MongoDB 模糊查询的三种实现方式-morphia实现

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MongoDB 模糊查询的三种实现方式-morphia实现如下：

方式一：正则表达式-单个值模糊查询

 

 

Pattern pattern = Pattern.compile("^" + treeIndex + ".*$", Pattern.CASE_INSENSITIVE);

assetInfoDao.createQuery().filter("treeIndexes", pattern);

方式二：正则表达式-多个值模糊查询

 

String[] categorys = category.split(",");

List patterns = Lists.newArrayList();

for (String treeIndex : categorys) {

Pattern pattern = Pattern.compile("^" + treeIndex + ".*$", Pattern.CASE_INSENSITIVE);

patterns.add(pattern);

}

query.filter("treeIndexes all", patterns);

方式三：contains模糊查询方式

 

if (StringUtils.isNotBlank(title)) {

query = query.field("title").contains(title);

}

方式四：精确查询

 

String[] categorys = category.split(",");

query.filter("treeIndexes all", Arrays.asList(categorys));

 

imports:

import java.util.regex.Pattern;

import com.google.code.morphia.DatastoreImpl;

import com.google.code.morphia.query.Query;

 

PS:

/**

*

Create a filter based on the specified condition and value.

*

* Note: Property is in the form of "name op" ("age >").

*

* Valid operators are ["=", "==","!=", "<>", ">", "<", ">=", "<=", "in", "nin", "all", "size", "exists"]

*

*

Examples:

* 

*

*

• {@code filter("yearsOfOperation >", 5)}

*

• {@code filter("rooms.maxBeds >=", 2)}

*

• {@code filter("rooms.bathrooms exists", 1)}

*

• {@code filter("stars in", new Long[]{3,4}) //3 and 4 stars (midrange?)}

*

• {@code filter("age >=", age)}

*

• {@code filter("age =", age)}

*

• {@code filter("age", age)} (if no operator, = is assumed)

*

• {@code filter("age !=", age)}

*

• {@code filter("age in", ageList)}

*

• {@code filter("customers.loyaltyYears in", yearsList)}

*

* 

*

You can filter on id properties if this query is

* restricted to a Class.

*/