Question

我想删除评论中的标签，例如此示例;

<?php

                    //this is a comment with tab
$tes = 1;
$this->HTTP_URL = str_replace('///', '//', $this->http.$this->serverName.$this->port.$_SERVER['PHP_SELF']);
if ($tes == 1) {
    //this is a comment with tab
    echo $tes;
}

            //this is a comment with tab
$tes = 2;

我用它测试了它;

sed -e's | / *。 * / || g'-e's | //. || g'test.php＆gt; test2.php

但结果是;

this is a comment with tab
$tes = 1;
', $this->http.$this->serverName.$this->port.$_SERVER['PHP_SELF']);
if ($tes == 1) {
this is a comment with tab
    echo $tes;
}

this is a comment with tab
$tes = 2;

我希望它看起来像;

//this is a comment with tab
$tes = 1;
$this->HTTP_URL = str_replace('///', '//', $this->http.$this->serverName.$this->port.$_SERVER['PHP_SELF']);
if ($tes == 1) {
//this is a comment with tab
    echo $tes;
}

//this is a comment with tab
$tes = 2;

甚至更好

/* this is a comment with tab */
$tes = 1;
$this->HTTP_URL = str_replace('///', '//', $this->http.$this->serverName.$this->port.$_SERVER['PHP_SELF']);
if ($tes == 1) {
/* this is a comment with tab */
    echo $tes;
}

/* this is a comment with tab */
$tes = 2;

我尝试过各种各样的方式，但仍未达到我想要的结果。谢谢。

...

我试过

gsed -i 's|^\s*//|//|' test.php

然后

gsed -i ':a;$!{N;ba};s/^/\x00/;tb;:b;s/\x00$//;t;s/\x00\(\/\*[^*]*\*\+\([^/*][^*]*\*\+\)*\/\)/\1\x00/;tb;s/\x00\/\/\([^\n]*\)/\/*\1\*\/\x00/;tb;s/\x00\(.\)/\1\x00/;tb' test.php

但结果是

<?php

/*this is a comment with tab*/
$tes = 1;
$this->HTTP_URL = str_replace('/*/', '//', $this->http.$this->serverName.$this->port.$_SERVER['PHP_SELF']);*/
if ($tes == 1) {
/*this is a comment with tab*/
    echo $tes;
}

/*this is a comment with tab*/
$tes = 2;

注意错误第4行

$this->HTTP_URL = str_replace('/*/', '//', $this->http.$this->serverName.$this->port.$_SERVER['PHP_SELF']);*/

不应该更换。需要另一种语法

Answer 1

sed '/^[[:space:]]*\/\//{s/^[[:space:]]*//}' yourfile.php

应该这样做。如果您想进行infile编辑，请执行

sed -i~ '/^[[:space:]]*\/\//{s/^[[:space:]]*//}' yourfile.php

将创建以〜开头的原始文件的备份。

Answer 2

这可能适合你（GNU sed）：

sed -i 's|^\s*//|//|' file

这将从前两个非空格字符为//的行的开头删除零个或多个空格或制表符（空格）。

如果只是要移除的标签：

sed -i s|^\t*//|//|' file

N.B。这可以替代什么都没有，即以//开头的行仍然会影响替换，所以最后的解决方案应该是：

sed -i 's|^\t\+//|//|' file

编辑：

sed -i 's|^\t\+//\(.*\)$|/*\1*/' file

或者更改//，无论如何：

sed -i 's|^\t\+//|//|;s|//\(.*\)|/*\1*/|' file

删除备注中的标签

2 个答案: