Question

这是我的代码行：

col_value = line_item[column].scan(/\d+./).join().to_i

当我在文本字段中输入30,000时，col_value为30。我希望它带来任何数字： 30000 30.5 30.55 30000

任何这些都是有效的......

扫描和/或连接是否有问题导致它返回30？使用下面建议的正则表达式仍然可以重新启动30，例如

col_value = line_item[column].scan(/\d+[,.]?\d+/).join().to_i

可能是“to_i”将“30,000”转换为30 ??

Answer 1

此正则表达式将匹配您所需的输出：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="col-xs-12 col-md-3 col-lg-3 servicesub" id="servicesub" >
					<ul>
						<li class="servicesubitem">
							<span class="subitem">Communications
								<br>
								&amp; Airtime</span><span class="fa1 fa-phone"> </span>
						</li>
						<div class="servicesubli">
							<ul>
								<li>
									VSAT
								</li>
								<li>
									IRIDIUM
								</li>
								<li>
									GSM
								</li>
								<li>
									INMARSAT
								</li>
								<li>
									IDIRECT
								</li>
							</ul>
						</div>
						<li class="servicesubitem">
							<span class="subitem">IT &amp;
								<br>
								Networking</span><span class="fa1 fa-sitemap"> </span>
						</li>
						<div class="servicesubli">
							<ul>
								<li>
									Built/Refit Consultancy
								</li>
								<li>
									Managed IT Support
								</li>
								<li>
									Networking (Wired &amp; Wireless)
								</li>
								<li>
									Backup &amp; Disaster Recovery
								</li>
								<li>
									Antivirus
								</li>
							</ul>
						</div>
                            </ul>
                        </div>

    <div id="serviceinfo">
    
</div>
                        <div style ="clear:both;"></div>
<div id = "footer">
    
</div>

此处\d+[,.]?\d*用作匹配的可选项。

DEMO

Answer 2

\d+(?:[,.]\d+)?

试试这个。这应该为你做。

Answer 3

是的，"30,000".to_i #=> 30"。请参阅String#to_i：“忽略有效数字末尾之外的无关字符。”

我建议您先删除逗号，然后应用正则表达式：

R = /
    \d+      # match >= 0 digits
    |        # or
    \d+\.\d+ # match > 0 digits, a decimal point, then > 0 digits
    /x       # extended mode

str = "30,000 30.5 30.55 30000 1. .1"
str1 = str.tr(',','')
  #=> "30000 30.5 30.55 30000 1. .1" 
a = str1.scan(R)
  #=> ["30000", "30", "5", "30", "55", "30000"] 
a.map(&:to_i)
  #=> [30000, 30, 5, 30, 55, 30000]

链接后，我们有：

str.tr(',','').scan(R).map(&:to_i)

如果是所需的解决方案：

  #=> [30000, 30, 5, 30, 55, 30000, 1, 0]

正则表达式需要修改如下：

R = /
    \d+      # match >= 0 digits
    |        # or
    \d+\.\d+ # match > 0 digits, a decimal point, then > 0 digits
    |        # or
    \d+\.    # match > 0 digits, then a decimal point
    |        # or
    \.\d+    # match a decimal point, then > 0 digits
    /x       # extended mode

正则表达式用小数或逗号引入数字

3 个答案: