php - 如何在php中使用sql变量传递sql查询

Question

我想在 php 中传递以下 sql 查询

SET @rank=0; SELECT @rank:=@rank+1 AS rank, `percentage_obtained`  FROM `result`  ORDER BY percentage_obtained DESC

如这里

mysql_query('SET @rank=0; SELECT @rank:=@rank+1 AS rank, `percentage_obtained`  FROM `result`  ORDER BY percentage_obtained DESC') or die(mysql_error());

但它抛出以下错误

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 'SELECT @rank:=@rank+1 AS rank, percentage_obtained FROM result ' at line 1

但是，当我直接在数据库内运行查询时，它会起作用。请帮忙。

Answer 1

mysql_query()不支持多个查询，因此需要将它们分开。 您也可以使用 mysqli_multi_query - 但请注意，您需要 mysqli 而不是 mysql

所以

mysql_query('SET @rank=0') or die(mysql_error());
mysql_query('SELECT @rank:=@rank+1 AS rank, `percentage_obtained`  FROM `result`  ORDER BY percentage_obtained DESC') or die(mysql_error());

应该处理这个问题