Question

我有这两个表：

表1

Username            Password
aswilliams40        pantsftw
liamvdheuvel        qwascoolzuiop
michaelsmith01      ilovejessica
rzajac              toocoolforscool
pooh_sweety_70      legendarymythic

表2

SELECT Username FROM Table1, Table2
    WHERE Table1.Username = Table2.Username AND
    Table2.Password LIKE '%cool%' AND
    Table1.Rate IS NOT NULL AND
    Table1.Opinion = 'Negative';

我想找到密码中包含＆＃39; cool＆＃39;在它的某个地方，如果用户对项目进行评级，那么该用户也必须始终对项目留下负面意见。未评级的项目无所谓。

我尝试将查询放在一起，但它确实不起作用：/

buildscript {

    repositories {
        jcenter()
    }
    dependencies {
        classpath 'com.android.tools.build:gradle:1.5.0'

        // NOTE: Do not place your application dependencies here; they belong
        // in the individual module build.gradle files
    }
}
allprojects {
    repositories {
        jcenter()
    }
}

android {
    buildToolsVersion '23.0.0'
}
dependencies {
}

所以基本上我希望唯一的输出是＆＃39; liamvdheuvel＆＃39;

Answer 1

gc.RgbFgColor = new Gdk.Color(204,204,204);

为了优化，你需要从table1中删除用户名（给这个表一个正确的名称，并将userID添加到table2（也正确命名）。）

SELECT DISTINCT Table1.Username 
FROM Table1
LEFT JOIN Table2
ON table1.Username=table2.Username
Where
Table2.Password LIKE '%cool%' AND
Table1.Rate IS NOT NULL AND
Table1.Opinion = 'Negative';

表1

Table2
------

UserId    **Username**         **Password**
1         aswilliams40        pantsftw
2         liamvdheuvel        qwascoolzuiop
3         michaelsmith01      ilovejessica
4         rzajac              toocoolforscool
5         pooh_sweety_70      legendarymythic

Answer 2

SELECT Table2.Username 
FROM   Table1, 
       Table2 
WHERE  Table1.Username = Table2.Username 
       AND Table2.Password LIKE '%cool%' 
       AND Table1.Rate IS NOT NULL 
       AND Table1.Opinion = 'Negative';

问题是您没有指定要获取“用户名”的表格。

如果您不想重复使用用户名，请添加DISTINCT：

SELECT DISTINCT Table2.Username 
FROM   Table1, 
       Table2 
WHERE  Table1.Username = Table2.Username 
       AND Table2.Password LIKE '%cool%' 
       AND Table1.Rate IS NOT NULL 
       AND Table1.Opinion = 'Negative';

如果您不希望用户有负面和正面意见，请尝试以下操作：

SELECT DISTINCT Table2.Username 
FROM   Table1, 
       Table2 
WHERE  Table1.Username = Table2.Username 
       AND Table2.Password LIKE '%cool%' 
       AND Table1.Rate IS NOT NULL 
       AND Table1.Opinion = 'Negative' 
       AND Table2.Username NOT IN (SELECT Table1.Username 
                                   FROM   Table1 
                                   WHERE  Table1.Opinion = 'Positive');

SQL新手查询问题

2 个答案: