有没有可能在SQL查询中的SELECT CASE的THEN之后编写具有select语句的查询?逻辑 例如何在SQL查询中SELECT CASE的THEN之后执行SELECT语句?
SELECT Id, CASE WHEN level=2 THEN
(select something here)
else 0 end as Grade FROM CLASS
有没有可能在SQL查询中的SELECT CASE的THEN之后编写具有select语句的查询?逻辑 例如何在SQL查询中SELECT CASE的THEN之后执行SELECT语句?
SELECT Id, CASE WHEN level=2 THEN
(select something here)
else 0 end as Grade FROM CLASS
你可以宣布第二选择为一个变量。
例如:
DECLARE @var as int
SET @var = (SELECT COUNT(*) FROM table)
SELECT
Id,
CASE WHEN level=2 THEN @var else 0 end as Grade
FROM CLASS
基础上,你可以运行东西沿着这些路线的PHP文件:创建数据库连接
开始:
<?php
$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
这if
语句检查表test
如果它不存在,则将其丢弃或创建表test
与现场id
这是一个INT
if (!$mysqli->query("DROP TABLE IF EXISTS test") || !$mysqli->query("CREATE TABLE test(id INT)")) {
echo "Table creation failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
然后如果条件尝试你在这里运行查询;这里他们有三个疑问。
首先 - 计算表中(其将等于0)
二总列数 - 值添加到表
第三 - 再次计数的总的行,其这次将输出“1”
// this is what you may be asking?
// they add to their sql query
$sql = "SELECT COUNT(*) AS _num FROM test; ";
$sql.= "INSERT INTO test(id) VALUES (1); ";
$sql.= "SELECT COUNT(*) AS _num FROM test; ";
确保查询已正确构建
if (!$mysqli->multi_query($sql)) {
echo "Multi query failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
然后运行该代码
do {
if ($res = $mysqli->store_result()) {
var_dump($res->fetch_all(MYSQLI_ASSOC)); // this will gather all records from the Database
$res->free();
}
} while ($mysqli->more_results() && $mysqli->next_result()); // You can find the links for documentation for more_results() and next_result() below
?>
的文档more_results()和next_results()
注意$sql
,你可以看到他们只是创建将一起运行,如果一切都是真的多个查询。
它会输出这样的:
array(1) {
[0]=>
array(1) {
["_num"]=>
string(1) "0"
}
}
array(1) {
[0]=>
array(1) {
["_num"]=>
string(1) "1"
}
}
请注意,这不是我的代码,我是从here
谢谢WJS,那很好 –