PHP提示Notice: Undefined offset 的解决方法

调试程序的时候总是出现错误提示：
Notice: Undefined offset: 1 in xxx.php on line 18
Notice: Undefined offset: 2 in xxx.php on line 18
Notice: Undefined offset: 3 in xxx.php on line 18
Notice: Undefined offset: 4 in xxx.php on line 18
这问题很常出现在数组中的,程序是能正确地运行下去,但是在屏幕上总会出现这样的提示:Notice: Undefined offset: ….. 网上普遍是采用抑制其显示的方法,即更改php.ini文件中error_repoting的参数为”EALL & Notice “,这样屏幕就能正常显示了.
问题是解决了,但是总想不透offset:接下去的数字(如 Notice: Undefined offset: 4 ….)是什么意思.还有,句子里的语法明明是正确的,为什么会出现警告.冷静地思考了好几遍并尝试了每种可能,终于找到了答案.offset:接下去的数字是出错的数组下标,一般是超出了数组的取值范围,如定义了数组$A[]有10个元数,如果出现了$A[10]就会出现错误(Notice: Undefined offset: 10 ….),因为数组的下标是从0开始的,所以这个数组的下标就只能是0~9.因此在出现这类问题时,不要急于用抑制显示的方法(更简单的可以在当前文件的最前面加上一句”error_reporting(填offset:接下去的那个数字);,一定要注意你所用的数组下标,仔细思考一下,问题一定会很快得到解决的 !也有可能是unset数组后再尝试读取其内容，php手册中有：
Just to confirm, USING UNSET CAN DESTROY AN ENTIRE ARRAY. I couldn’t find reference to this anywhere so I decided to write this.
The difference between using unset and using $myarray=array(); to unset is that obviously the array will just be overwritten and will still exist.
<?php
$myarray=array(“Hello”,”World”);
echo $myarray[0].$myarray[1];
unset($myarray);
//$myarray=array();
echo $myarray[0].$myarray[1];
echo $myarray;
?>
Output with unset is:
<?
HelloWorld
Notice: Undefined offset: 0 in C:webpagesdainsidermyarray.php on line 10
Notice: Undefined offset: 1 in C:webpagesdainsidermyarray.php on line 10
Output with $myarray=array(); is:
?>
<?
HelloWorld
Notice: Undefined offset: 0 in C:webpagesdainsidermyarray.php on line 10
Notice: Undefined offset: 1 in C:webpagesdainsidermyarray.php on line 10
Array
?>
这就是问题的根本原因了。