在数据库中找不到 Android SQLite ForeignKey 列答案

【问题标题】：Android SQLite ForeignKey Column not found in Database在数据库中找不到 Android SQLite ForeignKey 列
【发布时间】：2017-01-08 20:06:33
【问题描述】：

在程序成功运行并从 Android 设备监视器中提取 .db 文件后，我可以找到每个表的所有表和列，但 Foreign key 的列除外。有人知道为什么会这样吗？我的Query 有问题吗？这是我的代码：

 private static final int DATABASE_VERSION = 1;
private static final String DATABASE_NAME = "datacollector.db";

private static final String Table_Survey = "SURVEY";
private static final String Survey_KEY_ID = "id";
private static final String Survey_KEY_NAME = "Survey_Name";
private static final String KEY_Table_Name = "Table_Name";


private static final String Table_formDetails = "FormDetails";
private static final String formDetails_KEY_ID = "id";
private static final String formDetails_field_Type="Field_Type";
private static final String formDetails_label = "Label";
private static final String formDetails_Options = "Options";

public DbHelper(Context context) {
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}

@Override
public void onCreate(SQLiteDatabase db) {

    String CREATE_SURVEY_TABLE = "CREATE TABLE " + Table_Survey + "("
            + Survey_KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT," + Survey_KEY_NAME + " VARCHAR(30),"
            + KEY_Table_Name + " VARCHAR(50)" + ")";
    db.execSQL(CREATE_SURVEY_TABLE);


    String CREATE_FORM_DETAILS_TABLE = "CREATE TABLE " + Table_formDetails + "("
            + formDetails_KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT," + formDetails_field_Type + " VARCHAR(30),"
            + formDetails_Options + " VARCHAR(30),"
            + formDetails_label + " VARCHAR(30),"
            + " FOREIGN KEY ("+Survey_KEY_ID+") REFERENCES "+Table_Survey+"("+Survey_KEY_ID+"));";
    db.execSQL(CREATE_FORM_DETAILS_TABLE);



}

【问题讨论】：

标签： android sqlite android-studio foreign-key-relationship android-database

【解决方案1】：

您正在将同一个表列映射到外键。现在试试这个。

 private static final int DATABASE_VERSION = 1;
 private static final String DATABASE_NAME = "datacollector.db";
private static final String Table_Survey = "SURVEY";
private static final String Survey_KEY_ID = "id";
private static final String Survey_KEY_NAME = "Survey_Name";
private static final String KEY_Table_Name = "Table_Name";
private static final String Table_formDetails = "FormDetails";
private static final String formDetails_KEY_ID = "id";
private static final String formDetails_field_Type="Field_Type";
private static final String formDetails_label = "Label";
private static final String formDetails_Options = "Options";

public DbHelper(Context context) {
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}
@Override
public void onCreate(SQLiteDatabase db) {
    String CREATE_SURVEY_TABLE = "CREATE TABLE " + Table_Survey + "("
            + Survey_KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT," + Survey_KEY_NAME + " VARCHAR(30),"
            + KEY_Table_Name + " VARCHAR(50)" + ")";
    db.execSQL(CREATE_SURVEY_TABLE);
    String CREATE_FORM_DETAILS_TABLE = "CREATE TABLE " + Table_formDetails + "("
            + formDetails_KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT," + formDetails_field_Type + " VARCHAR(30),"
            + formDetails_Options + " VARCHAR(30),"
            + formDetails_label + " VARCHAR(30),"
             + forign_Survey_KEY_ID + "INTEGER DEFAULT 0,"
            + " FOREIGN KEY ("+forign_Survey_KEY_ID+") REFERENCES "+Table_Survey+"("+Survey_KEY_ID+"));";
    db.execSQL(CREATE_FORM_DETAILS_TABLE);
}

【讨论】：

forign_Survey_KEY_ID，你能解释一下这是什么吗？我这里出错了还是没有初始化？
您需要有一列来存储超级表的 id，因此您必须创建一列来存储它们。我为您的错误编辑了代码，未初始化 .立即查看

【解决方案2】：

因为你没有在CREATE_FORM_DETAILS_TABLE中添加survey_key_id

String CREATE_FORM_DETAILS_TABLE = "CREATE TABLE " + Table_formDetails + "("
            + formDetails_KEY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT," + formDetails_field_Type + " VARCHAR(30),"
            + formDetails_Options + " VARCHAR(30),"
            + formDetails_label + " VARCHAR(30),"
            + Survey_KEY_ID + " INTEGER,"
            + " FOREIGN KEY ("+Survey_KEY_ID+") REFERENCES "+Table_Survey+"("+Survey_KEY_ID+"));";
    db.execSQL(CREATE_FORM_DETAILS_TABLE);

【讨论】：

出现错误“列名重复：id（代码 1）：，编译时：CREATE TABLE FormDetails(id INTEGER PRIMARY KEY AUTOINCREMENT,Field_Type VARCHAR(30),Options VARCHAR(30),Label VARCHAR (30),id INTEGER, FOREIGN KEY (id) REFERENCES SURVEY(id));''
+ Survey_KEY_ID + " INTEGER," 在此处和外键中更改名称 ("+Survey_KEY_ID+")
只需更改声明 private static final String formDetails_KEY_ID = "id"; to private static final String formDetails_KEY_ID = "detail_id";