两个由英文句子组成的字符串S和D,其中D为目标串,拿S与D比较,要求: 1、两个串按英语单词进行比较,忽略单词间多余的空格,且单词区分大小写!
例:
D:='Good Morning!' S:='Good Morning! '
则 S=D 成立! 2、若串S掉了n个单词,则在相应掉词的位置插入'{n}'。
例:
D:='Convenient to wineries, hiking trails, State Parks and Sugar Loaf Observatory. Award winning income producing vineyard.'
S:=Convenient to wineries, hiking trails, State Loaf Observatory. Award winning producing vineyard.'
则结果S应为:
'Convenient to wineries, hiking trails, State {3} Loaf Observatory. Award winning {1} producing vineyard.' 3、若S中错了一个单词,则将错误的单词用括号‘()’括起来。
例:D:='Convenient to wineries, hiking trails, State Parks and Sugar Loaf Observatory. Award winning income producing vineyard.'
S:='Convenient to wineries, hiking trail, State Parks and Sugar Loaf Observatory. Award win income producing vineyard.'
则结果S应为:
'Convenient to wineries, hiking (trail), State Parks and Sugar Loaf Observatory. Award (win) income producing vineyard.' 4、若S中多了单词,则将多余的单词用[]括起来!
例:D:='Located in private mountain community of Yosemite Lakes Park.'
S:='Located in the private mountain community of Yosemite Lakes Park.'
则:S应为:
‘Located in [the] private mountain community of Yosemite Lakes Park.’ 5、标点符号的错误处理与上基本相同(将一个标点视为一个单词),处理时将正确答案结出。
例: D:='Good Morning!'
S:='Good Morning.'
则:结果S应为 'Good Morning!(.)' 应为‘!’,而不是'.'。
例: D:='wineries, hiking trails, State'
S:='wineries hiking trails State'
则结果S应为:' wineries,{} hiking trails,{} State'
例:D:='wineries hiking trails, State'
S:='wineries, hiking trails, State'
则结果为:'wineries[,] hiking trails, State'请结出处理的原代码,分不够再加!!!!!
例:
D:='Good Morning!' S:='Good Morning! '
则 S=D 成立! 2、若串S掉了n个单词,则在相应掉词的位置插入'{n}'。
例:
D:='Convenient to wineries, hiking trails, State Parks and Sugar Loaf Observatory. Award winning income producing vineyard.'
S:=Convenient to wineries, hiking trails, State Loaf Observatory. Award winning producing vineyard.'
则结果S应为:
'Convenient to wineries, hiking trails, State {3} Loaf Observatory. Award winning {1} producing vineyard.' 3、若S中错了一个单词,则将错误的单词用括号‘()’括起来。
例:D:='Convenient to wineries, hiking trails, State Parks and Sugar Loaf Observatory. Award winning income producing vineyard.'
S:='Convenient to wineries, hiking trail, State Parks and Sugar Loaf Observatory. Award win income producing vineyard.'
则结果S应为:
'Convenient to wineries, hiking (trail), State Parks and Sugar Loaf Observatory. Award (win) income producing vineyard.' 4、若S中多了单词,则将多余的单词用[]括起来!
例:D:='Located in private mountain community of Yosemite Lakes Park.'
S:='Located in the private mountain community of Yosemite Lakes Park.'
则:S应为:
‘Located in [the] private mountain community of Yosemite Lakes Park.’ 5、标点符号的错误处理与上基本相同(将一个标点视为一个单词),处理时将正确答案结出。
例: D:='Good Morning!'
S:='Good Morning.'
则:结果S应为 'Good Morning!(.)' 应为‘!’,而不是'.'。
例: D:='wineries, hiking trails, State'
S:='wineries hiking trails State'
则结果S应为:' wineries,{} hiking trails,{} State'
例:D:='wineries hiking trails, State'
S:='wineries, hiking trails, State'
则结果为:'wineries[,] hiking trails, State'请结出处理的原代码,分不够再加!!!!!
解决方案 »
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var
S, D: String;
begin
D := 'Good Morning!';
S := 'Good Morning! ';
D := StringReplace(D, #32, '', [rfReplaceAll]);
S := StringReplace(S, #32, '', [rfReplaceAll]);
if D = S then ShowMessage('相等');
end;—————————————————————————————————
宠辱不惊,看庭前花开花落,去留无意;毁誉由人,望天上云卷云舒,聚散任风。
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这样不行因为:D='a nurse'和S='an urse'会被视为相等!
2.用ExtractString将S,D分解为TStringList类型的单词串S1,S2;
3.对S1、S2进行比较;
2\3\4. 分解为单词,建一临时串,然后顺序比较S和D,遇到不等的时候做处理就行了.