IN子句和占位符

问题描述

我试图在Android的做下面的SQL查询：

I'm attempting to do the following SQL query within Android:

    String names = "'name1', 'name2";   // in the code this is dynamically generated

    String query = "SELECT * FROM table WHERE name IN (?)";
    Cursor cursor = mDb.rawQuery(query, new String[]{names});

不过，Android不正确的值替换问号。我可以做到以下几点，但是，这并没有对SQL注入保护：

However, Android does not replace the question mark with the correct values. I could do the following, however, this does not protect against SQL injection:

    String query = "SELECT * FROM table WHERE name IN (" + names + ")";
    Cursor cursor = mDb.rawQuery(query, null);

我

如何才能解决这个问题，可以使用IN子句？

How can I get around this issue and be able to use the IN clause?

推荐答案

一个字符串形式的？，？，？，？可以是动态创建的串并安全地投入的原始SQL查询（因为它是一个受限制的形式不包含外部数据），然后该占位符可作为正常

A string of the form "?, ?, ..., ?" can be a dynamically created string and safely put into the original SQL query (because it is a restricted form that does not contain external data) and then the placeholders can be used as normal.

考虑一个函数字符串makePlaceholders（INT LEN）返回 len个问题痕用逗号，然后分离：

Consider a function String makePlaceholders(int len) which returns len question-marks separated with commas, then:

String[] names = { "name1", "name2" }; // do whatever is needed first
String query = "SELECT * FROM table"
    + " WHERE name IN (" + makePlaceholders(names.length) + ")";
Cursor cursor = mDb.rawQuery(query, names);

只要确保准确地传递尽可能多的值的地方。主机的默认最大极限参数中的SQLite是999 - 至少在一个正常的版本，不知道安卓）

Just make sure to pass exactly as many values as places. The default maximum limit of host parameters in SQLite is 999 - at least in a normal build, not sure about Android :)

快乐编码。

下面是一个实现：

String makePlaceholders(int len) {
    if (len < 1) {
        // It will lead to an invalid query anyway ..
        throw new RuntimeException("No placeholders");
    } else {
        StringBuilder sb = new StringBuilder(len * 2 - 1);
        sb.append("?");
        for (int i = 1; i < len; i++) {
            sb.append(",?");
        }
        return sb.toString();
    }
}

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