是否可以有一个仅从a1bcdea1行开始解析a1bcdea1ABCa1DEFa1的正则表达式？
此grep命令不起作用：
$ cat txtfile
a1bcdea1ABCa1DEFa1
$ grep -oE "[A-Z,a-z]1.*?[A-Z,a-z]1" txtfile
a1bcdea1ABCa1DEFa1
我希望grep的输出仅为a1bcdea1。
编辑：
很明显，我可以在上面的行中使用grep -o“ a1bcdea1”，但是要考虑一个人是否有几千行，并且目标是每行匹配FIRST [A-Z,a-z]1.*?[A-Z,a-z]1。
答案 0 :(得分：3)
以下是使用using Mono.Cecil;
using Mono.Cecil.Cil;
using Mono.Cecil.Rocks;
using System;
using System.Linq;
using BindingFlags = System.Reflection.BindingFlags;
using Cecilifier.Runtime;
public class SnippetRunner
{
public static void Main(string[] args)
{
var assembly = AssemblyDefinition.CreateAssembly(new AssemblyNameDefinition("name", Version.Parse("1.0.0.0")), "moduleName", ModuleKind.Dll);
var t1 = new TypeDefinition("", "Foo", TypeAttributes.AnsiClass | TypeAttributes.BeforeFieldInit | TypeAttributes.NotPublic, assembly.MainModule.TypeSystem.Object);
assembly.MainModule.Types.Add(t1);
t1.BaseType = assembly.MainModule.TypeSystem.Object;
var Foo_ctor_ = new MethodDefinition(".ctor", MethodAttributes.Public | MethodAttributes.HideBySig | MethodAttributes.RTSpecialName | MethodAttributes.SpecialName, assembly.MainModule.TypeSystem.Void);
t1.Methods.Add(Foo_ctor_);
var il1 = Foo_ctor_.Body.GetILProcessor();
var Ldarg_02 = il1.Create(OpCodes.Ldarg_0);
il1.Append(Ldarg_02);
var Call3 = il1.Create(OpCodes.Call, assembly.MainModule.ImportReference(TypeHelpers.DefaultCtorFor(t1.BaseType)));
il1.Append(Call3);
var Ret4 = il1.Create(OpCodes.Ret);
il1.Append(Ret4);
var Foo_F_ = new MethodDefinition("F", MethodAttributes.Private | MethodAttributes.HideBySig, assembly.MainModule.TypeSystem.Void);
t1.Methods.Add(Foo_F_);
var il_Foo_F_ = Foo_F_.Body.GetILProcessor();
var lv_dictionary5 = new VariableDefinition(assembly.MainModule.ImportReference(typeof(System.Collections.Generic.Dictionary<,>)).MakeGenericInstanceType(assembly.MainModule.TypeSystem.String,assembly.MainModule.TypeSystem.String));
Foo_F_.Body.Variables.Add(lv_dictionary5);
var Newobj6 = il_Foo_F_.Create(OpCodes.Newobj, assembly.MainModule.ImportReference(TypeHelpers.ResolveMethod("System.Private.CoreLib", "System.Collections.Generic.Dictionary`2", ".ctor",System.Reflection.BindingFlags.Default|System.Reflection.BindingFlags.Instance|System.Reflection.BindingFlags.Public,"System.String,System.String")));
il_Foo_F_.Append(Newobj6);
var Stloc7 = il_Foo_F_.Create(OpCodes.Stloc, lv_dictionary5);
il_Foo_F_.Append(Stloc7);
// dictionary.Add("first", "1");
var Ldloc8 = il_Foo_F_.Create(OpCodes.Ldloc, lv_dictionary5);
il_Foo_F_.Append(Ldloc8);
var Callvirt9 = il_Foo_F_.Create(OpCodes.Callvirt, assembly.MainModule.ImportReference(TypeHelpers.ResolveMethod("System.Private.CoreLib", "System.Collections.Generic.Dictionary`2", "Add",System.Reflection.BindingFlags.Default|System.Reflection.BindingFlags.Instance|System.Reflection.BindingFlags.Public,"System.String,System.String", "System.String", "System.String")));
var Ldstr10 = il_Foo_F_.Create(OpCodes.Ldstr, "first");
il_Foo_F_.Append(Ldstr10);
var Ldstr11 = il_Foo_F_.Create(OpCodes.Ldstr, "1");
il_Foo_F_.Append(Ldstr11);
il_Foo_F_.Append(Callvirt9);
// dictionary.Add("second", "2");
var Ldloc12 = il_Foo_F_.Create(OpCodes.Ldloc, lv_dictionary5);
il_Foo_F_.Append(Ldloc12);
var Callvirt13 = il_Foo_F_.Create(OpCodes.Callvirt, assembly.MainModule.ImportReference(TypeHelpers.ResolveMethod("System.Private.CoreLib", "System.Collections.Generic.Dictionary`2", "Add",System.Reflection.BindingFlags.Default|System.Reflection.BindingFlags.Instance|System.Reflection.BindingFlags.Public,"System.String,System.String", "System.String", "System.String")));
var Ldstr14 = il_Foo_F_.Create(OpCodes.Ldstr, "second");
il_Foo_F_.Append(Ldstr14);
var Ldstr15 = il_Foo_F_.Create(OpCodes.Ldstr, "2");
il_Foo_F_.Append(Ldstr15);
il_Foo_F_.Append(Callvirt13);
var Ret16 = il_Foo_F_.Create(OpCodes.Ret);
il_Foo_F_.Append(Ret16);
assembly.Write(args[0]);
}
}
函数的gnu awk解决方案：
split
awk '(n = split($0, a, /[a-zA-Z]1/, b)) > 1 {print b[1] a[2] b[2]}' file
此a1bcdea1
命令拆分正则表达式awk上的每一行，并将拆分的令牌存储在数组/[a-zA-Z]1/中，并将定界符存储在数组a中。
答案 1 :(得分：3)
如何使用^ start anchor并限制使用的字符集：
grep -o '^[A-Za-z]1[A-Za-z]*1'
See this Bash demo或Regex Pattern at regex101
如果您希望两者之间有更多数字或其他字符，请转到with this
grep -oP '^[A-Za-z]1.*?[A-Za-z]1'
lazy匹配要求使用perl compatible模式。对于不在行首的情况，请转到with this
grep -oP '^.*?\K[A-Za-z]1.*?[A-Za-z]1'
\K resets是所报告比赛的开始，也是PCRE功能。