问题描述
我需要从一个api的响应中提取子元素，然后将其动态传递给下一个api请求.
I need to extract child elements from response of one api and then pass it dynamically to next api request.
假设我具有以下xml:
Suppose i have the below xml:
* def foo =
"""
<records>
<record index="1">a</record>
<record index="2">b</record>
<record index="3" foo="bar">c</record>
</records>
"""
我只想提取这个:
<record index="1">a</record>
<record index="2">b</record>
<record index="3" foo="bar">c</record>
我尝试了以下选项，但没有一个起作用:
I tried below options, but none of then worked:
* def chk = foo//records/*
* def chk = foo//records/child::*
* def chk = foo//records/descendant::*
* print chk
打印后，我得到以下信息，如果我缺少任何东西或其他任何方式，请提出建议.谢谢！
After printing, I get the below, please suggest if i'm missing anything or any other way to do it.Thank you!
13:51:07.046 INFO - [print] {
"records": {
"record": [
{
"_": "a",
"@": {
"index": "1"
}
},
{
"_": "b",
"@": {
"index": "2"
}
},
{
"_": "c",
"@": {
"foo": "bar",
"index": "3"
}
}
]
}
}
推荐答案
如果您需要做的只是获取一组元素并将它们填充到另一个XML模板中，则看起来像是简单的字符串替换即可解决问题:(我将原来较长的答案留在该答案的后面，以供参考)
looks like a simple string replace would do the trick if all you need to do is take a set of elements and stuff them into another XML template:(I'm leaving the original longer answer at the end of this answer for reference)
* def foo =
"""
<records>
<record index="1">a</record>
<record index="2">b</record>
<record index="3" foo="bar">c</record>
</records>
"""
* replace foo.<records> = ''
* replace foo.</records> = ''
# you can do def bar = read('template.txt') in your case
* text bar =
"""
<after new="namespace">
@@foo@@
</after>
"""
* replace bar.@@foo@@ = foo
* xml bar = bar
* match bar ==
"""
<after new="namespace">
<record index="1">a</record>
<record index="2">b</record>
<record index="3" foo="bar">c</record>
</after>
"""
是的，空手道本身不支持XML节点列表，通常不需要. XML具有这种特殊性，您总是需要一个父"或包装"元素.
Yes, Karate does not support an XML node-list natively, typically you don't need it. And XML has this peculiarity where you always need a "parent" or "wrapping" element.
如上所示，您确实拥有所需的所有信息.我仍然不清楚您需要将哪种XML传递给下一个API-但是无论如何，空手道中都有一种方法.这是2个选项.首先，遍历数据并手动发出XML:
As you see above, you do have all the information needed. I'm still not clear what kind of XML you need to pass to the next API - but whatever the case, there is a way in Karate. Here are 2 options. First, iterating over the data and manually emitting XML:
* def foo =
"""
<records>
<record index="1">a</record>
<record index="2">b</record>
<record index="3" foo="bar">c</record>
</records>
"""
* json records = foo.records.record
* def temp = <bar></bar>
* def fun =
"""
function(r, i){
karate.set('temp', '/bar/record[' + (i+1) + ']', r._);
karate.set('temp', '/bar/record[' + (i+1) + ']/@index', r['@'].index);
}
"""
* eval karate.forEach(records, fun)
* match temp ==
"""
<bar>
<record index="1">a</record>
<record index="2">b</record>
<record index="3">c</record>
</bar>
"""
请注意，karate.forEach仅在即将到来的0.8.0中可用，但是您可以尝试使用0.8.0.RC3
Note that karate.forEach is available only in the upcoming 0.8.0 but you can try 0.8.0.RC3
这是另一个有趣的选项，如果您需要做的就是更改XML父级，则可以使用字符串替换来解决问题:
Here's another interesting option, if all you need to do is change the XML parent, a string replace will do the trick:
* replace foo.<records> = '<bar>'
* replace foo.</records> = '</bar>'
* xml bar = foo
* match bar ==
"""
<bar>
<record index="1">a</record>
<record index="2">b</record>
<record index="3" foo="bar">c</record>
</bar>
"""
这篇关于如何在空手道框架中获取子元素的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！