Question

大家好，我试图从文件中对日志消息进行排序。我的日志文件如下所示 http://kopy.io/tA8bv 我希望得到像这样的东西

['debug'] => [
    ['logs'] => [
        ['log message'] => [
            '2016-04-10',
            '2016-04-10',
            '2016-04-10'
        ],
        ['log message'] => [
            '2016-04-10',
            '2016-04-10',
            '2016-04-10'
        ],
        ...
    ]
],
['Crit'] => [
    ['logs'] => [
        ['log message'] => [
            '2016-04-10',
            '2016-04-10',
            '2016-04-10'
        ],
        ['log message'] => [
            '2016-04-10',
            '2016-04-10',
            '2016-04-10'
        ],
        ...
    ]
]

这是我的代码，但它无法正常工作，省略了一些行

$file = fopen(self::DIRECTORY.$filename, 'rb');
while(($line = fgets($file)) !== false) {
    if($this->lineHasDate($line)) {
        $logDate = $this->getLogDate($line);
        $logType = $this->getTypeOfLog($line);
        $content .= $line;
        // $content ='';
    }
    while(!$this->lineHasDate($line) && ($line = fgets($file)) !== false) {
        // echo $line;
        $content .= $line;
    }
    $this->logs[$logType]['logs'][] = $content;
    $content = '';
}
fclose($file);

Answer 1

指定的代码 - 循环遍历文件循环文件， - 有点过于复杂，更容易出错。

尝试这样，在你的例子中它适用于我。它的读取和调试更简单，更简单：

$file = fopen (self::DIRECTORY.$filename, 'rb');
while (($line = fgets($file)) !== false)
{
    if ($this->lineHasDate($line))
    {
        $logDate = $this->getLogDate($line);
        $logType = $this->getTypeOfLog($line);

        $this->logs[$logType]['logs'][] = $line;
        $currentLog = &$this->logs[$logType]['logs'][count($this->logs[$logType]['logs']) - 1];
    }
    else
    {
        if (!isset($currentLog)) // In case of trash lines before logs begin
            continue;
        else
            $currentLog .= $line;
    }
}
fclose($file);

另外，当您将省略的行提供给它时，请测试$this->lineHasDate($line)返回true。

对日志消息进行排序

1 个答案: