在Java编程中,遇到“incompatible types: int cannot be converted to string”错误通常意味着你尝试将一个整数(int)类型直接赋值给了一个期望字符串(String)类型的变量或方法参数。这个错误表明类型不匹配,因为Java是强类型语言,要求严格的类型匹配。 以下是解决这个错误的几个步骤: 确认错误信息的上下文: 确保这个...
int qscore=Integer.parseInt(qs);将未知类型的值转化为int 型，需要用到Integer.parseInt()方法强制转化。另外你的逻辑也有问题。应该先判断ps 是否为空，若为空pscore=0; qs的判断也一样。你的代码应该这样写：int pscore;int qscore;if(ps.equals("")){ pscore=0;}else{ pscore=Integer...
charAt(0); if (firstChar < '0') { // Possible leading "+" or "-" if (firstChar == '-') { negative = true; limit = Integer.MIN_VALUE; } else if (firstChar != '+') throw NumberFormatException.forInputString(s); if (len == 1) // Cannot have lone "+" or "-" throw ...
问Integer to Roman“不兼容的类型: int无法转换为布尔型[在MainClass.java中]”EN根据罗马数字的规则...
Integer id=new Integer(-1);-1是int类型，不是integer类型的，肯定会爆类型不匹配的错，这样就好了 private
, value); } // Displays the following to the console: // Unable to convert '0xFFFFFFFFFFFFFFFF' to a long integer. 以下示例尝试将字符串数组中的每个元素解释为十六进制字符串,并将其转换为长整数。 C# 复制 运行 using System; public class Example { public static void Main() { string[] ...
*@params the {@codeString} to be converted to an * {@codeInteger}. *@exceptionNumberFormatException if the {@codeString} does not * contain a parsable integer. *@seejava.lang.Integer#parseInt(java.lang.String, int) */publicInteger(String s)throwsNumberFormatException {this.value = parseInt...
...but then you never cast it to an integer, so the input came back as a String, even though you meant to store an integer or something. So I'd check the call stack, find out what the line was that you're having trouble with, and just look at what that line might be trying ...
Integer id=new Integer(-1);-1是int类型，不是integer类型的，肯定会爆类型不匹配的错，这样就好了 类型
问修复SQL错误不兼容的类型: capture#1 of？无法转换为intEN版权声明：本文内容由互联网用户自发贡献，该...