sql - Mysql: How do you create a table with multiple primary keys? -

-

मेरे पास यह स्थिति है:

(तालिका वेतन )

आप एकाधिक प्राथमिक कुंजी के साथ एक तालिका कैसे बना सकते हैं? 

  तालिका वेतन बनाना (dep_id छोटी-छोटी ऑटो_इनक्रियम प्राथमिक कुंजी, emp_id छोटी-छोटी ऑटो_इनक्रिमेंट प्राथमिक कुंजी, bla varchar (20 ));

अगर मैं उपरोक्त कोड की कोशिश करता हूं तो मुझे एक त्रुटि मिलती है किसी भी विचार?

एक तालिका में केवल एक प्राथमिक कुंजी हो सकती है हालांकि, प्राथमिक कुंजी में कई कॉलम शामिल हो सकते हैं, उदाहरण के लिए 

  सब्सक्राइबर के वेतन (डिपाइड स्किलिंग अनअंसेंटेड नल, ए_आईड स्माललिंग अनसैंटेड नल, ब्लॉ VARCHAR (20), प्राइमरी की (डिपा_आईडी, एएनआईडी) ));

Comments

Post a Comment

Popular posts from this blog

ios - Form Sheet content disappearing -

-
Image
I am trying to apply a form sheet. Looks like this on the storyboard: App background only when I run Appears and everything else is hidden. I am presenting a form sheet using model sheets - I am working with the form sheet and I (WREGLR Regular) size. Why is this happening?
Read more

MySql variables and php -

-
I am getting an error with php in order to format mysql_query () in php to format this string What is the right way? SELECT SELECT SELECT (*) Agents @AgentCount; User_Agent_English, User_Angent_Oursal, US.___Gent_Aginal, COUNT (User_Agent_Percade) ACC, COUNT (*) / (AgentCount) Percentage by agents user_agent_preser order DEC limit limit 50; In PHP, how am i setting $ query $ query = "SELECT count (*) agents @ agentcount; SELECT user_agent_parsed, User_agent_original , COUNT (user_agent_parsed) ascount, COUNT (*) / (@AgentCount) as a percentage by user_agent_parced by URI date DESC limit 50 "; This exact query will work properly if I put it in MySQL directly through a command line session. Do I need to release two separate php calls from mysql_query () and store the results first? I get the following error: There is an error in your SQL syntax; Instead of using a sub-selection and instead of using the 'SELECT user_agent_parsed, user_agent_original, ne...
Read more

c++ - an enclosing-function local variable cannot be referenced in a lambda body unless if it is in capture list -

-
I have the json :: value object and I try to get the value in a straight but I'm about the capture list I get this error in I understand that this bracket holds the [] capture list in the phrase, but I can not understand how I can return a value in the Lambda function? Zero Jasondomain :: Set Value (JSN :: Value OBG) {weather.coord.lon = obj.at (L "Coordination"). (L "longitude") as_double () .; Weather.coord.lat = obj.at (L "Coordinate"). (L "Latitude"). As_double (); } Zero JsonDeneme :: getHttp () {// json :: value val; Http_client Client (U ("http://api.openweathermap.org/data/2.5/weather?q=Ankara,TR")); Client.request (methods :: GET) .then ([] (http_response response) - & gt; pplx :: task & lt; json :: value & gt; {if (response.status_code) == condition_code :: OK ) {Printf ("Received response status code:% u \ n", response.status_code ()); Return response .extract_json ();} return pplx :: task_fro...
Read more