在java.sql实例中使用SQL IN子句的最佳变通方法是什么?由于SQL注入攻击安全问题,不支持多值的PreparedStatement:一个?占位符表示一个值,而不是一个值列表。

考虑下面的SQL语句:

SELECT my_column FROM my_table where search_column IN (?)

使用preparedStatement。setString(1, "'A', 'B', 'C'");本质上是一种无用的尝试,试图解决使用原因?首先。

有什么可行的解决办法?


当前回答

你可以使用这个javadoc中提到的setArray方法:

PreparedStatement statement = connection.prepareStatement("Select * from emp where field in (?)");
Array array = statement.getConnection().createArrayOf("VARCHAR", new Object[]{"E1", "E2","E3"});
statement.setArray(1, array);
ResultSet rs = statement.executeQuery();

其他回答

下面是一个完整的Java解决方案来为你创建准备好的语句:

/*usage:

Util u = new Util(500); //500 items per bracket. 
String sqlBefore  = "select * from myTable where (";
List<Integer> values = new ArrayList<Integer>(Arrays.asList(1,2,4,5)); 
string sqlAfter = ") and foo = 'bar'"; 

PreparedStatement ps = u.prepareStatements(sqlBefore, values, sqlAfter, connection, "someId");
*/



import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.SQLException;
import java.util.ArrayList;
import java.util.List;

public class Util {

    private int numValuesInClause;

    public Util(int numValuesInClause) {
        super();
        this.numValuesInClause = numValuesInClause;
    }

    public int getNumValuesInClause() {
        return numValuesInClause;
    }

    public void setNumValuesInClause(int numValuesInClause) {
        this.numValuesInClause = numValuesInClause;
    }

    /** Split a given list into a list of lists for the given size of numValuesInClause*/
    public List<List<Integer>> splitList(
            List<Integer> values) {


        List<List<Integer>> newList = new ArrayList<List<Integer>>(); 
        while (values.size() > numValuesInClause) {
            List<Integer> sublist = values.subList(0,numValuesInClause);
            List<Integer> values2 = values.subList(numValuesInClause, values.size());   
            values = values2; 

            newList.add( sublist);
        }
        newList.add(values);

        return newList;
    }

    /**
     * Generates a series of split out in clause statements. 
     * @param sqlBefore ""select * from dual where ("
     * @param values [1,2,3,4,5,6,7,8,9,10]
     * @param "sqlAfter ) and id = 5"
     * @return "select * from dual where (id in (1,2,3) or id in (4,5,6) or id in (7,8,9) or id in (10)"
     */
    public String genInClauseSql(String sqlBefore, List<Integer> values,
            String sqlAfter, String identifier) 
    {
        List<List<Integer>> newLists = splitList(values);
        String stmt = sqlBefore;

        /* now generate the in clause for each list */
        int j = 0; /* keep track of list:newLists index */
        for (List<Integer> list : newLists) {
            stmt = stmt + identifier +" in (";
            StringBuilder innerBuilder = new StringBuilder();

            for (int i = 0; i < list.size(); i++) {
                innerBuilder.append("?,");
            }



            String inClause = innerBuilder.deleteCharAt(
                    innerBuilder.length() - 1).toString();

            stmt = stmt + inClause;
            stmt = stmt + ")";


            if (++j < newLists.size()) {
                stmt = stmt + " OR ";
            }

        }

        stmt = stmt + sqlAfter;
        return stmt;
    }

    /**
     * Method to convert your SQL and a list of ID into a safe prepared
     * statements
     * 
     * @throws SQLException
     */
    public PreparedStatement prepareStatements(String sqlBefore,
            ArrayList<Integer> values, String sqlAfter, Connection c, String identifier)
            throws SQLException {

        /* First split our potentially big list into lots of lists */
        String stmt = genInClauseSql(sqlBefore, values, sqlAfter, identifier);
        PreparedStatement ps = c.prepareStatement(stmt);

        int i = 1;
        for (int val : values)
        {

            ps.setInt(i++, val);

        }
        return ps;

    }

}

好吧,所以我不记得我之前是如何(或在哪里)这样做的,所以我来堆栈溢出来快速找到答案。我很惊讶我不能。

所以,很久以前我是用这样的语句来解决IN问题的:

where myColumn in (select regexp_substr(:myList,'[^,]+', 1, level) from dual connect by regexp_substr(:myList,'[^,]+', 1, level) not null)

将myList参数设置为逗号分隔的字符串:a,B,C,D…

注意:该参数必须设置两次!

尝试使用instr函数?

select my_column from my_table where  instr(?, ','||search_column||',') > 0

then

ps.setString(1, ",A,B,C,"); 

不可否认,这是一个肮脏的黑客,但它确实减少了sql注入的机会。在甲骨文工作。

我只是为此制定了一个特定于postgresql的选项。它有点像黑客,有自己的优缺点和局限性,但它似乎是有效的,而且不局限于特定的开发语言、平台或PG驱动程序。

当然,诀窍是找到一种方法,将任意长度的值集合作为单个参数传递,并让db将其识别为多个值。我的解决方案是从集合中的值构造一个带分隔符的字符串,将该字符串作为单个参数传递,并使用string_to_array()与PostgreSQL正确使用它所需的强制转换。

因此,如果你想搜索“foo”,“blah”和“abc”,你可以将它们连接成一个字符串,如:'foo,blah,abc'。下面是直接的SQL语句:

select column from table
where search_column = any (string_to_array('foo,blah,abc', ',')::text[]);

显然,您可以将显式强制转换更改为您想要的结果值数组——int、text、uuid等。因为函数接受一个字符串值(或者两个,如果你也想自定义分隔符),你可以在准备好的语句中作为参数传递它:

select column from table
where search_column = any (string_to_array($1, ',')::text[]);

这甚至足够灵活,可以支持like比较:

select column from table
where search_column like any (string_to_array('foo%,blah%,abc%', ',')::text[]);

Again, no question it's a hack, but it works and allows you to still use pre-compiled prepared statements that take *ahem* discrete parameters, with the accompanying security and (maybe) performance benefits. Is it advisable and actually performant? Naturally, it depends, as you've got string parsing and possibly casting going on before your query even runs. If you're expecting to send three, five, a few dozen values, sure, it's probably fine. A few thousand? Yeah, maybe not so much. YMMV, limitations and exclusions apply, no warranty express or implied.

但它确实有效。

SetArray是最好的解决方案,但它不适用于许多老司机。下面的解决方法可以在java8中使用

String baseQuery ="SELECT my_column FROM my_table where search_column IN (%s)"

String markersString = inputArray.stream().map(e -> "?").collect(joining(","));
String sqlQuery = String.format(baseSQL, markersString);

//Now create Prepared Statement and use loop to Set entries
int index=1;

for (String input : inputArray) {
     preparedStatement.setString(index++, input);
}

这个解决方案比其他难看的while循环解决方案好,其中查询字符串是通过手动迭代构建的