本篇博文爲mongodb的curd中一篇,前面介紹簡單的查詢使用,這一篇重點則放在插入數據;java
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首先是準備好基本環境,能夠參考博文git
MongoDB一個基本數據稱爲document,和mysql不同,沒有強制約束哪些字段,能夠隨意的插入,下面是一個簡單的插入演示github
private static final String COLLECTION_NAME = "demo"; @Autowired private MongoTemplate mongoTemplate; /** * 新增一條記錄 */ public void insert() { JSONObject object = new JSONObject(); object.put("name", "一灰灰blog"); object.put("desc", "歡迎關注一灰灰Blog"); object.put("age", 28); // 插入一條document mongoTemplate.insert(object, COLLECTION_NAME); JSONObject ans = mongoTemplate .findOne(new Query(Criteria.where("name").is("一灰灰blog").and("age").is(28)), JSONObject.class, COLLECTION_NAME); System.out.println(ans); }
使用的關鍵地方爲一行: mongoTemplate.insert(object, COLLECTION_NAME);
spring
執行後輸出結果爲以下sql
{"name":"一灰灰blog","_id":{"counter":12472353,"date":1548333180000,"machineIdentifier":14006254,"processIdentifier":17244,"time":1548333180000,"timeSecond":1548333180,"timestamp":1548333180},"age":28,"desc":"歡迎關注一灰灰Blog"}
一次插入多條記錄,傳集合進去便可mongodb
/** * 批量插入 */ public void insertMany() { List<Map<String, Object>> records = new ArrayList<>(); for (int i = 0; i < 3; i++) { Map<String, Object> record = new HashMap<>(4); record.put("wechart", "一灰灰blog"); record.put("blog", Arrays.asList("http://spring.hhui.top", "http://blog.hhui.top")); record.put("nums", 210); record.put("t_id", i); records.add(record); } // 批量插入文檔 mongoTemplate.insert(records, COLLECTION_NAME); // 查詢插入的內容 List<Map> result = mongoTemplate.find(new Query(Criteria.where("wechart").is("一灰灰blog")), Map.class, COLLECTION_NAME); System.out.println("Query Insert Records: " + result); }
返回結果以下:數組
Query Insert Records: [{t_id=0, wechart=一灰灰blog, _id=5c49b07cd5b7ee435cbe5022, blog=[http://spring.hhui.top, http://blog.hhui.top], nums=210}, {t_id=1, wechart=一灰灰blog, _id=5c49b07cd5b7ee435cbe5023, blog=[http://spring.hhui.top, http://blog.hhui.top], nums=210}, {t_id=2, wechart=一灰灰blog, _id=5c49b07cd5b7ee435cbe5024, blog=[http://spring.hhui.top, http://blog.hhui.top], nums=210}]
咱們但願在插入以前,判斷數據是否存在,若是不存在則插入;若是存在則更新;此時就能夠採用upsert來使用,通常三個參數spring-boot
mongoTemplate.upsert(Query query, Update update, String collectionName)
第一個爲查詢條件,第二個爲須要更新的字段,最後一個指定對應的collection,一個簡單的實例以下學習
/** * 數據不存在,經過 upsert 新插入一條數據 * * set 表示修改key對應的value * addToSet 表示在數組中新增一條 */ public void upsertNoMatch() { // addToSet 表示將數據塞入document的一個數組成員中 UpdateResult upResult = mongoTemplate.upsert(new Query(Criteria.where("name").is("一灰灰blog").and("age").is(100)), new Update().set("age", 120).addToSet("add", "額外增長"), COLLECTION_NAME); System.out.println("nomatch upsert return: " + upResult); List<JSONObject> re = mongoTemplate .find(new Query(Criteria.where("name").is("一灰灰blog").and("age").is(120)), JSONObject.class, COLLECTION_NAME); System.out.println("after upsert return should not be null: " + re); System.out.println("------------------------------------------"); }
輸出結果以下:
nomatch upsert return: AcknowledgedUpdateResult{matchedCount=0, modifiedCount=0, upsertedId=BsonObjectId{value=5c49b07ce6652f7e1add1ea2}} after upsert return should not be null: [{"add":["額外增長"],"name":"一灰灰blog","_id":{"counter":14491298,"date":1548333180000,"machineIdentifier":15099183,"processIdentifier":32282,"time":1548333180000,"timeSecond":1548333180,"timestamp":1548333180},"age":120}] ------------------------------------------
前面的demo是演示不存在,那麼存在數據呢?
