当我在maven test
本地运行时通过.但是当我在CI服务器上运行它时出现此错误.
Error Message
Could not open JPA EntityManager for transaction; nested exception is org.hibernate.exception.JDBCConnectionException: Unable to acquire JDBC Connection
Stacktrace
org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is org.hibernate.exception.JDBCConnectionException: Unable to acquire JDBC Connection
Caused by: org.hibernate.exception.JDBCConnectionException: Unable to acquire JDBC Connection
Caused by: com.mysql.jdbc.exceptions.jdbc4.CommunicationsException:
Communications link failure
The last packet sent successfully to the server was 0 milliseconds ago. The driver has not received any packets from the server.
Caused by: java.net.UnknownHostException: mysql …
Run Code Online (Sandbox Code Playgroud) 如果代码是
scanf("%s\n",message)
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VS
gets(message)
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有什么区别?似乎他们都得到了消息的输入.
声明就像 SELECT * FROM db.table group by id desc;
会引发这样的错误
15:02:24 SELECT * FROM db.table group by id desc LIMIT 0, 10 错误代码:1064。您的 SQL 语法有错误;检查与您的 MySQL 服务器版本相对应的手册,以获取在第 1 行 0.00014 秒附近的“desc LIMIT 0, 10”附近使用的正确语法
在 MySQL 8.0.13 上 Ubuntu 18.04 桌面 64 位
这在 Windows 或 CentOS 或 Ubuntu 中的 MySQL 5.7 上会很好。
我基本上知道,select 语句就像。
SELECT statement... [WHERE condition | GROUP BY `field_name(s)` HAVING condition] ORDER BY `field_name(s)` [ASC | DESC];
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那么这个 5.7 的问题是不发出错误吗?
或者在 SQL 标准上更复杂的东西?
在比特币维基 这样说:
0 - 拥有私人ECDSA密钥
18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725
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1 - 获取使用它生成的相应公钥 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
2 - 对公钥执行SHA-256哈希处理
600FFE422B4E00731A59557A5CCA46CC183944191006324A447BDB2D98D4B408
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但是当我运行sha256时 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
我没有
600FFE422B4E00731A59557A5CCA46CC183944191006324A447BDB2D98D4B408
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我得到的是
32511e82d56dcea68eb774094e25bab0f8bdd9bc1eca1ceeda38c7a43aceddce
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我究竟做错了什么?
glutInitDisplayMode(GLUT_SINGLE|GLUT_RGB);
但是当我将它设置为透明窗口时它会变成透明窗口
glutInitDisplayMode(GLUT_DEPTH | GLUT_DOUBLE | GLUT_RGBA);
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但是我需要这个例子在GLUT_DOUBLE
模式下使用一些绘图.
那么GLUT_DOUBLE
和之间的区别是GLUT_SINGLE
什么?
我知道在2.0-openGL中我们可以像这样绘制一条线.
glBegin(GL_LINES);
glVertex3f(20.0f,150.0f,0.0f);
glVertex3f(220.0f,150.0f,0.0f);
glVertex3f(200.0f,160.0f,0.0f);
glVertex3f(200.0f,160.0f,0.0f);
glEnd();
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但如何在现代openGL(3.0+)中做类似的事情
我已经阅读了使用现代OpenGL绘制圆形点,但答案不是关于某一点,因为我想绘制具有一定坐标的点的多边形,这不是很有用.
我使用这个代码,但它除了蓝色背景外什么都没显示.我错过了什么?
GLuint VertexArrayID;
glGenVertexArrays(1, &VertexArrayID);
glBindVertexArray(VertexArrayID);
static const GLfloat g_vertex_buffer_data[] = {
20.0f, 150.0f, 0.0f, 1.0f,
220.0f, 150.0f, 0.0f, 1.0f,
200.0f, 160.0f, 0.0f, 1.0f
};
GLuint vertexbuffer;
glGenBuffers(1, &vertexbuffer);
glBindBuffer(GL_ARRAY_BUFFER, vertexbuffer);
glBufferData(GL_ARRAY_BUFFER, sizeof(g_vertex_buffer_data), g_vertex_buffer_data, GL_STATIC_DRAW);
do{
// Clear the screen
glClear( GL_COLOR_BUFFER_BIT );
// 1rst attribute buffer : vertices
glEnableVertexAttribArray(0);
glBindBuffer(GL_ARRAY_BUFFER, vertexbuffer);
glVertexAttribPointer(
0, // attribute 0. No particular reason for 0, but must match the layout in the …
Run Code Online (Sandbox Code Playgroud) 我需要绘制一个立方体来指示OpenGL 3.3核心配置文件中的坐标.glutInitContextVersion (3, 3);
它可以正常工作,但在glutInitContextVersion (3, 3);
应用时它会变成全黑.这是绘图代码.
void display() {
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); // Clear color and depth buffers
glMatrixMode(GL_MODELVIEW); // To operate on model-view matrix
// Render a color-cube consisting of 6 quads with different colors
glLoadIdentity(); // Reset the model-view matrix
glTranslatef(1.5f, 0.0f, -7.0f); // Move right and into the screen
glBegin(GL_QUADS); // Begin drawing the color cube with 6 quads
// Top face (y = 1.0f)
// Define vertices in counter-clockwise (CCW) order with …
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