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Java SQLite删除where子句参数_Java_Android_Database_Sqlite_Where Clause - Fatal编程技术网

Java SQLite删除where子句参数

Java SQLite删除where子句参数,java,android,database,sqlite,where-clause,Java,Android,Database,Sqlite,Where Clause,我是Sqlite的新手。在此之前,我尝试使用trick\u id删除表中的一行 public void deleteTrick (int trick_id) { SQLiteDatabase db = this.getWritableDatabase(); db.delete(Constants.TRICKS_TABLE, Constants.TRICK_ID + " ?= ", new String[]{ String.valueOf(trick_id)}); db.c

我是Sqlite的新手。在此之前,我尝试使用trick\u id删除表中的一行

public void deleteTrick (int trick_id) {
    SQLiteDatabase db = this.getWritableDatabase();
    db.delete(Constants.TRICKS_TABLE, Constants.TRICK_ID + " ?= ", new String[]{ String.valueOf(trick_id)});
    db.close();
}
但现在我需要根据两个约束条件trick_id和opn_id删除行

如何使用“where子句”删除与作为参数传递给delete方法的相同opn_id和trick_id匹配的行

public void deleteTrick (int trick_id,int opn_id) {
    SQLiteDatabase db = this.getWritableDatabase();
    db.delete(Constants.TRICKS_TABLE, Constants.TRICK_ID + " ?= " WHERE Constants.OPN_ID + " ?= ", new String[]{ String.valueOf(trick_id), String.valueOf(opn_id)};
    db.close();
}
删除:

db.delete(TableName,ID_that_you_want_to_delete + " =?" , new String[]{your id that given for delete }) ;



db.delete(Constants.TRICKS_TABLE, Constants.TRICK_ID + " =? ",new String[]{String.valueOf(opn_id)})

您可以像这样使用delete方法

  public void deleteTrick() {
    SQLiteDatabase db = this.getWritableDatabase();
    db.delete(Constants.TRICKS_TABLE, null, null);
    db.close();
}

在这个代码中,第一个null是你的WHERE子句,第二个null是你的WHERE。如果我的答案正确,你可以接受。快乐编码:)

当你使用“id=和otherId=”时,它不起作用吗?你的delete命令看起来是随机生成的(或者被所谓的“Markov ape”破坏,如果你知道我在说什么的话)。