Sqlite“附近"iphone 语法错误 [英] Sqlite "near" syntax error on iphone

问题描述

我正在我的 iphone 应用程序中试验 sqlite 问题.读取部分没问题，但是当我尝试使用 UPADATE 命令写入数据库时​​(我也尝试使用 INSERT)，sql 语句返回错误 sqlite3_errmsg:

I am experimenting a problem with sqlite in my iphone app. Read part is ok, but when I try to write to a database using UPADATE command (I try with INSERT too), the sql sentence give me back the error sqlite3_errmsg:

Failed. Error is:  near "UP": syntax error

代码是:

-(void)ratesInfotoDb:(NSString *)idx rate:(NSString*)rate{

    query =@"UPDATE rules set rate = 5 where _id = 1994";

    sqlite3_stmt *statement;
    sqlite3_prepare_v2(_database, [query UTF8String], SQLITE_OPEN_READWRITE, &statement, nil);

    if(sqlite3_step(statement) == SQLITE_DONE ) {
        NSLog(@"element added");
        }
    else{
        NSLog( @"Failed. Error is:  %s", sqlite3_errmsg(_database));
    }

    sqlite3_finalize(statement);
    sqlite3_close(_database);

}

查询命令UPDATE rules set rate = 5 where _id = 1994"(直接在数据库上执行)有效，所以我认为查询是正确的.

The query command "UPDATE rules set rate = 5 where _id = 1994" (executed directly on a database), works, so I suppose query is corect.

有人可以帮我吗?

推荐答案

问题很简单.您将错误的值传递给 sqlite3_prepare_v2 函数.第三个参数应该代表查询字符串的长度.

The problem is simple. You are passing the wrong value to the sqlite3_prepare_v2 function. The 3rd parameter should represent the length of the query string.

宏 SQLITE_OPEN_READWRITE 的值为 2，因此您声明查询只有 2 个字节长，这是不正确的.

The macro SQLITE_OPEN_READWRITE has a value of 2 so you are stating that the query is only 2 bytes long which is not true.

更改对 sqlite3_prepare_v2 的调用，如下所示:

Change your call to sqlite3_prepare_v2 as follows:

sqlite3_prepare_v2(_database, [query UTF8String], -1, &statement, nil);

传递-1 告诉sqlite3_prepare_v2 获取查询的实际长度(可能通过使用strlen).

Passing -1 tells sqlite3_prepare_v2 to get the actual length of the query (probably by using strlen).

顺便说一句 - 始终检查 sqlite3_prepare_v2 的返回值并确保它返回 SQLITE_OK.

BTW - always check the return value of sqlite3_prepare_v2 and make sure it returns SQLITE_OK.

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