Php 无法从MySQL中提取值

我的代码有问题。我创建了一个类来搜索作业id或任务id。但是我无法在MySQL中检索信封id。该值应为TESTTEST

<?php
class Envelope {
    private $_db, $_data;
    public function __construct($taskidorjobid = null, $taskid = null) {
        $this -> _db = DB::getInstance();
        //$this -> _sessionName = Config::get('session/session_name');
        //$this -> _cookieName = Config::get('remember/cookie_name');
        if ($taskidorjobid == null && $taskid == null) {

        } elseif ($taskid == null) {
            $this -> findtaskid($taskidorjobid);
        } else {
            $this -> find($taskidorjobid, $taskid);
        }
    }

    public function findtaskid($taskid = null) {
        if ($taskid) {
            $field = 'id';
            $data = $this -> _db -> get('task', array($field, '=', $taskid));
            if ($data -> count()) {
                $this -> _data = $data -> first();
                return true;
            }
        }
        return false;
    }

    public function find($jobid = null, $taskid = null) {
        if ($jobid and $taskid) {
            //$field = 'id';
            //$data = $this -> _db -> get('task', array($field, '=', $task));
            $data = $this -> _db -> query("select * from document where job = " . $jobid . " and taskid = " . $taskid);
            if ($data -> count()) {
                $this -> _data = $data -> first();
                return true;
            }
        }
        return false;
    }

    public function findenvelopeid($taskid = null, $version = null) {

            $sql = "select envelope_id from document where taskid = ?, version = ?, latestflag = ?";
            $param = array($taskid, $version, "yes");


            $data = $this-> _db->query($sql, $param);



       //foreach ($data->results() as $row) {
        if ($data -> count()) {
                $this -> _data = $data -> first();
                $envelope_id = $this -> _data -> envelope_id;
                return $envelope_id;

        }
        return false;

       }
}        
?>

然后在我的另一组php文件中，我调用了函数findenvelopeid，但它不返回任何值。我可以将$tid作为$taskid进行回显

$envelope = new Envelope();
$envelopeid = $envelope->findenvelopeid($tid,"signed_online");
echo "Envelope ID: ".$envelopeid. "<br>";

我的编码器开发的查询功能是：

public function query($sql,$params=array(), $options=array()){
    $this->_error=false;
    if($this->_query=$this->_pdo->prepare($sql)){
        $x=1;
        if(count($params)){
            foreach($params as $param){
                //if(count($options)>=$x){
                    //var_dump("DB sql: " . $sql . "<br/> x: " . $x . ", param: " . $param . ", options:" . $options[$x-1]);
                //   $this->_query->bindValue($x,$param, $options[$x-1]);
                //}else{
                    //var_dump("DB sql: " . $sql . "<br/> x: " . $x . ", param: " . $param);
                   $this->_query->bindValue($x,$param);
                //}
                $x++;
            }
        }
        if($this->_query->execute()){
            $this->_results=$this->_query->fetchAll(PDO::FETCH_OBJ);
            $this->_count=$this->_query->rowCount();
        }else{
            //var_dump("<br /> errorcode " . $this->_pdo->errorCode() . "<br />error info:");
            //print_r($this->_pdo->errorInfo());
            $this->_error=true;
        }
    }
    return $this;
}

---

请帮忙

哎呀，我应该在where子句中使用AND而不是a。

这有点乱。为什么$db=db:：getInstance；当您的构造函数已经这样做并将其保存在$this->\u db上时，您已经知道，然后在方法代码中使用$this->\u db，如果您使用的是标记参数，即？您需要准备sql语句，然后->执行它，这真的很奇怪。因为我以前编写代码的程序员没有使用prepare和execute.like:public static函数getalldocumentsbytaskid$taskid{$db=db:：getInstance；$documents=array；$sql=select*从taskid=？和latestflag=？；$param=array$taskid，是；$data=$db->query$sql，$param；//$data=$db->getdocument，ArrayTaskKid，=，$taskid；如果$data{foreach$data->结果为$row{$documents[]=新文档$row->id；}}返回$documents；}