如何修复php.DEBUG:fsockopen():连接被拒绝了?

时间:2016-03-19 07:26:15

标签: php symfony fsockopen

我想使用symfony的服务器,首先关闭我的apache2。

;with DeltaCte as (
    SELECT DateTime, CurrValue, 
        LAG(CurrValue, 1, CurrValue + 1) OVER (ORDER BY DateTime) AS PrevVal, 
        LEAD(CurrValue, 1, CurrValue - 1) OVER (ORDER BY DateTime) AS NextVal
    FROM RawData
)
,GroupsCTE AS (
    select DateTime, CurrValue, CurrValue - PrevVal AS Delta1, CurrValue - NextVal AS Delta2, (ROW_NUMBER() OVER (ORDER BY DateTime) + 1) / 2 AS GroupNo
    FROM DeltaCte
    WHERE (CurrValue - PrevVal < 0) OR (NextVal - CurrValue <  0)
)
SELECT GroupNo, MIN(d) AS DateTimeMin, MAX(d) DateTimeMax, 
    MIN(v) AS CurrValueMin, MAX(v) CurrValueMax
from GroupsCTE
UNPIVOT (v FOR nValue IN ([CurrValue])) AS P1
UNPIVOT (d FOR nDate IN ([DateTime])) AS P2
GROUP BY GroupNo

我在symfony中的命令历史记录:

service apache2 stop

有些材料表示,端口8000不会听取问题的结果 如何正确设置symfony中的内置开发服务器?

1 个答案:

答案 0 :(得分:0)

如果您查看服务器命令(最新)中fsockopen的使用方式,您会发现它是used to make sure that nothing is listening on the port。它使用静音运算符,但如果你的php设置使用scream来覆盖它,它仍然可能会引发警告。

警告说 - &#34;那里什么都没有&#34;这是应该发生的事情。由于@静音运算符,不应仅提出警告。