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我正在尝试编写以下函数来显示 MySQL 表的内容。

$q=$_GET["q"];
function risk_allocation($db)  
  {    
    $result = $db->query("select r.risks as risks,r.risktype as risktype,j.job as job from risks r LEFT OUTER JOIN `jobsrisks` j on r.risks = j.risk and j.job=:q"); 
    $result ->bindParam(':q', $q, PDO::PARAM_INT);
    return $result; 
  }
        $allocationlist = risk_allocation($db);

然后我用以下方法调用函数：

while($row = $allocationlist->fetch(PDO::FETCH_ASSOC)) 
{
echo $line['risks'];
echo $line['risktype'];
}

我收到错误消息：

Fatal error: Call to a member function bindParam() on a non-object in /home/she/public_html/versionfour/getrisksperjob.php on line 11

第 11 行在哪里

$result ->bindParam(':q', $q, PDO::PARAM_INT);

我觉得这是一个简单的问题，因为我刚刚被介绍给 pdo，但任何帮助都一如既往地受到赞赏。

更新

根据建议的答案，我的代码现在如下执行查询并包含变量 q。但是仍然会发生相同的错误。感谢您迄今为止的帮助，有什么想法吗？

致命错误：在第 11 行的 /home/she/public_html/versionfour/getrisksperjob.php 中的非对象上调用成员函数 bindParam()

function risk_allocation($db,$q)  
  {    
    $result = $db->query("select r.risks as risks,r.risktype as risktype,j.job as job from risks r LEFT OUTER JOIN `jobsrisks` j on r.risks = j.risk and j.job=:q"); 
    $result ->bindParam(':q', $q, PDO::PARAM_INT);
    $result->execute();
    return $result; 
  }
  $allocationlist = risk_allocation($db,$q);
4

2 回答 2

3

您忘记执行查询，此外您需要准备而不是调用查询：

$result = $db->prepare("select r.risks as risks,r.risktype as risktype,j.job as job from risks r LEFT OUTER JOIN `jobsrisks` j on r.risks = j.risk and j.job=:q"); 
$result->bindParam(':q', $q, PDO::PARAM_INT);
$result->execute();
于 2012-10-29T08:42:44.317 回答
2

1.你forgot to execute query 

$result->execute();

2.$q将outside function so you will never get value of that inside function.$q 作为函数的第二个参数传递

$q=$_GET["q"];
function risk_allocation($db,$q)  
  {    
    $result = $db->prepare("select r.risks as risks,r.risktype as risktype,j.job as job from risks r LEFT OUTER JOIN `jobsrisks` j on r.risks = j.risk and j.job=:q"); 
    $result ->bindParam(':q', $q, PDO::PARAM_INT);
    $result->execute();
    return $result; 
  }
  $allocationlist = risk_allocation($db,$q);
于 2012-10-29T08:47:29.147 回答