是否可以使用此功能:
function getProjectData($uid, $what) {
$sql = "SELECT ? FROM projects WHERE fid = ?";
$stmt = $this->db->prepare($sql);
$stmt->execute(array($uid,$what);
return $stmt->fetch(PDO::FETCH_LAZY);
}
“uid”是用户ID,“what”将是我想要的列,例如“title”所以查询将是:
SELECT title FROM projects WHERE fid = 1
这可能吗?
答案 0 :(得分:1)
在PDO中无法替换表和列名称:
看看这里: Can PHP PDO Statements accept the table or column name as parameter?
答案 1 :(得分:1)
function getProjectData($uid) {
$sql = "SELECT * FROM projects WHERE fid = ?";
$stmt = $this->db->prepare($sql);
$stmt->execute(array($uid);
return $stmt->fetch();
}
这样称呼
$proj_data = getProjectData($uid);
然后获得所需的属性
$title = $proj_data[$what];