java - android how to build the apk that works for android 2.3 as minimum os level -

-

ऐसा मेरे मैनिफ़ेस्ट फ़ाइल को कैसे दिखता है 

  & lt; manifest xmlns: android = " Http://schemas.android.com/apk/res/android "package =" com.abc "android: versionCode =" 1 "एंड्रॉइड: versionName =" 1.0 "& gt; & Lt; उपयोग- sdk एंड्रॉइड: minSdkVersion = "8" एंड्रॉइड: targetSdkVersion = "17" / & gt;   
 जब मैं एंड्रॉइड प्रोजेक्ट पर & gt; गुण & gt; एंड्रॉइड मैं एंड्रॉइड 4.0 के रूप में न्यूनतम के रूप में निशाना नाम देखता हूं  
 मैं एंड्रॉइड 2.3 प्लेटफॉर्म में एपीके फाइल को प्रयोग करने योग्य बनाना चाहता हूं  
 फीचर्स और एपीआई का उपयोग कर रहा हूं:  
     
  •  Google मानचित्र v2, फ्यूज़ स्थान प्रदाता   
     मैं एंड्रॉइड के लिए काफी नया हूँ, मैंने   
     
     
     आपका ऐप एपीआई 17 वाले एंड्रॉइड वर्जन पर लक्षित है, लेकिन आपका ऐप जीबी (जिंजरब्रेड 2.3.x) के लिए समर्थित है, न्यूनतम एपीआई स्तरीय 10 तक का उल्लेख करके आपका ऐप चलाया जाना माना जाता है 2.3.x में और जब भी आप सभी सुविधाओं का उपयोग करते रहेंगे, तब तक जीबी जैसे फ्रैग्मेंट्स, एक्शनबार  
      टिप:   
     का उपयोग किया जा सकता है। सिर में दर्द नवीनतम कंपाइलर उदाहरण का उपयोग करता है: एपीआई 20 और उन उपकरणों पर लक्षित जो वर्तमान में रननंग (एपीआई 1 9)

Comments

Post a Comment

Popular posts from this blog

mysql - where clause in inner join query -

-
Please help me about this question: Look for functions ($ userpno) } {Echo $ userpno; $ This-> query = "SELECT task.employee_id, task.user_id, task.service_id, service.name AS servicename, service.description AS servicedescription, employee.name as employineame, employee.pic_path employment, employee.pic_path from work Work.user_id = '$ userpno' Join the INNER employee. Pno = task.employee_id Join INNER user user.pno = task.user_id service.service_id = task.service_id "INNER service; } query works perfectly without: WHERE task.user_id = '$ userpno' my Also try this method: WHERE task.user_id = $ userpno but it does not work. Error: Warning: mysqli_fetch_array () parameter 1 is expected to be mysqli_result, given in Boolean C: \ wamp \ www \ admin \ classes \ Task Php on line 22 Please tell me where I say where is it. Try: $ this-> query = "SELECT Task.employee_id, task.user_id, task.service_id, service.name AS s...
Read more

java - Why my included JSP file won't get processed correctly? -

-
I am trying to create Java web framework (and learning), and on the basis of 'Code Generator In the process of making the material view of the database in the process of development, I stumbled into a difficulty, which I do not know how to solve it. First of all, I want the following index.jsp : & lt; Body & gt; Use all pages to be created. & Lt;% @ include file = "header.jsp"% & gt; & Lt; Hours / & gt; & Lt;% @ included file = "body.jsp"%> & Lt; Hours / & gt; & Lt;% @ include file = "footer.jsp"% & gt; & Lt; / Body & gt; And, in body.jsp , I should have it like this: & lt; Jsp: include page = "$ {Application_modul}"%> Where application_modul is an attribute defined in its controller: request.setAttribute ("application_modul", "USER_ACCOUNT \\ ' View_user_account.jsp "); This file can be found properly, but the processed JSP does ...
Read more

php - MySQL Query for Advanced Search multiple criteria -

-
I am working through an advanced search query where users select the checkbox based on multiple criteria. I am starting because I am hoping that I will know whether I am on the right path. I know that the query is quite messy Edit: I realized that there are two questions to be needed, so I have strictly modified a question question table data tables ss_users user_id FIRST_NAME LAST_NAME admin_level user_approved 1 Nick Jones 0 1 2 Johnny Rocket 0 1 table ss_user_profile_status user_id photo_uploaded 1 1 2 1 table ss_user_photos selected user_id file name photo_id 1 1 photo_1 .jpg 1 2 2 photo_2.jpg 0 3 2 photo_2.jpg 1 table ss_general user_id city state zip neighborhood 1 Baltimore maryland 00125 hamsterdam 2 Lekland My 11542 treemont table ss_languages ​​user_id French German Italian Spanish 1 0 1 0 1 2 0 1 1 Table ss_experience Waldorf Kumon homeschooling user_id 1 0 1 0 2 0 0 1 My current The result is that all users, which should not be the case: ... Select ss_use...
Read more