iOS : Could not convert const char to NSString object

Question

我需要将 const char 类型的数据转换为 NSString 对象类型，因为 sqlite 数据库返回 char * 类型的数据。为此，我使用了 NSString 类方法 stringWithUTF8strin，但它是返回 NULL。它获取 sqlite 文本类型数据。

这是我的代码:

-(void)getAllData:(const char *)sqlQuery
{
name=[[NSMutableArray alloc]init];

department=[[NSMutableArray alloc]init];
image=[[NSMutableArray alloc]init];
testimonial=[[NSMutableArray alloc]init];
@try
{

NSArray *pathArray=NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
NSString *final_phone_path=[[pathArray objectAtIndex:0]stringByAppendingPathComponent:@"amarinf5_excel.sqlite"];
NSLog(@"%@",final_phone_path);

NSFileManager *myFileManager=[NSFileManager defaultManager];

if(![myFileManager fileExistsAtPath:final_phone_path])
{
    NSLog(@"file doesnot exist");
}
if( sqlite3_open([final_phone_path UTF8String], &dataBAse)!=SQLITE_OK)
{
   NSLog(@"could not be able to open the file");
}
const char *sql=sqlQuery;
sqlite3_stmt *statement;
if(sqlite3_prepare_v2(dataBAse, sql, -1, &statement, nil)!=SQLITE_OK)
{
    NSLog(@"Error message fail to prepare statement %s",sqlite3_errmsg(dataBAse));
}


while(sqlite3_step(statement)==SQLITE_ROW)
{
    const char *name1=(char *)sqlite3_column_text(statement, 0);
    const char *deprtment1=(char *)sqlite3_column_text(statement, 1);
    const char *image1= (char *)sqlite3_column_text(statement, 2);
    **const char *testimonial1=(char *)sqlite3_column_text(statement, 3);
    NSLog(@"%@",[NSString stringWithUTF8String:testimonial1]);**
    NSLog(@"%s",testimonial1);
    NSLog(@"%@",[NSString stringWithUTF8String:image1]);
    [name addObject:[NSString stringWithUTF8String:name1]];
    [department addObject:[NSString stringWithUTF8String:deprtment1]];
    [image addObject:[NSString stringWithUTF8String:image1]];

    //[testimonial addObject:[NSString stringWithUTF8String:testimonial1]];
    NSLog(@"%@",[NSString stringWithUTF8String:testimonial1]);
}
if (sqlite3_finalize(statement) != SQLITE_OK)
{
    NSLog(@"Failed to finalize data statement, normally error handling here.");
}
if (sqlite3_close(dataBAse) != SQLITE_OK)
{
    NSLog(@"Failed to close database, normally error handling here.");
}
}
@catch (NSException *exception) {
NSLog(@"An exception occurred: %@", [exception reason]);

}
@finally {

}

}

Answer 1

当我使用 stringWithFormat 方法绑定(bind) NSString 解决方案的不同方法时，下面是我的代码

 while(sqlite3_step(statement1)==SQLITE_ROW)
{
    const char *name1=(char *)sqlite3_column_text(statement1, 0);
    const char *deprtment1=(char *)sqlite3_column_text(statement1, 1);
    const char *image1= (char *)sqlite3_column_text(statement1, 2);
    const char *testimonial1=(char *)sqlite3_column_text(statement1, 3);


    [name addObject:[NSString stringWithUTF8String:name1]];
    [department addObject:[NSString stringWithUTF8String:deprtment1]];
    [image addObject:[NSString stringWithUTF8String:image1]];
    [testimonial addObject:[NSString stringWithFormat:@"%s",testimonial1]];

}