命令行PHP截断日期（“ c”）

Question

这是我的shell脚本PHP代码：

#!/usr/bin/php -q 
<?php

$user = get_current_user();
$line = date("c") . " - " . $user + "\r\n";
echo "---------------------\n";
echo "user => $user\n";
echo "---------------------\n";
echo "date('c') => " . date("c") . "\n";
echo "---------------------\n";
echo "date('Ymd') => " . date("Ymd") . "\n";
echo "---------------------\n";
echo "line => $line\n";
echo "---------------------\n";
echo date("c") . " - " . $user + "\n";
echo "---------------------\n";
echo date("c") . $user + "\n";
echo "---------------------\n";
echo date("c") . " - " . "\n";
echo "---------------------\n";

$ret = file_put_contents("/var/lib/foo/bar/test.txt", $line, FILE_APPEND);

echo "file_put_contents => $ret\n";
?>

运行它时，得到以下输出：

roffle:/var/lib/foo/bar # php Test.php
---------------------
user => wwwrun
---------------------
date('c') => 2014-07-27T16:39:34-04:00
---------------------
date('Ymd') => 20140727
---------------------
line => 2014
---------------------
2014---------------------
2014---------------------
2014-07-27T16:39:34-04:00 -
---------------------
file_put_contents => 4
roffle:/var/lib/foo/bar #

为什么$line与前两个echo date("c")调用一起被截断，为什么第三个对date("c")调用可以？

Answer 1

看起来您刚刚将+与混为一谈。 在相关行中进行串联。