我有一个字符串
Ex：“我们更喜欢可以回答的问题;不仅仅是讨论过”
现在我想从“;”拆分此字符串 喜欢 我们更喜欢可以回答的问题 和 不仅讨论了
这在DXL中是可行的。
我正在学习DXL，所以我不知道我们是否可以分手。
注意：这不是家庭作业。
答案 0 :(得分：3)
快速加入＆分裂我可以想出来。接缝工作正常。
int array_size(Array a){
int size = 0;
while( !null(get(a, size, 0) ) )
size++;
return size;
}
void array_push_str(Array a, string str){
int array_index = array_size(a);
put(a, str, array_index, 0);
}
string array_get_str(Array a, int index){
return (string get(a, index, 0));
}
string str_join(string joiner, Array str_array){
Buffer joined = create;
int array_index = 0;
joined += "";
for(array_index = 0; array_index < array_size(str_array); array_index++){
joined += array_get_str(str_array, array_index);
if( array_index + 1 < array_size(str_array) )
joined += joiner;
}
return stringOf(joined)
}
Array str_split(string splitter, string str){
Array tokens = create(1, 1);
Buffer buf = create;
int str_index;
buf = "";
for(str_index = 0; str_index < length(str); str_index++){
if( str[str_index:str_index] == splitter ){
array_push_str(tokens, stringOf(buf));
buf = "";
}else{
buf += str[str_index:str_index];
}
}
array_push_str(tokens, stringOf(buf));
delete buf;
return tokens;
}
答案 1 :(得分：3)
如果你只是拆分字符串，我就会这样做：
string s = "We prefer questions that can be answered; not just discussed"
string sub = ";"
int offset
int len
if ( findPlainText(s, sub, offset, len, false)) {
/* the reason why I subtract one and add one is to remove the delimiter from the out put.
First print is to print the prefix and then second is the suffix.*/
print s[0 : offset -1]
print s[offset +1 :]
} else {
// no delimiter found
print "Failed to match"
}
您也可以使用正则表达式参考DXL参考手册。如果你想用多个分隔符分割字符串，比如str =“this; is an; example”，那么最好使用正则表达式
答案 2 :(得分：2)
我很遗憾地发现了这篇文章。作为DXL的新手，我花了一些时间来应对同样的挑战。我注意到可用的实现具有不同的“拆分”字符串的规范。热爱Ruby language，我错过了一个至少接近Ruby version of String#split的实现。 也许我的发现对任何人都有帮助。
这是
的功能比较为了消除结构差异，所有实现都在函数中实现，返回Skip列表或Array。
请注意，所有实现都会返回不同的结果，具体取决于它们对“拆分”的定义：
string mellow yellow;分隔符ello
splitVariantA returns 1 elements: ["mellow yellow" ]
splitVariantB returns 2 elements: ["m" "llow yellow" ]
splitVariantC returns 3 elements: ["w" "w y" "" ]
splitVariantD returns 3 elements: ["m" "w y" "w" ]
string now's the time;分隔符
splitVariantA returns 3 elements: ["now's" "the" "time" ]
splitVariantB returns 2 elements: ["" "now's the time" ]
splitVariantC returns 5 elements: ["time" "the" "" "now's" "" ]
splitVariantD returns 3 elements: ["now's" "the" "time" ]
string 1,2,,3,4,,;分隔符,
splitVariantA returns 4 elements: ["1" "2" "3" "4" ]
splitVariantB returns 2 elements: ["1" "2,,3,4,," ]
splitVariantC returns 7 elements: ["" "" "4" "3" "" "2" "" ]
splitVariantD returns 7 elements: ["1" "2" "" "3" "4" "" "" ]
在我的机器上将模式1,2,,3,4,,的字符串,拆分10000次给出了这些时间：
splitVariantA() : 406 ms
splitVariantB() : 46 ms
splitVariantC() : 749 ms
splitVariantD() : 1077 ms
不幸的是，我的实施D是最慢的。令人惊讶的是，正则表达式实现C非常快。
// niol, modified
Array splitVariantA(string splitter, string str){
Array tokens = create(1, 1);
Buffer buf = create;
int str_index;
buf = "";
for(str_index = 0; str_index < length(str); str_index++){
if( str[str_index:str_index] == splitter ){
array_push_str(tokens, stringOf(buf));
buf = "";
}
else
buf += str[str_index:str_index];
}
array_push_str(tokens, stringOf(buf));
delete buf;
return tokens;
}
// PJT, modified
Skip splitVariantB(string s, string delimiter) {
int offset
int len
Skip skp = create
if ( findPlainText(s, delimiter, offset, len, false)) {
put(skp, 0, s[0 : offset -1])
put(skp, 1, s[offset +1 :])
