Question

我不知道如何做以下工作:

find all instances of /s[2-9])/ and replace the space with a <br /> tag.
Something as simple as this doesn t work:

$preg_num =   /s[2-9])/ ;
$preg_fix = " /<br>/[2-9]) ";
preg_replace($preg_num,$preg_fix,$notes);

我需要将“固定变量”的美元前值改为什么?

Answer 1

使用head头比采用背书更简单。

$preg_num =  /s(?=[2-9]))/ ;
$preg_fix =  <br/> ;
preg_replace($preg_num,$preg_fix,$notes);

这样,你只能用<代码><br/>取代空间。

Answer 2

替代体不被视为一种雷管,更像是一种字面的扼杀。如果你试图将原封顶的数位位位位数位位数位位位数位位数位数位位数位数位位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数,则位数位数位数位数位数位数的位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数位数。

$preg_num =  /s([2-9]))/ ;
$preg_fix = "<br />$1) ";
preg_replace($preg_num,$preg_fix,$notes);

详情:

preg_replace s replacement parameter documentation
Regular expression grouping: "Use Round Brackets for Grouping"

友情链接