Question

filemgr = [NSFileManager defaultManager];
sqlite3_stmt *insertstatement=nil;

NSString *cruddatabase = [self.GetDocumentDirectory stringByAppendingPathComponent:@"SalaryBook.sqlite"];

sqlite3_open([cruddatabase UTF8String], &SalaryBook);

const char *sql = "insert into contactDB (name, designation, address, phone ,email ,identification, picture, doj) ?, ?,?,?,?,?,?,?";
sqlite3_prepare_v2(SalaryBook, sql, 1, &insertstatement, NULL);

sqlite3_bind_text(insertstatement,0,[theContact.name UTF8String],-1,SQLITE_TRANSIENT);
sqlite3_bind_text(insertstatement,1,[theContact.designation UTF8String],-1,SQLITE_TRANSIENT);
sqlite3_bind_text(insertstatement,2,[theContact.address UTF8String],-1,SQLITE_TRANSIENT);
sqlite3_bind_text(insertstatement,3,[theContact.phoneNumber UTF8String],-1,SQLITE_TRANSIENT);

NSLog(@"%@",theContact.phoneNumber);

sqlite3_bind_text(insertstatement,4,[theContact.emailAddress UTF8String],-1,SQLITE_TRANSIENT);
sqlite3_bind_text(insertstatement,5,[theContact.identificationProof UTF8String],-1,SQLITE_TRANSIENT);

NSData *dataForPicture = UIImagePNGRepresentation(theContact.image);

sqlite3_bind_blob(insertstatement,6,[dataForPicture bytes],[dataForPicture length],SQLITE_TRANSIENT);
sqlite3_bind_text(insertstatement,7,[theContact.dateOfJoining UTF8String],-1,SQLITE_TRANSIENT);

sqlite3_step(insertstatement);
sqlite3_finalize(insertstatement);

sqlite3_close(SalaryBook);

我可以从数据库中读取但无法写入任何内容。

Answer 1

插入字符串的格式是.. INSERT INTO table_name（column1，column2，column3，...） VALUES（value1，value2，value3，...）

所以在字符串中替换它（双引号）＆amp;尝试... 插入contactDB（名称，名称，地址，电话，电子邮件，身份证明，图片，doj）值（？，？，？，？，？，？，？，？）

我无法添加到我的sqlite数据库。请指教

1 个答案: