I m使用php 5.2.13和Oracle数据库。
我试图隐藏错误信息,但确实不工作。
我试图......
在出现错误的网页上添加这一代码。
ini_set(display_errors,0);
在php.ini上设定了一个选项。
display_errors = Off
......以及我应该做些什么? 我认为我做得够了。
错误信息来自Oracle, 如同此。
ORA-01400 cannot insert NULL into .....
我想知道,是否有显示错误信息的任何单元。
感谢。
这并不是一种PHP错误,而是一种Oracle错误,你的一栏可能是无效的,你需要具体说明价值。 因此,你显示的“error”没有掩盖这一错误。
我必须同评论中的其他发言者一样强调,唯一好的错误是固定的错误。
如何在管理部使用
见:。
and: http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/statements_9014.htm#BGBDIGAH
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