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如何使用php在html表中显示MySQL数据库中的数据

更新时间:2023-01-31 09:36:44

文档 mysqli_query 在过程风格中使用时需要2个参数。我假设 $ connection 是您的mysqli链接尝试:

As stated in the docs. mysqli_query takes 2 parameters when used in a procedural style. I'm assuming $connection is your mysqli link Try:

$result = mysqli_query($connection, "SELECT * FROM projects ORDER BY project_title ASC");