Question

利用正则表达式提取字符串

如何在下面的字符串中：
“_+2_ U+47563137+34037717+00005324m+0063611d107_*L1++_,0.000_+3_ U+47563140+34037717+00005324m+0063611d107_*L1++_,0.000_+4_ U+47564547+34058752+00004847m+0373258d099_*L1_,0.000_+5_ U+47553513+34060607+00005356m+0513706d104_*L1_,0.000_+6_ U+47553363+34061758+00005692m+0525958d110_*L1_,0.000_+7_ U+47532851+34062538+00006537m+0892719d097_*L1_,0.000_+8_ U+47513555+34036646+00004864m+1723651d096_*L1++_,0.000_+9_ U+47463289+34035357+00005120m+1790136d096_*L1_,0.000_+10_ U+47462947+33953023+00005380m+2292212d111_*L1++_,0.000_+11_ U+47507774+33957622+00004876m+2520247d109_*L1_,0.000_+12_ U+47553337+33963307+00004599m+2861923d103_*L1_,0.000_+13_ U+47548966+33960215+00005976m+2822926d108_*L1++_,0.000_+14_ U+47533362+33959231+00005427m+2703549d098_*L1_,0.000_ ”
提取如下面的字符串：
2_ U+47563137+34037717+00005324m+0063611d107_*L1++
12_ U+47553337+33963307+00004599m+2861923d103_*L1
最好能直接提取到数组中：
点号 X Y Z 编码
2 47563137 34037717 00005324m L1++

Answer 1

reg=/[2_ U+47563137+34037717+00005324m+0063611d107_*L1++]/g;
String result=reg.exec(把上面那段字符串贴进来);
这样就提取出来了。然后你可以用String的split方法进行分割
String arr[]=result.split(/+/);//返回一个数组

追问

不好意思，没有详细的介绍一下问题，*号后面的编码长度是任意变化的
“_+2_ U+47563137+34037717+00005324m+0063611d107_*L1++_,0.000_+3_ U+47563140+34037717+00005324m+0063611d107_*F1++_,0.000_+4_ U+47564547+34058752+00004847m+0373258d099_*L1_,0.000_+5_ U+47553513+34060607+00005356m+0513706d104_*Q2_,0.000_+6_ U+47553363+34061758+00005692m+0525958d110_*_,0.000_+7_ U

Answer 2

# /bin/perl
$str='_+2_U+47563137+34037717+00005324m+0063611d107_*L1++_,0.000_+3_U+47563140+34037717+00005324m+0063611d107_*L1++_,0.000_+4_U+47564547+34058752+00004847m+0373258d099_*L1_,0.000_+5_U+47553513+34060607+00005356m+0513706d104_*L1_,0.000_+6_U+47553363+34061758+00005692m+0525958d110_*L1_,0.000_+7_U+47532851+34062538+00006537m+0892719d097_*L1_,0.000_+8_U+47513555+34036646+00004864m+1723651d096_*L1++_,0.000_+9_U+47463289+34035357+00005120m+1790136d096_*L1_,0.000_+10_U+47462947+33953023+00005380m+2292212d111_*L1++_,0.000_+11_U+47507774+33957622+00004876m+2520247d109_*L1_,0.000_+12_U+47553337+33963307+00004599m+2861923d103_*L1_,0.000_+13_U+47548966+33960215+00005976m+2822926d108_*L1++_,0.000_+14_U+47533362+33959231+00005427m+2703549d098_*L1_,0.000_';

@array = ($str =~ m/(\d+_U\+\d+\+\d+\+\d+m\+\d+d\d+_\*L1(?:\+\+)?)/g);
$n=join "\n",@array;
print "$n\n";

我试过了，可以的。

追问

你的不具有代表性，看这个能提取出来么？
“_+2_ U+47563137+34037717+00005324m+0063611d107_*L1++_,0.000_+3_ U+47563140+34037717+00005324m+0063611d107_*F1++_,0.000_+4_ U+47564547+34058752+00004847m+0373258d099_*L1_,0.000_+5_ U+47553513+34060607+00005356m+0513706d104_*F1_,0.000_+6_ U+47553363+34061758+00005692m+0525958d110_*F1_,0.000_+7_ U+47532851+34062538+00006537m+0892719d097_*_,0.000_+8_

追答

# /bin/perl
$str="_+2_ U+47563137+34037717+00005324m+0063611d107_*L1++_,0.000_+3_ U+47563140+34037717+00005324m+0063611d107_*F1++_,0.000_+4_ U+47564547+34058752+00004847m+0373258d099_*L1_,0.000_+5_ U+47553513+34060607+00005356m+0513706d104_*F1_,0.000_+6_ U+47553363+34061758+00005692m+0525958d110_*F1_,0.000_+7_ U+47532851+34062538+00006537m+0892719d097_*_,0.000_+8_";

@array = ($str =~ m/(\d+)_\s*U\+(\d+)\+(\d+)\+(\d+m)\+\d+d\d+_\*([LF]1(?:\+\+)?)/g);
print "点号 X Y Z 编码\n";
$m=0;
for(@array){
print "$_\t";
if(++$m == 5){
print "\n";
$m=0;
}
}

你应该好好学学正则表达式。
结果是：
点号 X Y Z 编码
2 47563137 34037717 00005324m L1++
3 47563140 34037717 00005324m F1++
4 47564547 34058752 00004847m L1
5 47553513 34060607 00005356m F1
6 47553363 34061758 00005692m F1

Answer 3

string.Format("","*"); //结果："*___"[长度为4] "_"表示一个空格

M：格式化后的长度
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