sql中如何匹配两个表的列

Question

我有这个代码$db = \Config\Database::connect();$query = $db->query("select * from g WHERE g_status = '0' ORDER BY g_date ASC Limit 10;");foreach ($query->getResult() as $g) {    $g_amount = $g->g_amount; }$query2 = $db->query("select * from g WHERE p_status = '0' ORDER BY p_date ASC Limit 10;");foreach ($query2->getResult() as $p) {    $p_amount = $p->p_amount; }if($p_amount ==  $g_amount){echo "do something";}else{echo "No match";}在这里，我尝试在表 g 和表 p 之间进行匹配...如果表 g 中的任何列 == 表 p 中的任何列，无论列数如何，请执行某些操作，但它总是回显“不匹配”我设置了“Limit 10”，以防表中的行数较多，它只会与“ordering”命令匹配前 10 行。请我需要一些帮助。

Answer 1

首先，以数组形式获取数据

$db = \Config\Database::connect();

$query = $db->query("select * from g WHERE g_status = '0' ORDER BY g_date ASC Limit 10;");

$g_results = $query->getResult('array'); 

$g_amounts = array_column($g_results,'g_amount');

$query2 = $db->query("select * from g WHERE p_status = '0' ORDER BY p_date ASC Limit 10;");

$p_results = $query2->getResult('array'); 

$p_amounts = array_column($p_results,'p_amount');

foreach(array_intersect($g_amounts,$p_amounts) as $amount){

   echo "do something";

}

Answer 2

为什么不使用 aJOIN看看它是否返回一些东西？不确定我的语法是否正确，我不经常进行 JOIN：

SELECT * FROM g g_table JOIN p p_table ON g_table.g_amount = p_table.p_amount WHERE g_table.status = '0' AND p_table.status = '0'