Home > Database > Mysql Tutorial > body text

Why Am I Getting a MySQLSyntaxErrorException with My Prepared Statement?

Mary-Kate Olsen
Release: 2024-11-01 08:25:02
Original
803 people have browsed it

Why Am I Getting a MySQLSyntaxErrorException with My Prepared Statement?

MySQL Prepared Statement Error: MySQLSyntaxErrorException

In Java programming, the MySQLSyntaxErrorException is thrown when there is an issue with the syntax of an SQL statement. One scenario where this error may occur is when using prepared statements with parameterized queries. Let's delve into a specific case and provide a solution.

The given code contains a select statement using a prepared statement to retrieve data from the ads_tbl table based on an ad_id. However, an error is encountered when executing the statement: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1. This indicates a problem with the syntax of the SQL statement.

Upon closer examination, the issue lies in the line:

<code class="java">rs = stmt.executeQuery(selectSQL);</code>
Copy after login

The correct way to execute a prepared statement is to omit the SQL statement itself as an argument to executeQuery(). This is because the prepared statement has already been set up with the appropriate parameters, and calling executeQuery() without a SQL statement will execute the prepared statement.

Therefore, the code should be modified to the following:

<code class="java">rs = stmt.executeQuery();</code>
Copy after login

By making this change, the SQL statement will be executed correctly using the prepared statement, and the error should be resolved.

The above is the detailed content of Why Am I Getting a MySQLSyntaxErrorException with My Prepared Statement?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!