/** * 只有一條數據匹配,upsert 即表示更新 */ public void upsertOneMatch() { // 數據存在,使用更新 UpdateResult result = mongoTemplate.upsert(new Query(Criteria.where("name").is("一灰灰blog").and("age").is(120)), new Update().set("age", 100), COLLECTION_NAME); System.out.println("one match upsert return: " + result); List<JSONObject> ans = mongoTemplate .find(new Query(Criteria.where("name").is("一灰灰blog").and("age").is(100)), JSONObject.class, COLLECTION_NAME); System.out.println("after update return should be one: " + ans); System.out.println("------------------------------------------"); }
輸出結果以下,注意下面的輸出數據的 _id
,正視前面插入的那條數據,兩個數據惟一的不一樣,就是age被修改了
one match upsert return: AcknowledgedUpdateResult{matchedCount=1, modifiedCount=1, upsertedId=null} after update return should be null: [{"add":["額外增長"],"name":"一灰灰blog","_id":{"counter":14491298,"date":1548333180000,"machineIdentifier":15099183,"processIdentifier":32282,"time":1548333180000,"timeSecond":1548333180,"timestamp":1548333180},"age":100}]
若是query條件命中多條數據,怎麼辦?會修改幾條數據呢?
/** * 兩條數據匹配時,upsert 將只會更新一條數據 */ public void upsertTwoMatch() { // 多條數據知足條件時,只會修改一條數據 System.out.println("------------------------------------------"); List<JSONObject> re = mongoTemplate .find(new Query(Criteria.where("name").is("一灰灰blog").and("age").in(Arrays.asList(28, 100))), JSONObject.class, COLLECTION_NAME); System.out.println("original record: " + re); UpdateResult result = mongoTemplate .upsert(new Query(Criteria.where("name").is("一灰灰blog").and("age").in(Arrays.asList(28, 100))), new Update().set("age", 120), COLLECTION_NAME); System.out.println("two match upsert return: " + result); re = mongoTemplate.find(new Query(Criteria.where("name").is("一灰灰blog").and("age").is(120)), JSONObject.class, COLLECTION_NAME); System.out.println("after upsert return size should be 1: " + re); System.out.println("------------------------------------------"); }
根據實際輸出進行查看,發現只有一條數據被修改;另一條保持不變,結果以下
------------------------------------------ original record: [{"name":"一灰灰blog","_id":{"counter":12472353,"date":1548333180000,"machineIdentifier":14006254,"processIdentifier":17244,"time":1548333180000,"timeSecond":1548333180,"timestamp":1548333180},"age":28,"desc":"歡迎關注一灰灰Blog"}, {"add":["額外增長"],"name":"一灰灰blog","_id":{"counter":14491298,"date":1548333180000,"machineIdentifier":15099183,"processIdentifier":32282,"time":1548333180000,"timeSecond":1548333180,"timestamp":1548333180},"age":100}] two match upsert return: AcknowledgedUpdateResult{matchedCount=1, modifiedCount=1, upsertedId=null} after upsert return size should be 1: [{"name":"一灰灰blog","_id":{"counter":12472353,"date":1548333180000,"machineIdentifier":14006254,"processIdentifier":17244,"time":1548333180000,"timeSecond":1548333180,"timestamp":1548333180},"age":120,"desc":"歡迎關注一灰灰Blog"}] ------------------------------------------
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