description TODO
* @Author Jiangnan Cui
* @Date 2022/11/18 7:51
* @Version 1.0
*/
public class StringToNumberTest2 {
public static void main(String[] args) {
String number = "12345";// 输出12345
// String number = "12345A";// 报错
// String number = "-12345";// 输出12345
// String number = "012345";// 输出12345
Optional result = Optional.ofNullable(number)
.filter(StringToNumberTest::isDigit)
.map(Integer::parseInt);// 先判断是否为空，再过滤出是数字的，最后进行提取
if (!result.isPresent()) {
System.out.println("Please provide a valid digit [0-9]");
} else {
System.out.println(result.get());
}
}
}
补充：
对于Integer类型，可以通过.intValue()方法转化为int类型。
其余String转基本数据类型方法应该类似，如果有差别，后续再补充。
参考链接：
源码：
/**
* Returns the string representation of the {@code int} argument.
*
* The representation is exactly the one returned by the
* {@code Integer.toString} method of one argument.
*
* @param i an {@code int}.
* @return a string representation of the {@code int} argument.
* @see java.lang.Integer#toString(int, int)
*/
@NotNull @Contract(pure=true)
public static String valueOf(int i) {
return Integer.toString(i);
}
/**
* Returns the string representation of the {@code Object} argument.
*
* @param obj an {@code Object}.
* @return if the argument is {@code null}, then a string equal to
* {@code "null"}; otherwise, the value of
* {@code obj.toString()} is returned.
* @see java.lang.Object#toString()
*/
public static String valueOf(Object obj) {
return (obj == null) ? "null" : obj.toString();
}
测试：
int num = 12345;
String s = String.valueOf(num);
System.out.println("s = " + s);// s = 12345
Integer num1 = 12345;
String s1 = String.valueOf(num1);
System.out.println("s1 = " + s1);// s1 = 12345
Integer num2 = null;
String s2 = String.valueOf(num2);
System.out.println("s2 = " + s2);// s2 = null
注意：valueOf()括号中的内容不能为空，否则会报空指针异常。
源码：
/**
* Returns a {@code String} object representing this
* {@code Integer}'s value. The value is converted to signed
* decimal representation and returned as a string, exactly as if
* the integer value were given as an argument to the {@link
* java.lang.Integer#toString(int)} method.
*
* @return a string representation of the value of this object in
* base 10.
*/
public String toString() {
return toString(value);
}
测试：
Integer num1 = 12345;
String s3 = num1.toString();
System.out.println("s3 = " + s3);// s3 = 12345
测试：
Integer num1 = 12345;
Integer num2 = null;
String s4 = num1 + "";
System.out.println("s4 = " + s4);// s4 = 12345
String s5 = num2 + "";
System.out.println("s5 = " + s5);// s5 = null
注意：第三种相对第一二种方法耗时较大。
参考链接：http://ych0108.iteye.com/blog/2174134
举例：
String.format("%010d", 25); // 25为int型
0代表前面要补的字符
10代表字符串长度
d表示参数为整数类型
测试代码：
// 将25保存成00025
int num = 25;
// 方法一：5前面不指定内容时，去除num位数后前面补空格
String format = String.format("%5d", num);
System.out.println("format = " + format);// format = 25
// 将空格替换为0
String str = format.replace(" ", "0");
System.out.println("str = " + str);// str = 00025
// 方法二：5前面指定内容0时，去除num位数后前面补0
String format1 = String.format("%05d", num);
System.out.println("format1 = " + format1);// format1 = 00025
注意：方法二只需要在数字前加0就可以进行填充，比方法一更加方便。