遍历两个数组，取出值的问题。。。泪奔中。

[madelaop5566]

我自己写的测试代码

            string shijianjilv = "■";
            string shijianjilvB = "□";
            string aa = "aaa";
            string bb = "bbb";
            string cc = "ccc";
            string dd = "ddd";
            string ee = "eee";


            string[] str1 = { aa, bb, cc, dd, ee };
            string[] str2 = { "aaa", "ccc" };

            foreach (string sb in str2)
            {
                if (aa == sb)
                {
                    aa = aa + shijianjilv;
                }
                else if (bb == sb)
                {

                    bb = bb + shijianjilv;
                }
                else if (cc == sb)
                {
                    cc = cc + shijianjilv;
                }
                else if (dd == sb)
                {
                    dd = dd + shijianjilv;
                }
                else if (ee == sb)
                {
                    ee = ee + shijianjilv;
                }
            }
            Response.Write(aa + "<br/>" + bb + "<br/>" + cc + "<br/>" + dd + "<br/>" + ee + "<br/>");

输出结果是：
aa=aaa ■
bb=bbb
cc=ccc ■
dd=ddd
ee=eee


本人太菜了，这是我自己写了一段代码，怎么判断让结果等于
aa=aaa ■
bb=bbb □
cc=ccc ■
dd=ddd □
ee=eee □

我这样写实要替换word文档里标签值的，非得单独取得各个变量的值才行啊。，搞一天也没解决.

[threenewbee]

            string aa = "aaa";
            string bb = "bbb";
            string cc = "ccc";
            string dd = "ddd";
            string ee = "eee";
            Dictionary<string, string> str1 = new Dictionary<string, string>()
            {
                { "aa", aa },
                { "bb", bb }, 
                { "cc", cc },
                { "dd", dd }, 
                { "ee", ee }
            };
            string[] str2 = { "aaa", "ccc" };

            str1 = str1.Select(x => new { Key = x.Key, Value = str2.Contains(x.Value) ? x.Value + " ■" : x.Value + " □" }).ToDictionary(x => x.Key, y => y.Value);
            string s = "";
            str1.ToList().ForEach(x => { s += x.Key + "=" + x.Value + "<br />\r\n"; });
            Response.Write(s);

[threenewbee]

这样也可以

            string aa = "aaa";
            string bb = "bbb";
            string cc = "ccc";
            string dd = "ddd";
            string ee = "eee";
            Dictionary<string, string> str1 = new Dictionary<string, string>()
            {
                { "aa", aa },
                { "bb", bb }, 
                { "cc", cc },
                { "dd", dd }, 
                { "ee", ee }
            };
            string[] str2 = { "aaa", "ccc" };

            str1 = str1.Select(x => new { Key = x.Key, Value = str2.Contains(x.Value) ? x.Value + " ■" : x.Value + " □" }).ToDictionary(x => x.Key, y => y.Value);
            aa = str1["aa"];
            bb = str1["bb"];
            cc = str1["cc"];
            dd = str1["dd"];
            ee = str1["ee"];
            Response.Write(aa + "<br/>" + bb + "<br/>" + cc + "<br/>" + dd + "<br/>" + ee + "<br/>");

[我姓区不姓区]

又发了一遍？
word标签不是string吧？你在代码里也不可能写1000个word标签的局部变量吧？
你可以把word标签放到数组里，而不是把标签里的字符串放数组里

[madelaop5566]

我不会啊 谢谢2楼的大哥 我回去试试

[我姓区不姓区]

        static void Main(string[] args)
        {
            string aa = "aaa";
            string bb = "bbb";
            string cc = "ccc";
            string dd = "ddd";
            string ee = "eee";

            string[] str2 = { "aaa", "ccc" };

            Replace(ref aa, str2);
            Replace(ref bb, str2);
            Replace(ref cc, str2);
            Replace(ref dd, str2);
            Replace(ref ee, str2);

            Console.WriteLine(aa);
            Console.WriteLine(bb);
            Console.WriteLine(cc);
            Console.WriteLine(dd);
            Console.WriteLine(ee);
        }

        const string shijianjilv = "■";
        const string shijianjilvB = "□";

        static void Replace(ref string str, string[] arr)
        {

            str += Array.IndexOf(arr, str) > -1 ? shijianjilv : shijianjilvB;
        }

[kid_wang]

你觉得string shijianjilvB = "□";你用到了吗？

[lbc2247117]

你代码写错了，你遍历的是str2，应该遍历str1才是你想要的结果。

[Tragedy]

[Quote=引用 7 楼 lbc2247117 的回复:]
你代码写错了，你遍历的是str2，应该遍历str1才是你想要的结果。
[/Quote]
你试试遍历str1能取到值吗？不可能。
string[] str2 = { "aaa", "ccc" };
lz你看看你的str2里面就2个元素，你遍历也就才遍历2次，你结果匹配的肯定就只有2个。

[kingdom_0]

 string shijianjilv = "■";
            string shijianjilvB = "□";
            string aa = "aaa";
            string bb = "bbb";
            string cc = "ccc";
            string dd = "ddd";
            string ee = "eee";


            string[] str1 = { aa, bb, cc, dd, ee };
            string[] str2 = { "aaa", "ccc" };

            foreach (string sb in str2)
            {
                if (aa == sb)
                {
                    aa = aa + shijianjilv;
                }
                else if (bb == sb)
                {

                    bb = bb + shijianjilvB ;
                }
                else if (cc == sb)
                {
                    cc = cc + shijianjilv;
                }
                else if (dd == sb)
                {
                    dd = dd + shijianjilvB ;
                }
                else if (ee == sb)
                {
                    ee = ee + shijianjilvB ;
                }
            }
            Response.Write(aa + "<br/>" + bb + "<br/>" + cc + "<br/>" + dd + "<br/>" + ee + "<br/>");

[缭绕飘渺]

[Quote=引用 6 楼 kid_wang 的回复:]

你觉得string shijianjilvB = "□";你用到了吗？
[/Quote]
对，+1

[chengjia2284]

[Quote=引用 6 楼 kid_wang 的回复:]
你觉得string shijianjilvB = "□";你用到了吗？
[/Quote]

+2
aa=aaa ■
bb=bbb □
cc=ccc ■
dd=ddd □
ee=eee □

想实现这样的效果。shijianjilvB 至少会用到吧