为什么我得到一个SQLITE_MISUSE:Out of Memory错误？

Question

我正在编写一个直接访问的iOS应用程序SQLite.我在Android上已经做了很多次这样的事情,所以我很难看到我的错误在哪里 - 但是我的插入返回SQLITE_MISUSE错误(代码21),消息"内存不足".以下是我采取的引导我插入的步骤.

一,表创建:

NSString *sql = @"CREATE TABLE IF NOT EXISTS UsersTable (lastName TEXT,id TEXT PRIMARY KEY NOT NULL,picture BLOB,firstName TEXT,age TEXT,email TEXT,sex TEXT,height TEXT,weight TEXT)";

//create the database if it does not yet exist
NSFileManager *filemgr = [NSFileManager defaultManager];
if ([filemgr fileExistsAtPath: path ] == NO)
{
    const char *dbpath = [path UTF8String];

    //This was if (sqlite3_open(dbpath, &store) == SQLITE_OK) , but it has not made a difference.
    if (sqlite3_open_v2(dbpath, &store, SQLITE_OPEN_READWRITE | SQLITE_OPEN_CREATE, NULL) == SQLITE_OK)
    {
        char *errMsg = NULL;
        const char *sql_stmt = [sql UTF8String];

        if (sqlite3_exec(store, sql_stmt, NULL, NULL, &errMsg) != SQLITE_OK)
        {
            NSLog(@"Failed to create table: %s", errMsg);
        }
        if (errMsg)
            sqlite3_free(errMsg);
    }
    else
    {
        NSLog(@"Failed to open/create database");
    }
}

接下来,插入(当前使用用户ID的电子邮件地址):

INSERT INTO UsersTable (id,lastName,firstName,email) VALUES ("jsmith@foobar.com","Smith","John","jsmith@foobar.com")

我正在为所有数据库交互使用一个选择器,因此上面的文本在这里传递:

-(int)execSQL:(NSString *)statement
{
    NSLog(@"%@",statement);

    const char *insert_stmt = [statement UTF8String];
    sqlite3_stmt *stmnt;

    sqlite3_prepare_v2(store, insert_stmt, -1, &stmnt, NULL);
    int result = sqlite3_step(stmnt);

    sqlite3_finalize(stmnt);

    if (result != SQLITE_OK)
    {
        NSLog(@"Error: %s", sqlite3_errmsg(store));//This prints "Error: out of memory"
    }
    return result;
}

我究竟做错了什么？

Answer 1

如果数据库不存在,您的打开例程仅创建/打开数据库.您的数据库可能已经存在,因此您的例程甚至没有打开它.

最重要的是,如果您尝试在不打开数据库的情况下调用SQLite函数,您将获得SQLITE_MISUSE返回代码(表示未以正确的顺序调用SQLite函数),并且sqlite3_errmsg将返回神秘的"内存不足"错误.

其他一些不相关的观察结果:

1. 你真的应该检查一下返回代码sqlite3_prepare:

   - (int)execSQL:(NSString *)statement
   {
       int result;
   
       NSLog(@"%@",statement);
   
       const char *insert_stmt = [statement UTF8String];
       sqlite3_stmt *stmnt;
   
       if ((result = sqlite3_prepare_v2(store, insert_stmt, -1, &stmnt, NULL)) != SQLITE_OK)
       {
           NSLog(@"%s: prepare failure '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
           return result;
       }
   
       if ((result = sqlite3_step(stmnt)) != SQLITE_DONE)
       {
           NSLog(@"%s: step failure: '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
       }
   
       sqlite3_finalize(stmnt);
   
       return result;
   }
   

   根据我的经验,许多常见的开发问题都与SQL本身有关,这是通过检查sqlite3_prepare_v2语句的返回代码来识别的.

2. 你真的不应该在一个构建你的SQL语句NSString.你打开自己的SQL注入攻击,或者考虑到更好的情况,如果有人的名字中有一个引号,只是一个SQL错误,例如The "Destroyer".您应该使用?占位符,然后使用sqlite3_bind_xxx函数绑定值.就像是:

   - (int)insertIdentifier:(NSString *)identifier
                  lastName:(NSString *)lastName
                 firstName:(NSString *)firstName
                     email:(NSString *)email
   {
       int result;
   
       const char *insert_stmt = "INSERT INTO UsersTable (id, lastName, firstName, email) VALUES (?, ?, ?, ?);";
       sqlite3_stmt *stmnt;
   
       if ((result = sqlite3_prepare_v2(store, insert_stmt, -1, &stmnt, NULL)) != SQLITE_OK)
       {
           NSLog(@"%s: prepare failure '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
           return result;
       }
   
       if ((result = sqlite3_bind_text(stmnt, 1, [identifier UTF8String], -1, NULL)) != SQLITE_OK)
       {
           NSLog(@"%s: bind #1 failure '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
           sqlite3_finalize(stmnt);
           return result;
       }
   
       if ((result = sqlite3_bind_text(stmnt, 2, [lastName UTF8String], -1, NULL)) != SQLITE_OK)
       {
           NSLog(@"%s: bind #2 failure '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
           sqlite3_finalize(stmnt);
           return result;
       }
   
       if ((result = sqlite3_bind_text(stmnt, 3, [firstName UTF8String], -1, NULL)) != SQLITE_OK)
       {
           NSLog(@"%s: bind #3 failure '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
           sqlite3_finalize(stmnt);
           return result;
       }
   
       if ((result = sqlite3_bind_text(stmnt, 4, [email UTF8String], -1, NULL)) != SQLITE_OK)
       {
           NSLog(@"%s: bind #4 failure '%s' (%d)", __FUNCTION__, sqlite3_errmsg(store), result);
           sqlite3_finalize(stmnt);
           return result;
       }
   
       if ((result = sqlite3_step(stmnt)) != SQLITE_DONE)
       {
           NSLog(@"%s: step failure: '%s'", __FUNCTION__, sqlite3_errmsg(store));
       }
   
       sqlite3_finalize(stmnt);
   
       return result;
   }
   

   然后您可以这样调用:

   [self insertIdentifier:@"jsmith@foobar.com"
                 lastName:@"Smith"
                firstName:@"John"
                    email:@"jsmith@foobar.com"];
   

3. 正如您所看到的,当您开始编写代码时,您正在适当地检查每个返回值,绑定每个变量等,您的SQLite代码会很快变得毛茸茸.我建议你考虑看FMDB.它是围绕SQLite函数的一个很好的,薄的包装器,它极大地简化了在Objective-C中编写SQLite代码的练习.