PDO:“无效参数号"(Invalid parameter number).当用相同的值替换多个参数时 [英] PDO: "Invalid parameter number" when substituting multiple parameters with same value

问题描述

如果参数在查询中多次出现，如何绑定我的参数，如下所示?

How do I bind my parameter if it appears multiple times in the query as follows?

$STH = $DBH->prepare("SELECT * FROM $table WHERE firstname LIKE :string OR lastname LIKE :string");

$STH->bindValue(':string', '%'.$string.'%', PDO::PARAM_STR);
$result = $STH->execute();

推荐答案

您为prepare语句提到了两个参数(具有相同的名称)，但是仅为第一个参数提供了一个值(这就是错误所在)

You mentioned two parameters (with the same name) for the prepare statement, yet you supply a value for the first parameter only (that's what the error was about).

不太确定PDO如何在内部解决相同参数名称的问题，但是您始终可以避免这种情况.

Not quite sure how PDO internally solved the same-parameter-name issue, but you can always avoid that.

两种可能的解决方案:

$sql = "select * from $table ".
       "where "
       "first_name like concat('%', :fname, '%') or ".
       "last_name  like concat('%', :lname, '%')";
$stmt= $DBH->prepare($sql);
$stmt->bindValue(':fname', $string, PDO::PARAM_STR);
$stmt->bindValue(':lname', $string, PDO::PARAM_STR);

---

$sql = "select * from $table ".
       "where "
       "first_name like concat('%', ?, '%') or ".
       "last_name  like concat('%', ?, '%')";
$stmt= $DBH->prepare($sql);
$stmt->bindValue(1, $string, PDO::PARAM_STR);
$stmt->bindValue(2, $string, PDO::PARAM_STR);

---

顺便说一句，您现有的方法仍然存在SQL注入问题.

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By the way, the existing way you have done still has SQL injection issues.

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