php - 嵌套查询sql，无法执行

Question

我想比较一些 float 值。要做，我的查询是这样的:

SELECT * FROM pricelist WHERE START_FREQUENCY >= '(
    SELECT START_FREQUENCY
    FROM pricelist
    WHERE START_FREQUENCY BETWEEN '0.49999' AND '0.50001'
)' AND STOP_FREQUENCY <= '(
    SELECT STOP_FREQUENCY
    FROM pricelist
    WHERE STOP_FREQUENCY BETWEEN '0.59999' AND '0.60001'
)'

但在我的 PHP 中它显示“无法执行查询”。我哪里做错了？请帮忙

编辑:

我有一个表，其中有两个字段，START_FREQUENCY 和 STOP_FREQUENCY。我正在尝试提取频率范围内的所有值。例如说

  START_FREQUENCY STOP_FREQUENCY
1            0.4            0.9
2            0.5            1.4
3            0.8            1.9

当用户输入 start frequency = 0.4 and stop frequency = 1.9 时，他应该得到 Row1、Row2 和 Row3 的结果。

Answer 1

不应引用您的内部 SELECT。试试这个:

SELECT * FROM pricelist WHERE START_FREQUENCY >= (
    SELECT START_FREQUENCY
    FROM pricelist
    WHERE START_FREQUENCY BETWEEN 0.49999 AND 0.50001
) AND STOP_FREQUENCY <= (
    SELECT STOP_FREQUENCY
    FROM pricelist
    WHERE STOP_FREQUENCY BETWEEN 0.59999 AND 0.60001
)

我还删除了您号码周围的引号； MySQL 会在您背后将它们转换为数字，但无需让数据库猜测并做更多工作。

您的版本具有以下结构:

SELECT *
FROM pricelist
WHERE START_FREQUENCY >= '...'0.49999' AND '0.50001'...'
  AND STOP_FREQUENCY  <= '...'0.59999' AND '0.60001'...'

并且 'string'0.49999' 无效。

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更新:听起来你只是把它复杂化了。我想你只是想要这个(使用你的 0.4 和 1.9):

select *
from pricelist
where START_FREQUENCY >= 0.4
  and STOP_FREQUENCY  <= 1.9