if (cupLyoStuWeighingLiquidLog.getStuMaximumWeight() == null && cupLyoStuWeighingLiquidLog.getStuMinimumWeight() == null) { // 第一次输入，设置最大值 cupLyoStuWeighingLiquidLog.setStuMaximumWeight(modifiedString); } else { double stuMaximumWeightNum = Double.parseDouble(cupLyoStuWeighingLiquidLog.getStuMaximumWeight()); double stuMinimumWeightNum = (cupLyoStuWeighingLiquidLog.getStuMinimumWeight() != null) ? Double.parseDouble(cupLyoStuWeighingLiquidLog.getStuMinimumWeight()) : Double.MAX_VALUE;
问题出在第二次输入逻辑中的 `cupLyoStuWeighingLiquidLog.setStuMaximumWeight(cupLyoStuWeighingLiquidLog.getStuMinimumWeight());` 这一行代码，当输入的值小于当前最小值时，不应该将当前最小值赋给最大值，而是应该保持最大值不变。以下是修正后的代码： ```java // 判断最大重量和最小重量是否都为 null if (cupLyoStuWeighingLiquidLog.getStuMaximumWeight() == null && cupLyoStuWeighingLiquidLog.getStuMinimumWeight() == null) { // 第一次输入，设置最大值 cupLyoStuWeighingLiquidLog.setStuMaximumWeight(modifiedString); } else { // 将最大重量转换为 double 类型 double stuMaximumWeightNum = Double.parseDouble(cupLyoStuWeighingLiquidLog.getStuMaximumWeight()); // 将最小重量转换为 double 类型，如果最小重量为 null 则设为 Double.MAX_VALUE double stuMinimumWeightNum = (cupLyoStuWeighingLiquidLog.getStuMinimumWeight() != null) ? Double.parseDouble(cupLyoStuWeighingLiquidLog.getStuMinimumWeight()) : Double.MAX_VALUE; // 如果输入值大于当前最大值 if (stuAnswersNum > stuMaximumWeightNum) { // 当前最大值变为最小值 cupLyoStuWeighingLiquidLog.setStuMinimumWeight(cupLyoStuWeighingLiquidLog.getStuMaximumWeight()); // 新输入值变为最大值 cupLyoStuWeighingLiquidLog.setStuMaximumWeight(modifiedString); } // 如果输入值小于当前最小值 else if (stuAnswersNum < stuMinimumWeightNum) { // 新输入值变为最小值 cupLyoStuWeighingLiquidLog.setStuMinimumWeight(modifiedString); } // 输入值介于最大值和最小值之间，保持不变 } ``` 额外需要注意的逻辑点 1. 在将字符串转换为 `double` 类型时，要确保字符串是有效的数字格式，否则会抛出 `NumberFormatException` 异常，可以添加异常处理代码来增强程序的健壮性。 2. 要注意 `Double.MAX_VALUE` 的使用，当最小重量初始为 `null` 时，使用 `Double.MAX_VALUE` 作为初始值，在后续逻辑中要确保这个初始值不会影响正常的比较逻辑。 3. 代码中 `modifiedString` 和 `stuAnswersNum` 要确保其来源和类型是正确的，`modifiedString` 应该是可以转换为 `double` 类型的字符串，`stuAnswersNum` 应该是正确的输入值。 [2025-07-30 16:59:32 | AI写代码神器 | 621点数解答]
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- dim caizhi1 as string dim caizhi2 as string dim caizhi as string dim dianzulv1 as double dim dianzulv2 as double dim midu1 as double dim midu2 as double dim zhijing as double dim jiejubi as double dim jiemianji as double dim zhouchang as double dim dianzusichangdu as double dim zhongliang as double dim quanshu as double dim jieju as double dim dianzulv as double dim midu as double dim zuzhi as double private sub command1_click() dim xlapp as object dim xlworkbook as object dim xlworkshee（32点数解答 | 2024-10-31 15:19:51）207
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- #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct sqlist //单链表结构体 { int data; struct sqlist *next; //指针域 }sqlist; void initlist (sqlist *l, sqlist *m, sqlist *n, sqlist *h) { //todo list yes l->next = null; m->next = null; n->next = null; h->next = null; } void emp(sqlist *l) { if (l->next==null) //todo list)yes printf("链表为空表。\n"); else printf("链表为非空表。\n"); } void length(sqlist *l) { sqlist *p; int length = 0; p=l->next; while(p!=null) { length++; p=p->next; } //todo list（345点数解答 | 2024-10-23 23:11:22）180
- CREATE TABLE `sys_role` ( `id` bigint(20) NOT NULL AUTO_INCREMENT COMMENT 'ID', `parent_id` bigint(20) DEFAULT '0' COMMENT '上级ID（0表示没有上级）', `role_name` varchar(30) DEFAULT '' COMMENT '角色名称', `sort` int(11) DEFAULT '1' COMMENT '排序', `status` tinyint(1) DEFAULT NULL COMMENT '状态：0无效 1有效', `remarks` varchar(100) DEFAULT NULL COMMENT '备注描述', `create_time` timestamp NULL DEFAULT NULL COMMENT '创建时间', `create_by` bigint(20) DEFAULT NULL COMMENT '创建人', `update_time` timestamp NULL DEFAU（168点数解答 | 2025-04-10 14:39:47）126
- ```cpp #include <iostream> using namespace std; struct node { int data; node* link; node(int x) : data(x), link(null) {} }; // 查找最大节点及其前一个节点 void findmaxandprev(node* list, node*& maxnode, node*& prevmax) { node* curr = list; maxnode = list; prevmax = null; node* prev = null; while (curr!= null) { if (curr->data > maxnode->data) { maxnode = curr; prevmax = prev; } prev = curr; curr = curr->link; } } // 将最大节点移到链表末尾 void movemaxtoend(node*& list) { node* maxnode = null; node* prevmax = null;（549点数解答 | 2024-10-14 22:55:13）222
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- #define _crt_secure_no_warnings #include <iostream> #include <cstdlib> #include <cmath> #include <vector> #include <cstdio> using namespace std; // 你的代码... const int nx = 784, nb = 500, ny = 10;//输入层x，隐藏层b，输出层y节点数 const double eta = 0.1;//学习率 struct node { double val{}; double bias{}; vector<double> weight; } x[nx], b[nb], y[ny];//输入层x，隐藏层b，输出层y double g[ny], e[nb];//用于反向传播 double trainx[nx], trainy[ny];//训练数据 double sigmoid(double x) { return 1.0 / (1.0 + exp(-x)); } double（34点数解答 | 2024-09-06 08:43:52）269
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- #include <stdio.h> #include <math.h> #define pi 3.14159265358979323846 double meridianarclength(double b) { double a = 6378140.0; double e2 = 0.00669438002290; double m = a * (1 - e2) * (1 - e2 * sin(b) * sin(b)) * pow(1 - e2 * sin(b) * sin(b), 1.5); double n = a / sqrt(1 - e2 * sin(b) * sin(b)); return m + n * sin(b) * cos(b) * pow(tan(b),2) /2 + n * sin(b) * pow(cos(b), 3) * (5 - pow(tan(b), 2) + 9 * e2 * pow(cos(b), 2)) * pow(tan(b), 4) / 24; } double calculatebaselatitude(double x, double y)（565点数解答 | 2024-11-21 13:08:38）169