2012-01-08 77 views
16

我經歷的SOQL文檔的實體,但找不到查詢來獲取一個實體的所有現場數據說,帳戶,像銷售人員SOQL:查詢來獲取所有字段上

select * from Account [ SQL syntax ] 

在SOQL中是否有上述語法來獲取帳戶的所有數據,或者唯一的方法是列出所有字段(儘管有很多字段需要查詢)

回答

16

您必須指定字段if您希望構建動態的describeSObject調用返回關於對象的所有字段的元數據,因此您可以從中構建查詢。

+2

感謝您的答覆。你介意分享一個例子,從describeSObject構建查詢。 – Sukhhhh 2012-01-08 19:26:28

24

創建地圖是這樣的:

Map<String, Schema.SObjectField> fldObjMap = schema.SObjectType.Account.fields.getMap(); 
List<Schema.SObjectField> fldObjMapValues = fldObjMap.values(); 

然後你就可以通過fldObjMapValues迭代創建SOQL查詢字符串:

String theQuery = 'SELECT '; 
for(Schema.SObjectField s : fldObjMapValues) 
{ 
    String theLabel = s.getDescribe().getLabel(); // Perhaps store this in another map 
    String theName = s.getDescribe().getName(); 
    String theType = s.getDescribe().getType(); // Perhaps store this in another map 

    // Continue building your dynamic query string 
    theQuery += theName + ','; 
} 

// Trim last comma 
theQuery = theQuery.subString(0, theQuery.length() - 1); 

// Finalize query string 
theQuery += ' FROM Account WHERE ... AND ... LIMIT ...'; 

// Make your dynamic call 
Account[] accounts = Database.query(theQuery); 

superfell是正確的,沒有辦法直接做一個SELECT *。然而,這個小代碼配方將工作(嗯,我沒有測試過,但我認爲它看起來沒問題)。可以理解的是Force.com想要一個多租戶架構,其中資源僅規定爲明確需要 - 不容易做SELECT *當實際需要通常隻字段的子集。

+0

謝謝亞當。當你與superfell同意,接受他的回答:-) – Sukhhhh 2012-01-09 10:02:00

+0

當然Sukhhhh :) – Adam 2012-01-09 16:14:24

6

我使用Force.com Explorer和模式過濾器中,您可以點擊旁邊的表名的複選框,它會選擇所有字段,並插入到你的查詢窗口 - 我用這個作爲一個快捷方式typeing這一切 - 從查詢窗口複製並粘貼。希望這可以幫助。

3

爲了防止有人尋找一個C#的做法,我可以使用反射,並拿出如下:

public IEnumerable<String> GetColumnsFor<T>() 
{ 
    return typeof(T).GetProperties(System.Reflection.BindingFlags.Public | System.Reflection.BindingFlags.Instance) 
     .Where(x => !Attribute.IsDefined(x, typeof(System.Xml.Serialization.XmlIgnoreAttribute))) // Exclude the ignored properties 
     .Where(x => x.DeclaringType != typeof(sObject)) // & Exclude inherited sObject propert(y/ies) 
     .Where(x => x.PropertyType.Namespace != typeof(Account).Namespace) // & Exclude properties storing references to other objects 
     .Select(x => x.Name); 
} 

這似乎爲我測試過(與對象一起列匹配由API測試生成)。從那裏,它是關於創建查詢:

/* assume: this.server = new sForceService(); */ 

public IEnumerable<T> QueryAll<T>(params String[] columns) 
    where T : sObject 
{ 
    String soql = String.Format("SELECT {0} FROM {1}", 
     String.Join(", ", GetColumnsFor<T>()), 
     typeof(T).Name 
    ); 
    this.service.QueryOptionsValue = new QueryOptions 
    { 
     batchsize = 250, 
     batchSizeSpecified = true 
    }; 
    ICollection<T> results = new HashSet<T>(); 
    try 
    { 
     Boolean done = false; 
     QueryResult queryResult = this.service.queryAll(soql); 
     while (!finished) 
     { 
      sObject[] records = queryResult.records; 
      foreach (sObject record in records) 
      { 
       T entity = entity as T; 
       if (entity != null) 
       { 
        results.Add(entity); 
       } 
      } 
      done &= queryResult.done; 
      if (!done) 
      { 
       queryResult = this.service.queryMode(queryResult.queryLocator); 
      } 
     } 
    } 
    catch (Exception ex) 
    { 
     throw; // your exception handling 
    } 
    return results; 
} 
+0

好男人布拉德,是工作! – Shazoo 2017-07-24 11:58:58

1

對我來說這是今天第一次與Salesforce和我在Java中想出了這一點:

/** 
* @param o any class that extends {@link SObject}, f.ex. Opportunity.class 
* @return a list of all the objects of this type 
*/ 
@SuppressWarnings("unchecked") 
public <O extends SObject> List<O> getAll(Class<O> o) throws Exception { 
    // get the objectName; for example "Opportunity" 
    String objectName= o.getSimpleName(); 

    // this will give us all the possible fields of this type of object 
    DescribeSObjectResult describeSObject = connection.describeSObject(objectName); 

    // making the query 
    String query = "SELECT "; 
    for (Field field : describeSObject.getFields()) { // add all the fields in the SELECT 
     query += field.getName() + ','; 
    } 
    // trim last comma 
    query = query.substring(0, query.length() - 1); 

    query += " FROM " + objectName; 

    SObject[] records = connection.query(query).getRecords(); 

    List<O> result = new ArrayList<O>(); 
    for (SObject record : records) { 
     result.add((O) record); 
    } 
    return result; 
} 
+3

請解釋我們正在查看的內容,而不是僅僅發佈一段代碼。謝謝。 – Andrew 2014-09-24 18:21:16