pdo 不绑定值答案

【问题标题】：pdo not binding valuepdo 不绑定值
【发布时间】：2012-10-10 18:41:31
【问题描述】：

请尝试从 mysqli 转移到 pdo 但运行此代码

          <?php
  $sql = "select concat(register.fname ,' ' ,register.lname) as name,register.matric   as matric,register.username as uname,register.sex as sex,register.phone as phone ,register.passport as passport ,register.email as email,register.level as level,faculty.fac_name as fac,dept.dept_name as dept,diary.diary as diary,diary.date_added as added from register,faculty,dept,diary where register.user_id = :_id  and diary.username = :name limit 1";
$sth = $dbh->prepare($sql);
$sth->bindValue(':id', $id, PDO::PARAM_INT);
$sth->bindParam(':name', $username);
$sth->execute();


    ?>

显示此错误 (!) 警告：PDOStatement::bindValue() [pdostatement.bindvalue]: SQLSTATE[HY093]: Invalid parameter number: parameter was not defined in C:\wamp\www\uni\det\viewuser .php 在第 14 行

【问题讨论】：

标签： php pdo mysqli

【解决方案1】：

在查询中，您在绑定值 :id 中将其命名为 :_id。

【讨论】：

【解决方案2】：

您有:_id，请在您的bindValue 调用中更改它

【讨论】：

【解决方案3】：

您将变量$id 绑定到:id，但在语句中您需要:_id。

$sth->bindValue(':_id', $id, PDO::PARAM_INT);

或者换个说法：

  $sql = "select concat(register.fname ,' ' ,register.lname) as name,register.matric   as matric,register.username as uname,register.sex as sex,register.phone as phone ,register.passport as passport ,register.email as email,register.level as level,faculty.fac_name as fac,dept.dept_name as dept,diary.diary as diary,diary.date_added as added".
           "from register,faculty,dept,diary ".
          " where register.user_id = :id  and diary.username = :name limit 1";

【讨论】：