在 PDO 中实现 LIKE 查询 [英] implement LIKE query in PDO

问题描述

我在 PDO 中实现 LIKE 时遇到问题

I am running problems in implementing LIKE in PDO

我有这个查询:

$query = "SELECT * FROM tbl WHERE address LIKE '%?%' OR address LIKE '%?%'";
$params = array($var1, $var2);
$stmt = $handle->prepare($query);
$stmt->execute($params);

我检查了 $var1 和 $var2 它们包含我想要搜索的两个词，我的 PDO 工作正常，因为我的一些查询 SELECT INSERT 他们工作，只是我不熟悉LIKE 这里的PDO.

I checked the $var1 and $var2 they contain both the words I want to search, my PDO is working fine since some of my queries SELECT INSERT they work, it's just that I am not familiar in LIKE here in PDO.

结果没有返回.我的 $query 在语法上是否正确?

The result is none returned. Do my $query is syntactically correct?

推荐答案

您必须在 $params 中包含 % 符号，而不是在查询中:

You have to include the % signs in the $params, not in the query:

$query = "SELECT * FROM tbl WHERE address LIKE ? OR address LIKE ?";
$params = array("%$var1%", "%$var2%");
$stmt = $handle->prepare($query);
$stmt->execute($params);

如果您查看之前代码中生成的查询，您会看到类似 SELECT * FROM tbl WHERE address LIKE '%"foo"%' OR address LIKE '%"bar"%' 的内容，因为准备好的语句在已经引用的字符串中引用您的值.

If you'd look at the generated query in your previous code, you'd see something like SELECT * FROM tbl WHERE address LIKE '%"foo"%' OR address LIKE '%"bar"%', because the prepared statement is quoting your values inside of an already quoted string.

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