1

我不断收到一条错误消息，上面写着“调用类型 NSString 不是函数或函数指针。我已经被困了好几天了。我知道我的问题在哪里吗？

- (void)saveData
{

    sqlite3_stmt    *statement;
    const char *dbpath = [databasePath UTF8String];

    if (sqlite3_open(dbpath, &testdb1) == SQLITE_OK)
    {


        NSString *insertSQL = [NSString stringWithFormat: @"INSERT INTO TESTDB1"
                               (email, username, password, age) VALUES (\"%@\", \"%@\", \"%@\", \"%@\")",
                                email.text, username.text, password.text, age.integer];


        const char *insert_stmt = [insertSQL UTF8String];
        sqlite3_prepare_v2(testdb1, insert_stmt, -1, &statement, NULL);

        if (sqlite3_step(statement) == SQLITE_DONE)
        {
           status.text = @"User added";
           email.text = @"";
           username.text = @"";
           password.text = @"";
           age.text = @"";
        } else {
           status.text = @"Failed to add user";
        }
          sqlite3_finalize(statement);
          sqlite3_close(testdb1);
    }

}
4

2 回答 2

2 
NSString *insertSQL = [NSString stringWithFormat: @"INSERT INTO TESTDB1"
                   (email, username, password, age)

您过早地关闭引号；如果你仔细观察这个表达式，你会注意到

@"INSERT INTO TESTDB" (email, username, password, age)

所以编译器看到你正在尝试将 NSString 作为函数调用。将格式字符串文字更改为

NSString *insertSQL = [NSString stringWithFormat: @"INSERT INTO TESTDB1 (email, username, password, age) VALUES (\"%@\", \"%@\", \"%@\", \"%@\")",
                    email.text, username.text, password.text, age.integer];

你应该可以走了。

P.s.：在发布您的问题之前，请务必查看预览- 它的格式非常糟糕。

于 2012-10-06T16:47:30.650 回答
1

您在 stringWithFormat 调用的第二行开头缺少引号。

于 2012-10-06T16:48:22.593 回答