python findall只是第一次出现

Question

我有这个xml：

有多个＆＃34;成分＆＃34;标签。我需要遍历每个级别：

import sys
from xml.etree import ElementTree as et

base="<ss><cod>cod1</cod><measure><m>1</m></measure><constituents><cod>const1</cod><measure><m>2</m></measure><constituents><cod>const1_1</cod><measure><m>3</m></measure><constituents><cod>const3</cod><measure><m>4</m></measure></constituents></constituents><constituents><cod>const1_2</cod><measure><m>3</m></measure><constituents><cod>const3</cod><measure><m>42</m></measure></constituents></constituents></constituents></ss>"
tsString = et.fromstring(base)
ss=tsString.getiterator('ss')       
for r in ss:
    measure = risp.findall('.//constituents') #(1) get const1, const1_1, const3, const1_2, const3_2, only needed is const1
    for c in measure:
        measure1 = c.findall('.//constituents') #(2) get const1_1, const3, const1_2, const3_2, only needed are const1_1, const1_2
        ....

但是findall会返回每个成员的出现。 我需要（1）findall只返回＆＃34; const1＆＃34;测量，（2）仅返回＆＃34; const1_1＆＃34;和＆＃34; const1_2＆＃34; （3）＆＃34; const3＆＃34;，＆＃34; const3_2＆＃34;

如何修复2个findall？

Answer 1

只需省略.//部分即可在当前节点中执行非递归搜索：

for r in ss:
    measure = r.findall('constituents')
    for c in measure:
        measure1 = c.findall('constituents')

Answer 2

.//获得所有孩子。只需获得下一步＆＃34;步骤＆＃34;。

import sys
from xml.etree import ElementTree as et

base="<ss><cod>cod1</cod><measure><m>1</m></measure><constituents><cod>const1</cod><measure><m>2</m></measure><constituents><cod>const1_1</cod><measure><m>3</m></measure><constituents><cod>const3</cod><measure><m>4</m></measure></constituents></constituents><constituents><cod>const1_2</cod><measure><m>3</m></measure><constituents><cod>const3</cod><measure><m>42</m></measure></constituents></constituents></constituents></ss>"
tsString = et.fromstring(base)
ss=tsString.getiterator('ss')
for r in ss:
    measure = r.findall('./constituents') #(1) get const1, const1_1, const3, const1_2, const3_2, only needed is const1
    for t in measure: #for test 
            print t[0].text # for test 
    for c in measure:
        measure1 = c.findall('./constituents') #(2) get const1_1, const3, const1_2, const3_2, only needed are const1_1, const1_2
        for t in measure1: # for test
            print t[0].text # for test