ios - Printing value of a optional variable includes the word "Optional" in Swift -

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इस सवाल का पहले से ही एक उत्तर है: < / P>

  • 6 उत्तर
  • 3 जवाब

.मैं आईओएस और स्विफ्ट के लिए नया हूँ। अपने आवेदन में मैं एक वैकल्पिक मूल्य को मुद्रित करने की कोशिश कर रहा हूं और यह प्रिंट करता है "वैकल्पिक वेरिएबल) "मैं यह शब्द कैसे वैकल्पिक 

  var bDay = StringUtils.convertDateToString (जन्मदिन, प्रारूप: Constants.BIRTHDAY_FORMAT) को हटा सकता हूँ उम्र = self.clientDetail? .getAge () println ("उम्र ..... \ (उम्र)") bDay + = "\ (उम्र)"

कंसोल में आउटपुट

< पूर्व> आयु ..... वैकल्पिक (2 9)

मैं एक यूआईएलबल को इस वैरिएबल को आवंटित करने की कोशिश कर रहा हूं लेकिन स्क्रीन पर यह दिखाई देता है कि 17 सितंबर, 1986 वैकल्पिक (29)

मेरा उद्देश्य इस वैकल्पिक को दूर करना है शब्द और इसे सितंबर 17, 1986 (2 9)

पहले से धन्यवाद

वैकल्पिक चेनिंग का यहां प्रयोग किया जाता है: 

  उम्र दें = self.clientDetail। .getAge ()

तो वापसी getAge () वैकल्पिक मान है वैकल्पिक बाध्यकारी प्रयास करें: 

  यदि उम्र = उम्र {println ("उम्र ..... \ (उम्र)")}

या बस खोलना आयु उम्र! के साथ, लेकिन यह आपके ऐप को क्रैश कर देगा यदि age शून्य है।


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