php eval() 代码中出现意外的 $end
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Unexpected $end in eval()'d code
提问by andrewtweber
I hate to ask such a specific question, but I'm getting an error I can't figure out. This is in a cron job which runs on the hour. I'm creating an array of tasks, each of which has a date check which is supposed to be eval()'d.
我讨厌问这样一个具体的问题，但我遇到了一个我无法弄清楚的错误。这是在每小时运行的 cron 作业中。我正在创建一个任务数组，每个任务都有一个日期检查，应该是 eval()'d。
$todo = array();
$todo[] = array( "date('z')%3 == 0", "Task 1" );
$todo[] = array( "date('N') == 1", "Task 2" );
foreach( $todo as $task )
{
if( eval($task[0]) ) {
echo $task[1];
}
}
For some reason the eval() line is giving me this error. Note that I am getting this error for both tasks.
出于某种原因， eval() 行给了我这个错误。请注意，我在这两个任务中都收到此错误。
Parse error: syntax error, unexpected $end in /file.php(21) : eval()'d code on line 1
Any suggestions? I tried searching for this but couldn't find anything. Thank you.
有什么建议？我尝试搜索此内容，但找不到任何内容。谢谢你。
回答by mario
evalonly accepts statements, not expressions. You need to convert your tests with:
eval只接受语句，不接受表达式。您需要将测试转换为：
if (eval("return $task[0];")) {
回答by Tomero Indonesia
You are missing ';' on end of string evaluation. Eval function able to process statements or expressions.
你缺少';' 在字符串评估结束时。Eval 函数能够处理语句或表达式。
Example:
例子：
$value = 7;
eval("$value+=2;");
echo $value;
