I am using Apache 2.4 with PHP 5.4.
我正在使用Apache 2.4和PHP 5.4。
I am executing the following:
我正在执行以下操作：
# retrieve caller name
$calledby = debug_backtrace(); print_r($calledby);
$caller = (strlen($calledby[1]['class'])) ? $calledby[1]['class'] : $calledby[0]['class'];
# arguments are required
if (!func_num_args()) { return; }
# fill variables with argument contents if exists
$variables = (func_num_args() == 0) ? NULL : (is_array(func_get_arg(0)) ? func_get_arg(0) : NULL);
from within a private method of a class.
来自一个类的私有方法。
If I comment out the $caller = and $variables = lines then it works.
如果我注释掉$ caller =和$ variables =行，那么它可以工作。
If I change my code to the following it also works (AS DEFINED)
如果我将我的代码更改为以下代码也可以（AS DEFINED）
if (strlen($calledby[1]['class'])) $caller = $calledby[1]['class'];
else $caller = $calledby[0]['class'];
Should I place a bug report or am I doing something wrong with a new 5.4 syntax quirk?
我应该发布错误报告还是我在使用新的5.4语法怪癖时做错了什么？
Thanks In Advance!
提前致谢！
[UPDATE] I executed the file using the CLI and the script produced the desired result.
[更新]我使用CLI执行文件，脚本生成了所需的结果。
1 个解决方案
#1
0
I have try with php 5.4 config and have no problem You should see the apache / php log
我尝试使用php 5.4配置并没有问题你应该看到apache / php日志
#1
0
I have try with php 5.4 config and have no problem You should see the apache / php log
我尝试使用php 5.4配置并没有问题你应该看到apache / php日志