Sudzc生成的代码正在为反序列化节点编写字典。如果我使用NSLog(@“子节点:%@”,[[[element children] objectAtIndex:0] stringValue]);它会在每次通过时写出正确的项目。当我尝试在代码中检索结果时,只有最后一个可用(杰克逊3)。我究竟做错了什么?
// Deserializes the element in a dictionary.
+(id)deserializeAsDictionary:(CXMLNode*)element {
if([element childCount] == 1) {
CXMLNode* child = [[element children] objectAtIndex:0];
if([child kind] == CXMLTextKind)
{
NSLog(@"The Child Node: %@", [[[element children] objectAtIndex:0] stringValue]);
return [[[element children] objectAtIndex:0] stringValue];
}
}
NSMutableDictionary* d = [NSMutableDictionary dictionary];
for(CXMLNode* child in [element children]) {
id v = [Soap deserialize:child];
if(v == nil) { v = [NSNull null]; }
[d setObject:v forKey:[child name]];
}
return d;
}
的NSLog:
2012-04-19 14:13:07.802 Management[3043:10703] Hopefully Child: Allen
2012-04-19 14:13:07.803 Management[3043:10703] Hopefully Child: 1
2012-04-19 14:13:07.804 Management[3043:10703] Hopefully Child: John
2012-04-19 14:13:07.804 Management[3043:10703] Hopefully Child: 2
2012-04-19 14:13:07.805 Management[3043:10703] Hopefully Child: Jackson
2012-04-19 14:13:07.805 Management[3043:10703] Hopefully Child: 3
XML:
<TC diffgr:id="TC1" msdata:rowOrder="0">
<CSHR_POS_NAME>Allen</CSHR_POS_NAME>
<CSHR_NUM>66</CSHR_NUM>
</TC>
<TC diffgr:id="TC2" msdata:rowOrder="1">
<CSHR_POS_NAME>John</CSHR_POS_NAME>
<CSHR_NUM>2</CSHR_NUM>
</TC>
<TC diffgr:id="TC3" msdata:rowOrder="2">
<CSHR_POS_NAME>Jackson</CSHR_POS_NAME>
<CSHR_NUM>3</CSHR_NUM>
</TC>
答案 0 :(得分:2)
解决了(改变了soap.m):
[d setObject:v forKey:[child name]];
NSString* key = [child name];
id check = [d objectForKey:key];
if( check != nil ) {
NSInteger next = 1;
key = [NSString stringWithFormat:@"%@%d", [child name], next];
check = [d objectForKey:key];
while( check != nil ) {
next++;
key = [NSString stringWithFormat:@"%@%d", [child name], next];
check = [d objectForKey:key];
}
[d setObject:v forKey:key];
}
[d setObject:v forKey:[child name]];
答案 1 :(得分:1)
一旦我得到足够的重复点,我会,但我注意到代码以
开头和结尾 [d setObject:v forKey:[child name]];
对我来说,我必须删除初始行,并为我修复它,所以代码看起来像这样:
// Deserializes the element in a dictionary.
+(id)deserializeAsDictionary:(CXMLNode*)element {
if([element childCount] == 1) {
CXMLNode* child = [[element children] objectAtIndex:0];
if([child kind] == CXMLTextKind) {
return [[[element children] objectAtIndex:0] stringValue];
}
}
NSMutableDictionary* d = [NSMutableDictionary dictionary];
for(CXMLNode* child in [element children]) {
id v = [Soap deserialize:child];
if(v == nil) { v = [NSNull null]; }
//[d setObject:v forKey:[child name]]; //seems to be duped (maybe my bad)
//Extended by iDev on StackOverflow
//http://stackoverflow.com/questions/10235496/sudzc-deserializeasdictionary-over-written-dictionary/10358458#10358458
NSString* key = [child name];
id check = [d objectForKey:key];
if( check != nil ) {
NSInteger next = 1;
key = [NSString stringWithFormat:@"%@%04d", [child name], next];
check = [d objectForKey:key];
while( check != nil ) {
next++;
key = [NSString stringWithFormat:@"%@%04d", [child name], next];
check = [d objectForKey:key];
}
[d setObject:v forKey:key];
}
[d setObject:v forKey:[child name]];
//End Extension
}
return d;
}
答案 2 :(得分:1)
我发现你还必须放入一个其他块:
if( check != nil ) {
NSInteger next = 1;
key = [NSString stringWithFormat:@"%@%04d", [child name], next];
check = [d objectForKey:key];
while( check != nil ) {
next++;
key = [NSString stringWithFormat:@"%@%04d", [child name], next];
check = [d objectForKey:key];
}
[d setObject:v forKey:key];
} else {
[d setObject:v forKey:[child name]];
}
//End Extension
否则'd'中的元素将被覆盖,因为setObject被调用两次。
答案 3 :(得分:0)
我不确定你是如何应用你的修补程序的,你是在替换 deserializeAsDictionary 中的整个代码还是追加到代码的末尾?
看到代码行有一个for循环
[d setObject:v forKey:[child name]];
找到了,所以我猜你只是扩展了它,这样你就可以在这里扩展它而不是关闭for循环,这是对的吗?
答案 4 :(得分:0)
我遇到上述代码的问题 - 它每次都覆盖了第一个条目 - 即我会得到一个包含4个项目的列表,第一个和第四个将被复制。
在通过代码(爱的递归代码)之后经过了大量的精力充沛,我找到了我认为的问题;
我的代码如下:
if( check != nil ) {
NSInteger next = 1;
key = [NSString stringWithFormat:@"%@%04d", [child name], next];
check = [d objectForKey:key];
while( check != nil ) {
next++;
key = [NSString stringWithFormat:@"%@%04d", [child name], next];
check = [d objectForKey:key];
}
[d setObject:v forKey:key];
}
else
{
[d setObject:v forKey:[child name]];
}