我有一个SQLite数据库,但是当我尝试动态添加它不起作用的项目时。
这是添加脚本:
-(void) addPatientToDatabase:(Patient *)newPatient {
NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
NSString *documentsPath = [paths objectAtIndex:0];
NSString *filePath = [documentsPath stringByAppendingPathComponent:@"cities.sqlite"];
sqlite3 *database;
if(sqlite3_open([filePath UTF8String], &database) == SQLITE_OK) {
const char *sqlStatement = "insert into patients (firstName, surname, dob, homeNumber, mobileNumber, email, address, image) VALUES (?, ?, ?, ?, ?, ?, ?, ?)";
sqlite3_stmt *compiledStatement;
if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
sqlite3_bind_text(compiledStatement, 1, [newPatient.patientName UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(compiledStatement, 2, [newPatient.patientSurname UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(compiledStatement, 3, [newPatient.patientDoB UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(compiledStatement, 4, [newPatient.patientHomeNumber UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(compiledStatement, 5, [newPatient.patientMobileNumber UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(compiledStatement, 6, [newPatient.patientEmail UTF8String], -1, SQLITE_TRANSIENT);
sqlite3_bind_text(compiledStatement, 7, [newPatient.patientAddress UTF8String], -1, SQLITE_TRANSIENT);
NSData *dataForPicture = UIImagePNGRepresentation(newPatient.patientPicture);
sqlite3_bind_blob(compiledStatement, 8, [dataForPicture bytes], [dataForPicture length], SQLITE_TRANSIENT);
}
if(sqlite3_step(compiledStatement) == SQLITE_DONE) {
sqlite3_finalize(compiledStatement);
}
}
sqlite3_close(database);
}
我添加了一些断点并注意到如果if语句中有断点:
if(sqlite3_step(compiledStatement) == SQLITE_DONE) {
然后它没有被拿起。
提前致谢
答案 0 :(得分:0)
当step返回除SQLITE_DONE以外的任何内容时,请尝试调用:sqlite3_errmsg(database);
它将返回指向错误消息(http://www.sqlite.org/c3ref/errcode.html)的指针。显示或记录该消息,以找出错误的信息。