C 不知道如何获取数据库吗?
我完全是sqlite和gtk的初学者。我正在用code::blocks做我的第一个项目。 我在获取数据库时遇到问题我的代码是:C 不知道如何获取数据库吗?,c,sqlite,gtk,codeblocks,C,Sqlite,Gtk,Codeblocks,我完全是sqlite和gtk的初学者。我正在用code::blocks做我的第一个项目。 我在获取数据库时遇到问题我的代码是: #include <stdio.h> #include <stdlib.h> #include <gtk/gtk.h> #include "sqlite3.c" #include <sqlite3.h> static int callback(void *data, int argc, char **argv, char
#include <stdio.h>
#include <stdlib.h>
#include <gtk/gtk.h>
#include "sqlite3.c"
#include <sqlite3.h>
static int callback(void *data, int argc, char **argv, char **azColName){
int i;
fprintf(stderr, "%s: ", (const char*)data);
for(i=0; i<argc; i++){
printf("%s = %s\n", azColName[i], argv[i] ? argv[i] : "NULL");
}
printf("\n");
return 0;
}
int main(int argc, char *argv[]) {
sqlite3 *db;
char *zErrMsg = 0;
int rc;
char *sql;
const char* data = "Callback function called";
/* Open database */
rc = sqlite3_open("sign_in.db", &db);
if( rc ){
fprintf(stderr, "Can't open database: %s\n", sqlite3_errmsg(db));
exit(0);
}else{
fprintf(stderr, "Opened database successfully\n");
}
GtkWidget *window;
GtkWidget *table;
GdkColor color;
GtkWidget *label1;
GtkWidget *label2;
GtkWidget *label3;
GtkWidget *entry1;
GtkWidget *entry2;
GtkWidget *button1;
gtk_init(&argc, &argv);
window = gtk_window_new(GTK_WINDOW_TOPLEVEL);
gtk_window_set_position(GTK_WINDOW(window), GTK_WIN_POS_CENTER);
gtk_window_set_title(GTK_WINDOW(window), "LOG IN");
gtk_container_set_border_width(GTK_CONTAINER(window), 30);
gtk_widget_modify_bg(window, GTK_STATE_NORMAL, &color);
table = gtk_table_new(3, 2, FALSE);
gtk_container_add(GTK_CONTAINER(window), table);
label1 = gtk_label_new("---------------------------------- LOGIN AND ENJOY OUR FACILITY -----------------------------------");
label2 = gtk_label_new("USERNAME :");
label3= gtk_label_new("PASSWORD :");
button1 = gtk_button_new_with_label("LOG IN");
gtk_table_attach(GTK_TABLE(table), label1, 0, 1, 0, 1,
GTK_FILL |GTK_EXPAND | GTK_SHRINK, GTK_FILL |GTK_EXPAND | GTK_SHRINK, 5, 5);
gtk_table_attach(GTK_TABLE(table), label2, 0, 1, 1, 2,
GTK_FILL | GTK_EXPAND |GTK_SHRINK, GTK_FILL | GTK_EXPAND |GTK_SHRINK, 5, 5);
gtk_table_attach(GTK_TABLE(table), label3, 0, 1, 2, 3,
GTK_FILL | GTK_EXPAND |GTK_SHRINK, GTK_FILL | GTK_EXPAND |GTK_SHRINK, 5, 5);
entry1 = gtk_entry_new();
entry2 = gtk_entry_new();
gtk_table_attach(GTK_TABLE(table), entry1, 1, 2, 1, 2,
GTK_FILL |GTK_EXPAND | GTK_SHRINK, GTK_FILL |GTK_EXPAND | GTK_SHRINK, 5, 5);
gtk_table_attach(GTK_TABLE(table), entry2, 1, 2, 2, 3,
GTK_FILL |GTK_EXPAND | GTK_SHRINK, GTK_FILL |GTK_EXPAND | GTK_SHRINK, 5, 5);
gtk_table_attach(GTK_TABLE(table), button1, 1, 2, 3, 4,
GTK_FILL |GTK_EXPAND | GTK_SHRINK, GTK_FILL |GTK_EXPAND | GTK_SHRINK, 5, 5);
gtk_widget_show(table);
gtk_widget_show(label1);
gtk_widget_show(label2);
gtk_widget_show(label3);
gtk_widget_show(entry1);
gtk_widget_show(entry2);
