Question

我有<代码>ArrayList<Hashtable<String, String>>。

this is the hash table.

Hashtable<String, String> table = new Hashtable<String, String>();

清单将包含许多散列表格。我必须驳斥名单,并将内容列入每个表头索引。能够在一条法典中做到这一点? 我想使用一个变量,而不是通过使用许多漏洞和变量使这一变化复杂化。我希望我很清楚。 plz帮助。

Answer 1

You will have to iterate over the list, and in each iteration you have to get the element at the first index. It s not quite clear to me what you mean with "element at the first index of a Hashtable", however, since the order of elements in a Hashtable or HashMap is not guaranteed. It would make sense, though, if you were looking for the value mapped to a specific key.

int i = 0;
for ( HashMap<String,String> map : listOfMaps )
{
    array[ i ] = System.out.println(map.get( "name" ));
    i++;
}

或

for ( int i = 0; i < listOfMaps.length; i++ )
{
    array[ i ] = map.get("name");
}

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