Question

我有一个问题：

SELECT DISTINCT phone
  FROM contacts
  WHERE (phone='123456' OR phone='456789' OR phone='789789')
     AND deleted=0 AND user=1

这将显示联系人表中的所有行，并且不会删除它们。如何返回联系人表中未找到的值？

注意： 我的表格中没有找到的字段。这就是我要归还的内容
例如：

|   Phone   | deleted | found |
|-----------|---------|-------|
|  123456   |    0    |  YES  |
|  456789   |    1    |  YES  |
|  789789   |    0    |   NO  | <---

谢谢，尼科斯

Answer 1

最简单的解决方案：检查PHP上的结果并验证不存在的手机。

修改

如果您想将其作为一个查询运行，您可以创建此语句：

SELECT t.phone FROM ( SELECT 123456 as phone UNION ALL SELECT 456789 as phone UNION ALL SELECT 789789 as phone ) as t LEFT JOIN contacts as c ON t.phone = c.phone WHERE c.phone IS NULL

但我不确定它能否快速起作用

Answer 2

在PHP中执行此操作不是更容易吗？

// The list of phone numbers to search
$listAll = array('123456', '456789', '789789', );

// Use data from $listAll to compose this query:
$query = "
  SELECT DISTINCT phone
  FROM contacts
  WHERE phone IN ('123456', '456789', '789789')
     AND deleted=0 AND user=1
";

// Run the query, get the list of phone numbers in $listFound
// Let's suppose it produces:
//     $listFound = array('123456', '456789');

// Get the list of phone numbers that are missing from table
$listMissing = array_diff($listAll, $listFound);

Answer 3

试试这个：

    select DISTINCT phone 
    from contacts 
    WHERE (phone='123456' OR phone='456789' OR phone='789789') 
    AND deleted = 0 
    AND user = 1
    AND found = 0

Answer 4

试试这个：

SELECT distinct phone 
FROM contacts 
WHERE phone NOT IN (123456,456789) AND
deleted = 0 AND 
user=1

MySQL查询，从表中选择，并返回未找到的值

4 个答案: