PostgreSQL 输入结束时的语法错误 [英] Syntax error at end of input in PostgreSQL

问题描述

我在 MySQL 和 PostgreSQL 中都使用了下一条 SQL 语句，但在 PostgreSQL 中失败

db.Query(`SELECT COUNT(*) as N FROM email WHERE address = ?`, email)

出现此错误:

pq: F:"scan.l" M:"输入结束时的语法错误" S:"ERROR" C:"42601" P:"50" R:"scanner_yyerror" L:"993"

有什么问题吗?PostgreSQL 中的错误信息非常神秘.

解决方案

您还没有提供有关语言/环境的任何详细信息，但我还是会尝试胡乱猜测:

MySQL 的预处理语句本身使用 ? 作为参数占位符，但 PostgreSQL 使用 $1、$2 等.尝试替换 ? 和 $1 看看它是否有效:

WHERE 地址 = $1

<块引用>

PostgreSQL 中的错误信息非常神秘.

总的来说，我发现 Postgres 错误消息比竞争产品(嗯，MySQL 和特别是 Oracle)要好，但是在这种情况下，您已经设法使解析器变得异常混乱.:)

I have used the next SQL statement in both MySQL and PostgreSQL, but it fails in PostgreSQL

db.Query(`SELECT COUNT(*) as N FROM email WHERE address = ?`, email)

with this error:

pq: F:"scan.l" M:"syntax error at end of input" S:"ERROR" C:"42601" P:"50" R:"scanner_yyerror" L:"993"

What's the problem? The error messages in PostgreSQL are very cryptic.

解决方案

You haven't provided any details about the language/environment, but I'll try a wild guess anyway:

MySQL's prepared statements natively use ? as the parameter placeholder, but PostgreSQL uses $1, $2 etc. Try replacing the ? with $1 and see if it works:

WHERE address = $1

The error messages in PostgreSQL are very cryptic.

In general, I've found that Postgres error messages are better than competing products (ahem, MySQL and especially Oracle), but in this instance you've managed to confuse the parser beyond sanity. :)

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