[英]Help with recursion and returning value
This is an extension of a previously posted question. 这是先前发布的问题的扩展。 I'm trying to recursively build a string. 我试图递归地建立一个字符串。 I need to change the function below to do this - each recursion of the function generates the desired string, but I need to concat these together and return the whole string. 我需要更改下面的函数来执行此操作-函数的每次递归都会生成所需的字符串,但是我需要将它们合并在一起并返回整个字符串。 'related' is passed into the function as an empty string and I thought the way I was using string.Format would append each recursion to the 'related' string? “相关”作为一个空字符串传递到函数中,我认为我使用string.Format的方式会将每个递归追加到“相关”字符串上吗? Apparently not. 显然不是。
Not sure how... 不确定如何...
private string getRelatedNews(Taxonomy taxData, string related, string contentTitle)
{
foreach (TaxonomyItemData item in taxData.TaxonomyItems)
{
if (taxData.TaxonomyName.Equals(contentTitle) && taxData.TaxonomyItemCount != 0)
{
related = string.Format("{0}<li><a href='{1}'\">{2}</a></li>", related, item.Link, item.Name);
}
}
// Show all its sub categories
foreach (TaxonomyData cat in taxData.Taxonomy)
{
getRelatedNews(cat, related, contentTitle);
}
return(related);
}
foreach (TaxonomyData cat in taxData.Taxonomy)
{
getRelatedNews(cat, related, contentTitle);
}
should be 应该
foreach (TaxonomyData cat in taxData.Taxonomy)
{
related = getRelatedNews(cat, related, contentTitle);
}
because strings are immutable. 因为字符串是不可变的。
Well try this... 好吧试试这个...
related = getRelatedNews(cat, related, contentTitle);
I am not sure about your logic and flow of the program... then also I think recursive function must be called like this.... 我不确定您的程序逻辑和流程...然后我还认为必须这样调用递归函数....
private string getRelatedNews(Taxonomy taxData, string related, string contentTitle)
{
foreach (TaxonomyItemData item in taxData.TaxonomyItems)
{
if (taxData.TaxonomyName.Equals(contentTitle) && taxData.TaxonomyItemCount != 0)
{
related = string.Format("{0}<li><a href='{1}'\">{2}</a></li>", related, item.Link, item.Name);
}
}
// Show all its sub categories
foreach (TaxonomyData cat in taxData.Taxonomy)
{
related = getRelatedNews(cat, related, contentTitle);
}
return(related);
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.