//JUST little bit modified the answer of giuseppe  which returns array of table columns
+(NSMutableArray*)tableInfo:(NSString *)table{

    sqlite3_stmt *sqlStatement;

    NSMutableArray *result = [NSMutableArray array];

    const char *sql = [[NSString stringWithFormat:@"PRAGMA table_info('%@')",table] UTF8String];

    if(sqlite3_prepare(md.database, sql, -1, &sqlStatement, NULL) != SQLITE_OK)

    {
        NSLog(@"Problem with prepare statement tableInfo %@",
                [NSString stringWithUTF8String:(const char *)sqlite3_errmsg(md.database)]);

    }

    while (sqlite3_step(sqlStatement)==SQLITE_ROW)
    {
        [result addObject:
          [NSString stringWithUTF8String:(char*)sqlite3_column_text(sqlStatement, 1)]];
    }

    return result;
}

我知道这是一个旧的帖子，但最近我需要同样的帖子，并找到了一个简洁的方法:

SELECT c.name FROM pragma_table_info('your_table_name') c;

这是一个老问题，但这里有一个替代答案，检索SQLite数据库中的所有列，并为每列提供关联表的名称:

WITH tables AS (SELECT name tableName, sql 
FROM sqlite_master WHERE type = 'table' AND tableName NOT LIKE 'sqlite_%')
SELECT fields.name, fields.type, tableName
FROM tables CROSS JOIN pragma_table_info(tables.tableName) fields

返回这样的结果:

{
    "name": "id",
    "type": "integer",
    "tableName": "examples"
}, {
    "name": "content",
    "type": "text",
    "tableName": "examples"
}

对于包含标识符和字符串内容的简单表。

如果您正在搜索任何特定的列，您可以使用Like语句

ex:

SELECT * FROM sqlite_master where sql like('%LAST%')

function getDetails(){
var data = [];
dBase.executeSql("PRAGMA table_info('table_name') ", [], function(rsp){
    if(rsp.rows.length > 0){
        for(var i=0; i<rsp.rows.length; i++){
            var o = {
                name: rsp.rows.item(i).name,
                type: rsp.rows.item(i).type
            } 
            data.push(o);
        }
    }
    alert(rsp.rows.item(0).name);

},function(error){
    alert(JSON.stringify(error));
});             
}

当你运行sqlite3命令行时，输入:

sqlite3 -header

也会给你想要的结果吗