Flex空應用程序SQLStatement返回null

Question

我的SQL語句查詢返回null。讓我瘋狂的是，我可以讓loadSchema工作並打印出每張表的所有表和列。但是我無法成功執行簡單的查詢。任何想法，爲什麼這可能會發生？

public function db():void { 
    conn = new SQLConnection(); 
    conn.addEventListener(SQLEvent.OPEN, openHandler); 
    dbFile = File.applicationStorageDirectory.resolvePath("DB.sql"); 
    conn.open(dbFile); 
} 

private function openHandler(event:SQLEvent):void { 
    trace("The database was created successfully"); 
    makeTables(); 
    trace("Make Tables Called"); 
} 

做桌子方法：

private function makeTables():void{ 
     var createGraduateTable:SQLStatement = new SQLStatement(); 
     createGraduateTable.sqlConnection = conn; 
     createGraduateTable.text = graduateLoader.data; 

     var createGraduateAgeTable:SQLStatement = new SQLStatement(); 
     createGraduateAgeTable.sqlConnection = conn; 
     createGraduateAgeTable.text = graduateAgeLoader.data; 

     var createMarathonTable:SQLStatement = new SQLStatement(); 
     createMarathonTable.sqlConnection = conn; 
     createMarathonTable.text = marathonLoader.data; 

     try { 
      createGraduateTable.execute(); 
      createGraduateAgeTable.execute(); 
      createMarathonTable.execute(); 
      trace("Tables Created"); 
      // Describe Database 
      conn.loadSchema(SQLTableSchema); 
      for each(var SQLTableStmt:SQLTableSchema in conn.getSchemaResult().tables){ 
       trace("Table: " + SQLTableStmt.name); 
       for each (var SQLColm:SQLColumnSchema in SQLTableStmt.columns){ 
        trace("Column name: " + SQLColm.name); 
       } 
      } 
      queryDB(); 

     }catch(error:SQLError){ 
      trace("Error message:", error.message); 
      trace("Details:", error.details); 
     } 
    } 

我試着查詢使用監聽器的，並嘗試catch塊，但沒有任何工程。

queryDB方法：

public function queryDB(tableName:String = null):void{ 
     var query:SQLStatement = new SQLStatement(); 
     query.sqlConnection = conn; 
     query.text = "SELECT * FROM GRADUATE"; 
       //query.addEventListener(SQLEvent.RESULT, queryResult); 
       //query.addEventListener(SQLErrorEvent.ERROR, queryError); 
     try{ 
      query.execute(); 
      var result:SQLResult = query.getResult(); 
      if (result != null){ 
       trace("Results: " + result.data); 
      }else{ 
       trace("Result NULL"); 
      } 

     }catch(e:SQLError){ 

     } 
    } 

    private function queryResult(e:SQLEvent):void{ 
     var result:SQLResult = e.target.getResult(); 
     trace("Results: " + result.data); 
     conn.close(); 
    } 

    private function queryError(e:SQLErrorEvent):void{ 
     trace("Error message:", e.error.message); 
     trace("Query Error: " + e.error.details); 
    } 

就像我說的，奇怪的是，當我執行這一點，我會打印出所有的表名和列正確，但result.data始終爲空。任何幫助是極大的讚賞。

以下是用於創建GRADUATE_AGE表的graduate_age.sql文件的前20行。

-- DROP TABLE IF EXISTS GRADUATE_AGE; 
CREATE TABLE IF NOT EXISTS GRADUATE_AGE (
    GPA REAL, 
    CREDITS_ATTEMPTED REAL, 
    CREDITS_PASSED REAL, 
    CURRENT_CREDITS REAL, 
    AGE REAL, 
    GENDER TEXT 
); 
INSERT INTO GRADUATE_AGE VALUES (2.7, 138.5, 138.5, 7, 25.74, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (3.57, 146, 78, 9, 22.09, 'F'); 
INSERT INTO GRADUATE_AGE VALUES (3.63, 131, 131, 14, 23.21, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (3.86, 152, 119, 14, 26.56, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (3.03, 115.5, 115.5, 15, 26.3, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (2.25, 39, 39, 16, 23.8, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (0, 48, 0, 14, 22.45, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (3.02, 63.5, 51.5, 14.5, 25.4, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (3.14, 40, 40, 14, 32.93, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (2.73, 62, 62, 14, 21.35, 'M'); 
INSERT INTO GRADUATE_AGE VALUES (3.39, 144, 81, 17, 24.32, 'M'); 

在迴應@ crooksy88我替換insert（）方法調用queryDB（）方法調用，並給它下面的聲明：

private function insert():void{ 
     trace("Attempt Insert"); 
     var insert:SQLStatement = new SQLStatement(); 
     insert.sqlConnection = conn; 
     var insertStr:String = ""; 
     insertStr += "INSERT INTO GRADUATE_AGE (GPA, CREDITS_ATTEMPTED, CREDITS_PASSED, CURRENT_CREDITS, AGE, GENDER) VALUES (2.7, 138.5, 138.5, 7, 25.74, 'M');"; 
     insertStr += "INSERT INTO GRADUATE_AGE (GPA, CREDITS_ATTEMPTED, CREDITS_PASSED, CURRENT_CREDITS, AGE, GENDER) VALUES (3.57, 146, 78, 9, 22.09, 'F');"; 
     insert.text = insertStr; 
     insert.addEventListener(SQLEvent.RESULT, quereyResult); 
     insert.addEventListener(SQLErrorEvent.ERROR, quereyError); 
     insert.execute(); 
    } 

private function insertResult(e:SQLEvent):void{ 
     trace("Insert Complete: " + e.target.text); 
     queryDB(); 
    } 

我仍然得到空了我的結果。快把我逼瘋！它的確很愚蠢。

Answer 1

也許去將是第一個使用非常簡單的SQL語句關閉以創建該表並插入一條記錄的方式。如果這樣的作品，你可以建立在那。

我會先嚐試從SQL語句中刪除所有下劃線，並可能將您的表和列命名爲小寫字母，以便您可以更輕鬆地將它們與SQL關鍵字區分開來。

在你的SQL語句中，還有一些看起來不正確的東西。

您正在創建名爲GRADUATE_AGE的表，但試圖插入到GRADUATE中。

CREATE TABLE IF NOT EXISTS graduate (id INTEGER PRIMARY KEY AUTOINCREMENT, gap REAL, creditsAttempted REAL, creditsPassed REAL, currentCredits REAL, age INTEGER, gender TEXT) 

INSERT INTO graduate (gap, creditsAttempted, creditsPassed, currentCredits, age, gender) VALUES (2.7, 138.5, 138.5, 7, 25.74, 'M') 

Answer 2

它可能會更好，你會插入GRADUATE_AGE和不進GRADUATE;）