Question

NSString *query =  [[NSString alloc] initWithFormat:@"SELECT * FROM objects ", catIdS];
sqlite3_stmt *statement;
    if (sqlite3_prepare_v2(database, [query UTF8String], -1, &statement, nil) == SQLITE_OK) {
        while (sqlite3_step(statement) == SQLITE_ROW) {
            char *nameChar = sqlite3_column_text(statement, 1);

            NSString *title = [[NSString alloc] initWithUTF8String:nameChar];
       }

       sqlite3_finalize(statement);
    }

语义问题＆＃34;使用const char类型的表达式初始化char *会丢弃限定符＆＃34;在char *nameChar = sqlite3_column_text(statement, 1)

行

如果将char *nameChar更改为char *nameChar 语义问题＆＃34;将unsigned char *发送到const char *＆＃34;类型的参数;将指针转换为具有不同符号的整数类型

Answer 1

施放它应该可以解决您的问题。

char *nameChar = (char *)sqlite3_column_text(statement, 1);

或

const unsigned char *nameChar = sqlite3_column_text(statement, 1);

请查看方法的签名

SQLITE_API const unsigned char *sqlite3_column_text(sqlite3_stmt*, int iCol);

Answer 2

使用它像这样，它可能会有所帮助

    NSString *title = [[NSString alloc] initWithUTF8String:(char *)sqlite3_column_text(statement, 1)];

使用const char类型的表达式初始化char *

2 个答案: