[英]Regex to validate url AND does not end in semicolon or space
我有以下正則表達式:
var regex = /https:\/\/company\.zoom\.us\/j\/.*\?pwd=.*/i;
如何修改此正則表達式以確保 url 字符串不以分號或空格結尾?
https://company.zoom.us/j/xxxxx?pwd=zzzzz //Should pass
https://company.zoom.us/j/xxxxx?pwd=zzzzz //Should fail (ends in a space after the z)
https://company.zoom.us/j/xxxxx?pwd=zzzzz; //Should fail (ends in a semicolon)
https://company.zoom.us/j/xxxxx?pwd=zzzzz; //Should fail (ends in a space after the semicolon)
如果您不想在任何地方允許空格並且不以;
結尾;
然后從允許任何字符的正則表達式中刪除.*
。
你可以只使用:
/^https:\/\/company\.zoom\.us\/j\/\S*\?pwd=[^\s;]*$/
變化:
\\S:
:匹配 0 個或多個任何非空白字符[^\\s;]*
匹配任何不是;
字符而不是空格
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