}
return skp
}
// Brett, modified
Skip splitVariantC (string s, string delim) {
Skip skp = create
int i = 0
Regexp split = regexp "^(.*)" delim "(.*)$"
while (split s) {
string temp_s = s[match 1]
put(skp, i++, s[match 2])
s = temp_s
}
put(skp, i++, s[match 2])
return skp
}
Skip splitVariantD(string str, string pattern) {
if (null(pattern) || 0 == length(pattern))
pattern = " ";
if (pattern == " ")
str = stringStrip(stringSqueeze(str, ' '));
Skip result = create;
int i = 0; // index for searching in str
int j = 0; // index counter for result array
bool found = true;
while (found) {
// find pattern
int pos = 0;
int len = 0;
found = findPlainText(str[i:], pattern, pos, len, true);
if (found) {
// insert into result
put(result, j++, str[i:i+pos-1]);
i += pos + len;
}
}
// append the rest after last found pattern
put(result, j, str[i:]);
return result;
}
答案 3 :(得分：1)
实际工作：
如果字符串中不存在分隔符，此解决方案将根据需要分割多次，或者不分割。
这是我使用的，而不是传统的“拆分”命令。 它实际上跳过了数组的创建，只是循环遍历数组中的每个字符串，并在每个字符串上调用“someFunction”。
string s = "We prefer questions that can be answered; not just discussed"
// for this example, ";" is used as the delimiter
Regexp split = regexp "^(.*);(.*)$"
// while a ";" exists in s
while (split s) {
// save the text before the last ";"
string temp_s = s[match 1]
// call someFunction on the text after the last ";"
someFunction(s[match 2])
// remove the text after the last ";" (including ";")
s = temp_s
}
// call someFunction again for the last (or only) string
someFunction(s)
抱歉找到一个旧帖子;我只是没有发现其他答案有用。
答案 4 :(得分：0)
也许有人会找到方便的融合解决方案。它基于分隔符在Skip中拆分字符串，实际上长度可以超过一个。
Skip splitString(string s1, string delimit)
{
int offset, len
Skip splited = create
while(findPlainText(s1, delimit, offset, len, false))
{
put(splited, s1[0:offset-1], s1[0:offset-1])
s1 = s1[offset+length(delimit):length(s1)-1]
}
if(length(s1)>0)
{
put (splited, s1, s1)
}
return splited
}
答案 5 :(得分：0)
我尝试过了，为我努力了...
string s = "We prefer questions that can be answered,not just discussed,hiyas"
string sub = ","
int offset
int len
string s1=s
while(length(s1)>0){
if ( findPlainText(s1, sub, offset, len, false)) {
print s1[0 : offset -1]"\n"
s1= s1[offset+1:length(s1)]
}
else
{
print s1
s1=""
}
}
答案 6 :(得分：0)
这是一个更好的实现。这是通过搜索关键字来对字符串进行递归拆分。
pragma runLim, 10000
string s = "We prefer questions that can be answered,not just discussed,hiyas;
Next Line,Var1,Nemesis;
Next Line,Var2,Nemesis1;
Next Line,Var3,Nemesis2;
New,Var4,Nemesis3;
Next Line,Var5,Nemesis4;
New,Var5,Nemesis5;"
string sub = ","
int offset
int len
string searchkey=null
string curr=s
string nxt=s
string searchline=null
string Modulename=""
string Attributename=""
string Attributevalue=""
while(findPlainText(curr,"Next Line", offset,len,false))
{
int intlen=offset
searchkey=curr[offset:length(curr)]
if(findPlainText(searchkey,"Next Line",offset,len,false))
{
curr=searchkey[offset+1:length(searchkey)]
}
if(findPlainText(searchkey,";",offset,len,false))
{
searchline=searchkey[0:offset]
}
int counter=0
while(length(searchline)>0)
{
if (findPlainText(searchline, sub, offset, len, false))
{
if(counter==0)
{
Modulename=searchline[0 : offset -1]
counter++
}
else if(counter==1)
{
Attributename=searchline[0 : offset -1]
counter++
}
searchline= searchline[offset+1:length(searchline)]
}
else
{
if(counter==2)
{
Attributevalue=searchline[0:length(searchline)-2]
counter++
}
searchline=""
}
}
print "Modulename="Modulename " Attributename=" Attributename " Attributevalue= "Attributevalue "\n"
}