Question

我想创建一个数据库查询字符串，应该是这样的，

(SELECT COUNT(*) FROM states WHERE country_id = 121 ) + 
(SELECT COUNT(*) FROM cities WHERE country_id = 121 ) + 
(SELECT COUNT(*) FROM areas WHERE country_id = 121) + 
(SELECT COUNT(*) FROM states WHERE country_id = 122) + 
(SELECT COUNT(*) FROM cities WHERE country_id = 122) + 
(SELECT COUNT(*) FROM areas WHERE country_id = 122)

因为我使用了这样的代码。

$id = array('121','122');
$table = array('states','cities','areas');
$loopCount = count($id) * count($table);
$queryString = array();
for($i=0;$i<$loopCount;$i++)
{
    $queryString[] = "(SELECT COUNT(*) FROM $table[$i] WHERE country_id = $id[$i] ) + ";
}

我知道上面的代码是完全错误的，实现代码以获得所需结果的正确方法是什么？

更新：

我想得到总数来自country_id而不是个人，可以存在于1到3个表中，以下查询似乎对我来说很好，但如果你有更好的解决方案，请告诉我< / p>

SELECT( 
      (SELECT COUNT(*) FROM states WHERE country_id IN(121,122)) + 
      (SELECT COUNT(*) FROM cities WHERE country_id IN(121,122) ) +
      (SELECT COUNT(*) FROM areas WHERE country_id IN(121,122) )
      );

Answer 1

您可以使用JOIN：

执行此操作

$id = array('121', '122');

jon_darkstar的数量（谢谢！）：

"SELECT COUNT(DISTINCT(state_id)) + COUNT(DISTINCT(city_id)) + COUNT(DISTINCT(area_id)) as desired_sum

使用相同的cities

加入另一个名为country_id的表格

LEFT JOIN cities ON states.country_id = cities.country_id

使用相同的areas

加入另一个名为country_id的表格

LEFT JOIN areas ON states.country_id = areas.country_id

过滤您在数组中提供的country_id的结果，使用名为implode()的php函数对其进行内嵌。

WHERE states.country_id IN (" . implode(', ', $id) . ")"

或者，在$cid将成为你的country_id的foreach循环中（请参阅下面的jon_darkstar评论）

WHERE states.country_id = " . $cid . ""

一个查询：

"SELECT * FROM states, 
        COUNT(DISTINCT(state_id)) + 
        COUNT(DISTINCT(city_id)) + 
        COUNT(DISTINCT(area_id)) AS desired_sum
        JOIN `cities` ON states.country_id = cities.country_id
        JOIN `areas` ON states.country_id = areas.country_id
        WHERE states.country_id IN (" . implode(', ', $id) . ")";

Answer 2

试试这段代码：

<?php
$a=array();
$a[]=array('states','121');
$a[]=array('cities','121');
$a[]=array('areas','121');
$a[]=array('states','122');
$a[]=array('cities','122');
$a[]=array('areas','122');

$r=array();
foreach($a as $v)
    $r[]='(SELECT COUNT(*) FROM '.$v[0].' WHERE country_id = '.$v[1].' )';
$r=implode(' + ',$r);
?>

但我强烈建议不要采用这种方法来获得你想要做的事情。

Answer 3

注意到最后几条评论，试一试

$countryCodes = array(121, 122);
$countryCodeString = implode($countryCodes, ', ');

$sql = 
  "SELECT COUNT(DISTINCT(S.id)) + COUNT(DISTINCT(C.id)) + COUNT(DISTINCT(A.id)) as desired_sum 
   FROM states S
        JOIN cities C ON S.country_id = C.country_id
        JOIN areas A ON A.country_id = S.country_id
   WHERE s.country_id in ($countryCodes)";

$res = mysql_query($sql);
$arr = mysql_fetch_assoc($res);
$val = $arr['desired_sum'];

因此，这将给出一个整数，即$countryCodes中指定的任何国家/地区内任何“单位”（城市，州或地区）的总数。它确实假设每个国家至少有一个州。（即 - 如果国家X中有城市和区域但没有州，则不计算这些城市和地区）。还假设states，cities，areas的主键是state_id，city_id和area_id（不仅仅是id ）

Answer 4

您可以减少查询次数，但我不认为尝试UNION会更好。我的建议是改进这样的查询：

SELECT COUNT(*) AS number FROM states WHERE country_id IN (121, 122) GROUP BY country_id

现在每个表有一个查询。在php方面你可以这样做：

$id = array('121','122');
$table = array('states','cities','areas');
foreach($table as $tableName) {
    $queryString[] = "SELECT COUNT(*) AS number FROM {$tableName} WHERE country_id IN (". implode(',', $id) .") GROUP BY country_id"
}

使用循环将此数组转换为字符串的逻辑

4 个答案: