Android SQLite：约束失败错误代码 19

Question

我执行了以下sql：

更新记录集out_ payment = 4 where out_ payment = 4;

记录的out_ payment 被定义为Integer 并引用Payment(id)

我可以确保4 是付款表的ID 之一；

但我仍然遇到约束失败......

07-31 10:20:36.014: ERROR/Database(19085): Error updating out_payment=4 using UPDATE record_table SET out_payment=? WHERE out_payment = 4
07-31 10:20:45.964: ERROR/AndroidRuntime(19085): FATAL EXCEPTION: main
07-31 10:20:45.964: ERROR/AndroidRuntime(19085): android.database.sqlite.SQLiteConstraintException:   error code 19: constraint failed

代码如下：

values.clear();
values.put(RecordSchema.ID_OUT_PAYMENT, oldid);
selection = RecordSchema.ID_OUT_PAYMENT + " = " + oldid + "";
this.db.update(Table.RECORD, values, selection, null);

架构如下：

db.execSQL("CREATE TABLE IF NOT EXISTS " + TABLE.RECORD
+ " (" + RecordSchema.ID + " INTEGER PRIMARY KEY"
+ "," + RecordSchema.AMOUNT + " TEXT NOT NULL"
+ "," + RecordSchema.ID_CATEGORY + " INTEGER NOT NULL"
+ "," + RecordSchema.ID_SUBCATEGORY + " INTEGER"
+ "," + RecordSchema.DATE + " DATE NOT NULL"
+ "," + RecordSchema.ID_IN_PAYMENT + " INTEGER"
+ "," + RecordSchema.ID_OUT_PAYMENT + " INTEGER"

+ ",FOREIGN KEY(" + RecordSchema.ID_CATEGORY + ") REFERENCES " 
+ IsAiZanTable.CATEGORY + "(" + CategorySchema.ID + ") ON UPDATE CASCADE"

+ ",FOREIGN KEY(" + RecordSchema.ID_SUBCATEGORY + ") REFERENCES "
+ IsAiZanTable.SUBCATEGORY + "(" + SubcategorySchema.ID + ") ON UPDATE CASCADE"

+ ",FOREIGN KEY(" + RecordSchema.ID_IN_PAYMENT + ") REFERENCES "
+ IsAiZanTable.PAYMENT + "(" + PaymentSchema.ID + ") ON UPDATE CASCADE"

+ ",FOREIGN KEY(" + RecordSchema.ID_OUT_PAYMENT + ") REFERENCES "
+ IsAiZanTable.PAYMENT + "(" + PaymentSchema.ID + ") ON UPDATE CASCADE"
                + ");");

db.execSQL("CREATE TABLE IF NOT EXISTS " +Table.PAYMENT
+ " (" + PaymentSchema.ID + " INTEGER PRIMARY KEY"
+ "," + PaymentSchema.KIND + " INTEGER NOT NULL"
+ "," + PaymentSchema.NAME + " TEXT NOT NULL UNIQUE"
+ "," + PaymentSchema.TOTAL + " TEXT NOT NULL"
+ "," + PaymentSchema.HIDDEN + " INTEGER NOT NULL"
+ ");");

任何人都可以帮助我解决这个问题吗？

实际上我想将 id 从 4 更新为 5 我可以确保4和5都是付款表的现有ID。

但是又出现了同样的问题，所以我更新4到4也是一样。 我想如果我能解决4到4的问题，我就应该解决4到5的问题。

非常感谢您的回答！！

Answer 1

我已经找到答案了。
我的团队之一运行以下命令

db.execSQL("PRAGMA foreign_keys = OFF;");

，然后将 0 放入记录表的 ID_OUT_PAYMENT 中。
0 不是 PAYMENT 表的存在 ID。

因此，当我更新 ID_OUT_PAYMENT 时，它将触发约束失败。

所以如果我想让上面的sql成功运行。
我也需要运行sql。

 db.execSQL("PRAGMA foreign_keys = OFF;");

非常感谢大家回复我！！

谢谢！