iphone - How to Get the Direct Access Index for NSDictionary of NSDictionary -


यहां कोड है: 

  NSMutableDictionary * circuit_step = [NSMutableDictionary dictionary]; NSMutableDictionary * step_info = [NSMutableDictionary शब्दकोश]; [Step_info setObject: @ "खोज" के लिए की: @ "खोज-प्रकार"]; [Step_info setObject: @ "छोटा" forKey: @ "खोज-प्रारूप"]; [Step_info setObject: @Key के लिए "सर्दियों": @ "खोज-मौसम"]; [Step_info setObject: @ "पेड़" के लिए की: @ "खोज-स्थान"]; ** [सर्किट_स्टेप सेट ऑब्जेक्ट: सर्किट_स्टेप फॉरके: @ "01"]; ** [step_info setObject: @ "खोज" के लिए की: @ "सर्च-टाइप"]; [Step_info setObject: @ "सूक्ष्म" forKey: @ "खोज-प्रारूप"]; [Step_info setObject: @Key के लिए "गर्मी" @: @ "खोज-मौसम"]; [Step_info setObject: @ "झील द्वारा" के लिए: @ "खोज-स्थान"]; ** [circuit_step setObject: circuit_step forKey: @ "02"]; **

क्या शब्दकोश circuit_step कुंजी "01" पर एक सीधी पहुँच करने के लिए कोड है और शब्दकोश step_info कुंजी "खोज स्थान" NSLog के लिए उपयुक्त स्वरूप में?

<

के बारे में

क्या 
  NSLog (@, [:: @ "01"] objectForKey @ "खोज-स्थान" [circuit_step objectForKey] "मूल्य% @ है") / प्री> 
 इसके अलावा, आपका कोड सभी गलत है यहां एक निश्चित संस्करण है:  <पूर्व>  NSMutableDictionary * circuit_step = [NSMutableDictionary dictionary]; अगर (सर्किट_स्टेप! = शून्य) {NSMutableDictionary * step_info = शून्य; Step_info = [NSMutableDictionary शब्दकोश]; अगर (step_info! = शून्य) {[step_info setObject: @ "खोज" के लिए की: @ "खोज-प्रकार"]; [Step_info setObject: @ "छोटा" forKey: @ "खोज-प्रारूप"]; [Step_info setObject: @Key के लिए "सर्दियों": @ "खोज-मौसम"]; [Step_info setObject: @ "पेड़" के लिए की: @ "खोज-स्थान"]; [सर्किट_स्टेप सेट ऑब्जेक्ट: step_info forKey: @ "01"]; } Step_info = [NSMutableDictionary शब्दकोश]; अगर (step_info! = शून्य) {[step_info setObject: @ "खोज" के लिए की: @ "खोज-प्रकार"]; [Step_info setObject: @ "सूक्ष्म" forKey: @ "खोज-प्रारूप"]; [Step_info setObject: @Key के लिए "गर्मी" @: @ "खोज-मौसम"]; [Step_info setObject: @ "झील द्वारा" के लिए: @ "खोज-स्थान"]; [सर्किट_स्टेप सेट ऑब्जेक्ट: step_info forKey: @ "02"]; }}

आप सर्किट_स्टेप में सही ऑब्जेक्ट सेट नहीं कर रहे थे और आप एक शब्दकोश पुनः उपयोग कर रहे थे ताकि आप '2' के मान के साथ एक ही शब्दकोश की ओर इशारा करते हुए दो प्रविष्टियों के साथ समाप्त हो जाएंगे।


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