What I'm after is to replace a space with a dash - BUT only if that space is has a character after it. Reason being is I have an array of strings some of which have spaces inserted after the final word or character. (this is out of my control). e.g In the example below I have used %20 to show where a space is string1 = farmer%20jones string2 = farmer%20jim%20
我所追求的是用短划线替换空格 - 但前提是该空格后面有一个字符。原因是我有一个字符串数组,其中一些字符串在最后的单词或字符后面插入了空格。 (这是我无法控制的)。例如,在下面的示例中,我使用%20来显示空格是string1 =农民%20jones string2 =农民%20jim%20
I have the following regex preg_replace('/\s./', '-', $string);
我有以下正则表达式preg_replace('/ \ s。/',' - ',$ string);
I think I'm half way there, but the above searches for a space preceding a character and replaces that with a -
我想我已经走了一半,但上面搜索了一个字符前面的空格并用 - 替换了 -
What I get with the above regex is string1 = farmer-jones string2 = farmer-jim-
我用上面的正则表达式得到的是string1 = farmer-jones string2 = farmer-jim-
What I want is: string1 = farmer-jones string2 = farmer-jim
我想要的是:string1 = farmer-jones string2 = farmer-jim
I don't want that trailing - to be added.
我不希望添加尾随。
Any help much appreciated
任何帮助非常感谢
1 个解决方案
#1
2
You can use negative lookahead here:
您可以在此处使用否定前瞻:
$repl = preg_replace('/\h+(?!$)/', '-', $string);
-
\h+will match 1 or more horizontal whitespace. -
(?!$)will assert that next position is not end of line.
\ h +将匹配1个或更多水平空格。
(?!$)将声明下一个位置不是行尾。
#1
2
You can use negative lookahead here:
您可以在此处使用否定前瞻:
$repl = preg_replace('/\h+(?!$)/', '-', $string);
-
\h+will match 1 or more horizontal whitespace. -
(?!$)will assert that next position is not end of line.
\ h +将匹配1个或更多水平空格。
(?!$)将声明下一个位置不是行尾。