如何使用参数占位符将列名值作为SQL参数传递？

目标是使此工作正常进行：

var sql = "SELECT * FROM Condos WHERE @0 LIKE @1";
var sqlData = db.Query(sql,choice,"%"+searchString+"%");

choice 是一个变量，它将存储 column name

@0是column name（并且我没有成功将其作为参数传递）

@1是一个search string（我对此没有问题）

已经阅读并尝试了很多东西：

低于1不会引发任何错误，但不会带来任何数据

var sql = "SELECT * FROM Condos WHERE @0 LIKE @1";
var sqlData = db.Query(sql,choice,"%"+searchString+"%");

低于一投 Column name not valid = '@0'

var sql = "SELECT * FROM Condos WHERE [@0] LIKE @1"
var sqlData = db.Query(sql,choice,"%"+searchString+"%");

低于一投 Column name not valid = 'choice'

var sql "SELECT * FROM Condos WHERE choice LIKE @0");
var sqlData = db.Query(sql,"%"+searchString+"%");

低于一投 Must declare scalar variable "@choice"

var sql "SELECT * FROM Condos WHERE @choice LIKE @0");
var sqlData = db.Query(sql,"%"+searchString+"%");

低于1不会引发任何错误，但不会带来任何数据

var sql = "SELECT * FROM Condos WHERE '@choice' LIKE @0";
var sqlData = db.Query(sql,"%"+searchString+"%");

低于1不会引发任何错误，但不会带来任何数据

var sql = "SELECT * FROM Condos WHERE '@choice' LIKE @0";
var sqlData = db.Query(sql,"%"+searchString+"%");

低于1不会引发任何错误，但不会带来任何数据

var sql = "SELECT * FROM Condos WHERE '"+choice+"' LIKE @0";
var sqlData = db.Query(sql,"%"+searchString+"%");

下一个：BIG CRASH

var sql = "SELECT * FROM Condos WHERE '"+@choice+"' LIKE @0";
var sqlData = db.Query(sql,"%"+searchString+"%");

低于一投 Column name not valid = 'NameShort'

这正是正确的列名

var sql = "SELECT * FROM Condos WHERE ['"+choice+"'] LIKE @0";
var sqlData = db.Query(sql,"%"+searchString+"%");

下一个：BIG CRASH

var sql = "SELECT * FROM Condos WHERE ['"+@choice+"'] LIKE @0";
var sqlData = db.Query(sql,"%"+searchString+"%");

帮助 ！！！！！！！！！！！！！！