如何使用RegEx排除[]内的符号

Question

我正在使用PHP preg_match_all,这是我到目前为止所能获得的....

[A-Za-z+\W]+\s[\d]

唯一的问题是我需要\ W不是一个".

所以我试过了:

[A-Za-z+[^\dA-Za-z"]\s?]+\s[\d]


[A-Za-z+]\s?+[^A-Za-z\d"]?\s[\d]

除其他外,它只是失败,我真的无法弄清楚为什么.

编辑:

这是整个RegEx;

([A-Z][a-z]+\s){1,5}\s?[^a-zA-Z\d\s:,.\'\"]\s?
[A-Za-z+\W]+\s[\d]{1,2}\s[A-Z][a-z]+\s[\d]{4}

我把它分成两行,第二行从我发布的内容开始.

试图匹配的模式:

    India – Adulterated Tea Powder Seized 18 April 2011
    India – Importer of Haldiram’s Petha Sweet Cubes Issuing Voluntary Recall 26 April 2011
    India – Undeclared Gluten Found in Sweets by Canadian Authorities 27 April 2011
    India – Adulteration Found in Edible Oils 28 April 2011
    India – Viral Disease Affects Chili Crop in Goa 28 April 2011
NOT ---->   Chili – India: Goa”. 8 April 2011
    Ivory Coast – Potential Cocoa Quality Decline despite Sufficient Surplus 11 April 2011
    Japan – Sanuki Kanzume Co. and Failure to Comply with FDA Standards 27 April 2011
    Madagascar – Toxic Sardines 14 April 2011
    Madagascar – Update: Toxic Sardines 26 April 2011

Answer 1

您显示的模式匹配所有字母和非字符.唯一没有包含在模式中的是数字,你也想要与双引号不匹配.

[^\d\"_]+\s\d

编辑:

我可能是错的,但是从示例输入来看,您似乎只是想匹配所有没有双引号的行.如果这样的事情更容易,我甚至将日期与字符串的其余部分分开.如果您不需要对sting/date进行分组,则只需删除所有括号.

<?php
error_reporting(E_ALL);
$str = "    India - Adulterated Tea Powder Seized 18 April 2011
    India - Importer of Haldiram’s Petha Sweet Cubes Issuing Voluntary Recall 26 April 2011
    India - Undeclared Gluten Found in Sweets by Canadian Authorities 27 April 2011
    India - Adulteration Found in Edible Oils 28 April 2011
    India - Viral Disease Affects Chili Crop in Goa 28 April 2011
    Chili - India: Goa\". 8 April 2011
    Ivory Coast - Potential Cocoa Quality Decline despite Sufficient Surplus 11 April 2011
    Japan - Sanuki Kanzume Co. and Failure to Comply with FDA Standards 27 April 2011
    Madagascar - Toxic Sardines 14 April 2011
    Madagascar - Update: Toxic Sardines 26 April 2011";
preg_match_all("/^([^\"]+?)(\d?\d\s[a-z]+\s\d{4})$/im", $str, $m);
echo '<pre>'.print_r($m, true).'</pre>';
?>