com.yahoo.squidb.sql.Query.hasTable()方法的使用及代码示例

x33g5p2x  于2022-01-29 转载在 其他  
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本文整理了Java中com.yahoo.squidb.sql.Query.hasTable方法的一些代码示例,展示了Query.hasTable的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。Query.hasTable方法的具体详情如下:
包路径:com.yahoo.squidb.sql.Query
类名称:Query
方法名:hasTable

Query.hasTable介绍

暂无

代码示例

代码示例来源:origin: yahoo/squidb

  1. private Query inferTableForQuery(Class<? extends AbstractModel> modelClass, Query query) {
  2. if (!query.hasTable() && modelClass != null) {
  3. SqlTable<?> table = getSqlTable(modelClass);
  4. if (table == null) {
  5. throw new IllegalArgumentException("Query has no FROM clause and model class "
  6. + modelClass.getSimpleName() + " has no associated table");
  7. }
  8. query = query.from(table); // If argument was frozen, we may get a new object
  9. }
  10. return query;
  11. }

代码示例来源:origin: com.yahoo.squidb/squidb

  1. private Query inferTableForQuery(Class<? extends AbstractModel> modelClass, Query query) {
  2. if (!query.hasTable() && modelClass != null) {
  3. SqlTable<?> table = getSqlTable(modelClass);
  4. if (table == null) {
  5. throw new IllegalArgumentException("Query has no FROM clause and model class "
  6. + modelClass.getSimpleName() + " has no associated table");
  7. }
  8. query = query.from(table); // If argument was frozen, we may get a new object
  9. }
  10. return query;
  11. }

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