gtk_widget_show(button1);
gtk_widget_show(window);
/* Create SQL statement */
sql="SELECT * from user where username='char *entry1' and password='char *entry2 '";
if(sqlite3_exec(db, sql, callback, (void*)data, &zErrMsg))
{
int count=0;
while(sqlite3_exec(db, sql, callback, (void*)data, &zErrMsg)+1)
{
count++;
}
if(count==1)
fprintf(stdout, "username and password is correct\n");
if(count>1)
fprintf(stdout, "Dulitcate user\n");
if(count<1)
fprintf(stdout, "username and password is not correct\n");
}
/* Execute SQL statement */
rc = sqlite3_exec(db, sql, callback, (void*)data, &zErrMsg);
if( rc != SQLITE_OK ){
fprintf(stderr, "SQL error: %s\n", zErrMsg);
sqlite3_free(zErrMsg);
}else{
fprintf(stdout, "Operation done successfully\n");
}
sqlite3_close(db);
g_signal_connect(G_OBJECT(button1), "clicked",
G_CALLBACK(gtk_main_quit), G_OBJECT(window));
g_signal_connect(window, "destroy",
G_CALLBACK(gtk_main_quit), NULL);
我在上述代码中遇到了以下问题:
在SELECT语句中,我知道它完全错误,请更正它。
在sqlite3_execdb、sql、回调、void*data和zErrMsg++的while循环中
/* Create SQL statement */
sql="SELECT * from user where username='char *entry1' and password='char *entry2 '";
if(sqlite3_exec(db, sql, callback, (void*)data, &zErrMsg))
{
int count=0;
while(sqlite3_exec(db, sql, callback, (void*)data, &zErrMsg)++)
{
count++;
}
if(count==1)
fprintf(stdout, "username and password is correct\n");
if(count>1)
fprintf(stdout, "Dulitcate user\n");
if(count<1)
fprintf(stdout, "username and password is not correct\n");
}
/* Execute SQL statement */
rc = sqlite3_exec(db, sql, callback, (void*)data, &zErrMsg);
if( rc != SQLITE_OK ){
fprintf(stderr, "SQL error: %s\n", zErrMsg);
sqlite3_free(zErrMsg);
}else{
fprintf(stdout, "Operation done successfully\n");
}
sqlite3_close(db);
请帮帮我。我真的需要有人帮忙。提前
如果要在查询返回数据的位置处理返回的数据,sqlite3_exec不合适,并且还有一些其他缺点
对于查询,应该始终使用,然后在循环中调用。
要将变量值放入语句中,请使用参数标记?以及功能。
要读取返回的值,请使用函数,但在本例中,您实际上不想读取任何值
char *user = "Supu";
char *password = "secret";
sqlite3_stmt *stmt;
const char *sql = "SELECT username, password FROM user WHERE username = ? AND password = ?";
rc = sqlite3_prepare_v2(db, sql, -1, &stmt, NULL);
if (rc != SQLITE_OK) {
fprintf(stderr, "error: %s, %s\n", sql, sqlite3_errmsg(db));
} else {
sqlite3_bind_text(stmt, 1, user, -1, SQLITE_TRANSIENT);
sqlite3_bind_text(stmt, 2, password, -1, SQLITE_TRANSIENT);
while ((rc = sqlite3_step(stmt)) == SQLITE_ROW) {
printf("returned row: user = %s, password = %s\n",
sqlite3_column_text(stmt, 0),
sqlite3_column_text(stmt, 1));
count++;
}
if (rc != SQLITE_DONE)
fprintf(stderr, "error: %s\n", sqlite3_errmsg(db));
sqlite3_finalize(stmt);
}
1.在需要从代码中输入的select语句中使用分隔符。。像字符串a=Select+entry1+。。exec行上的错误是什么?Thanx但是它给出了错误:::binary+的无效操作数有'char*'和'struct GtkWidget',对于exec它不接受++操作数..但是你能告诉我…为什么我们使用?而不是entry1和entry2。我有点困惑。你想在点击按钮的时候检查数据库,是吗?是的……没错。。。如果用户名和密码不在我创建的数据库中,则当单击按钮时,它将不会进入下一个窗口,否则它将继续。有关获取数据的问题已得到回答。要询问GTK事件处理程序,请提出一个新问题。