遍历匹配并replaceVBA RegEx

是否有可能迭代通过VBA RegEx匹配，并用一个ID值给定的特定数据进行replace？

例如，

 <a id="a-UP:124" {REPLACEITEMHERE} .../a> 

与我的模式是这样的：

 <a id="a-([\w\d]+:[\w\d]+)" ({REPLACEITEMHERE}) 

所以我有多个“replace项目”，每个都是UP:124的值。

这在VBA RegEx可能吗？ 只是在我经历一个更繁琐的过程之前想知道！ 谢谢！

更新（根据评论者的要求，更多细节 – 希望这可以让我更清楚我在找什么！我知道如何创build模式，它更加反复遍历结果，然后对每个发现进行replace麻烦，谢谢！）：

这是我正在使用的RegEx模式：

 <a id="a-([\w\d]+:[\w\d]+)"[^{]+({FILE})[^{]+({PERCENT})[^{]+({COLOR}) 

设置如下：

 .Global = True .IgnoreCase = True .MultiLine = False 

我想要的replace模式是检查第一个捕获组$1的值是什么，然后用我已经存储的适当值replace值{FILE} {PERCENT} {COLOR} （组$2 ， $3和$4 ）在一个class级。

 <path style="fill:#d40000;fill-opacity:1;filter:url(#filter5248)" d="m 168.04373,162.08375 c -4.7586,-5.00473 -8.65201,-9.35811 -8.65201,-9.67419 0,-0.81973 18.30811,-16.3921 25.16949,-21.40847 7.11903,-5.20474 16.462,-10.93031 17.83606,-10.93031 0.56369,0 3.81291,5.56174 7.22048,12.35942 l 6.19558,12.35941 -7.13301,3.9009 c -7.96536,4.3561 -21.53264,13.83148 -27.5305,19.22729 -2.16466,1.94738 -4.05237,3.47876 -4.19491,3.40307 -0.14254,-0.0757 -4.15257,-4.2324 -8.91118,-9.23712 z" id="path5246" inkscape:connector-curvature="0" transform="matrix(0.8,0,0,-0.8,0,792)" /> <a id="a-UP:115E" xlink:href="{FILE}" xlink:title="UP:115E {PERCENT}%"> <path id="UP:115E" style="fill:{COLOR};fill-opacity:1;stroke:none" d="m 272.81031,529.10942 c 0.32799,18.973 -0.6558,38.48935 0.49159,57.12295 13.02609,-0.33792 26.60749,0.66479 39.29456,-0.4916 -0.32799,-18.973 0.6558,-38.48935 -0.49159,-57.12294 -13.01823,0.33523 -26.61862,-0.66099 -39.29456,0.49159 z" inkscape:connector-curvature="0" transform="matrix(0.8,0,0,-0.8,0,792)" /> </a> <a id="a-UP:115D" xlink:href="{FILE}" xlink:title="UP:115D {PERCENT}%"> <path id="UP:115D" style="fill:{COLOR};fill-opacity:1;stroke:none" d="m 314.75946,529.10942 c 0.32799,18.973 -0.6558,38.48935 0.4916,57.12295 9.11694,0.926 18.85965,-1.04961 27.69299,0.721 -0.31086,4.08011 6.71077,4.04524 8.35706,1.67141 -0.0756,-1.75206 -3.96676,-2.62149 0,-2.32687 8.75271,2.70871 7.9153,-4.7371 7.43942,-11.04442 -0.32811,-15.47719 0.65596,-31.4979 -0.49159,-46.63566 -14.41803,0.33385 -29.41334,-0.65954 -43.48948,0.49159 z" inkscape:connector-curvature="0" transform="matrix(0.8,0,0,-0.8,0,792)" /> </a> </g></svg> 

Answer 1

我认为你可以做这样的简单的例子（给你inputstring很难设置testing）

• 将Global变为False可以为每个replace使用一个Regexp （另一种方法是使用Global = True ，然后针对不同的第一个匹配运行不同的Regexps ）
• 使用Do循环来testing是否保留有效的Regexp
• testing第一submatch ，然后使用Select Case运行不同的例程来replace子submatches $2-$4 （我存储在一个简单的数组中）

码

 Sub TestSub() Dim strIn As String Dim objRegex As Object Dim objRegMC As Object Dim objRegM As Object Dim vArray(1 To 3, 1 To 2) vArray(1, 1) = "No 1a" vArray(2, 1) = "No 2a" vArray(3, 1) = "No 3a" vArray(1, 2) = "number 1b" vArray(2, 2) = "number 2b" vArray(3, 2) = "number 3b" Set objRegex = CreateObject("vbscript.regexp") strIn = "a12stuff notme b34other missthis" With objRegex .Pattern = "([az]{1})(\d)(\d)([az]+)" .Global = False .IgnoreCase = True .MultiLine = False Do While .test(strIn) Set objRegMC = .Execute(strIn) For Each objRegM In objRegMC Select Case objRegM.submatches(0) Case "a" strIn = .Replace(strIn, "$1" & vArray(1, 1) & vArray(2, 1) & vArray(3, 1)) Case "b" strIn = .Replace(strIn, "$1" & vArray(1, 2) & vArray(2, 2) & vArray(3, 2)) End Select Next Loop End With MsgBox strIn End Sub