将命名参数与 PDO 一起用于 LIKE-php问题

Using named parameters with PDO for LIKE(将命名参数与 PDO 一起用于 LIKE)

本文介绍了将命名参数与 PDO 一起用于 LIKE的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在尝试使用 LIKE 在我的数据库中搜索 name 字段.如果我像这样手工"制作 SQL:

I am trying to search the name field in my database using LIKE. If I craft the SQL 'by hand` like this:

$query = "SELECT * 
"
        . "FROM `help_article` 
"
        . "WHERE `name` LIKE '%how%'
"
        . "";
$sql = $db->prepare($query);
$sql->setFetchMode(PDO::FETCH_ASSOC);
$sql->execute();

然后它会返回'how'的相关结果.
然而，当我把它变成一个准备好的语句时:

Then it will return relevant results for 'how'.
However, when I turn it into a prepared statement:

$query = "SELECT * 
"
        . "FROM `help_article` 
"
        . "WHERE `name` LIKE '%:term%'
"
        . "";
$sql->execute(array(":term" => $_GET["search"]));
$sql->setFetchMode(PDO::FETCH_ASSOC);
$sql->execute();

我总是得到零结果.

我做错了什么?我在代码的其他地方使用了准备好的语句，它们工作正常.

What am I doing wrong? I am using prepared statements in other places in my code and they work fine.

推荐答案

绑定的 :placeholders 不能用单引号括起来.这样它们就不会被解释，而是被视为原始字符串.

The bound :placeholders are not to be enclosed in single quotes. That way they won't get interpreted, but treated as raw strings.

当你想使用一个作为 LIKE 模式时，然后将 % 与值一起传递:

When you want to use one as LIKE pattern, then pass the % together with the value:

$query = "SELECT * 
          FROM `help_article` 
          WHERE `name` LIKE :term ";

$sql->execute(array(":term" => "%" . $_GET["search"] . "%"));

哦，实际上你需要先清理这里的输入字符串(addcslashes).如果用户在参数中提供了任何无关的 % 字符，则它们将成为 LIKE 匹配模式的一部分.请记住，整个 :term 参数作为字符串值传递，并且该字符串中的所有 % 都成为 LIKE 子句的占位符.

Oh, and actually you need to clean the input string here first (addcslashes). If the user supplies any extraneous % chars within the parameter, then they become part of the LIKE match pattern. Remember that the whole of the :term parameter is passed as string value, and all %s within that string become placeholders for the LIKE clause.

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