[英]Room error: no viable alternative at input?
I am trying to update the table using @Query annotation of room library, below is my code ( In Dao interface ) : 我正在尝试使用@Query注释的房间库来更新表,下面是我的代码(在Dao界面中):
@Query("UPDATE table_name SET table_name.col1 = :val1 WHERE table_name.col2 = :val2")
void updateValue(long val1, long val2);
Complete error string as below : 完整的错误字符串如下:
Error:(11, 10) error: no viable alternative at input 'UPDATE table_name SET table_name.'
Here is entity class : 这是实体类:
@Entity(tableName = "table_name")
public class SampleTable {
@PrimaryKey
@ColumnInfo(name = "_id")
private Long Id;
@ColumnInfo(name = "col1")
private Long column1;
@ColumnInfo(name = "col2")
private Long column2;
public Long getId() {
return Id;
}
public void setId(Long id) {
Id = id;
}
public Long getColumn1() {
return column1;
}
public void setColumn1(Long column1) {
this.column1 = column1;
}
public Long getColumn2() {
return column2;
}
public void setColumn2(Long column2) {
this.column2 = column2;
}
}
What's wrong with my code ? 我的代码出了什么问题?
This error could also happen in this case: 在这种情况下也可能发生此错误:
If you are passing a list, don't forget to add the parentheses enclosing your list variable, otherwise you will get the same error: 如果要传递列表,请不要忘记添加括在列表变量中的括号,否则会出现相同的错误:
@Query("select name from mytable where name in (:myList)")
LiveData<List<String>> findNames(List<String> myList);
Try changing your statement to: 尝试将您的陈述更改为:
UPDATE table_name SET col1 = :val1 WHERE col2 = :val2.
The error message makes it feel like Room is tripping over the prefix. 错误消息使得感觉Room正在跳过前缀。
This is a bug in Room, at least through
1.0.0-alpha8
.
这是Room中的一个错误,至少通过
1.0.0-alpha8
。
Track
this issue to see when it gets fixed.
跟踪
此问题以查看何时修复。
That's actually not valid SQLite syntax, as it turns out . 事实证明 ,这实际上不是有效的SQLite语法。 Table prefixes go on columns in
SELECT
statements, not UPDATE
statements . 表前缀在
SELECT
语句中的列上, 而不是在UPDATE
语句中 。
In my case a message was error: no viable alternative at input 'is'
(and a query name was mentioned). 在我的情况下,一条消息是
error: no viable alternative at input 'is'
(并且提到了查询名称)。 I found that the query didn't have necessary spaces between words like here: 我发现查询在这里的单词之间没有必要的空格:
"from car" + // Add a space here.
"where brand_id is null" +
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