Question

我有两张桌子。

First tabel called:(data):
---------------------------------------- 
id    link    number    isik    status
---------------------------------------- 
1    /link    78788     56677     55

Second table called:(test)
----------------------------------------
id     kood     status
----------------------------------------
1      56677     111

两张桌子中唯一相似的东西是isik和kood

如何从第一个表格中获取isik(First table) = kood(Second table)所有的行？

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT data * FROM data INNER JOIN test ON data.isik = test.kood";
$result = $conn->query($sql);

if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc()) {
        echo "Numb: " . $row["number"]. " - Name: " . $row["isik"]. " " . $row["link"]. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?>

但我得到0结果

Answer 1

您可以使用内部联接将表连接在一起，并仅从第一个表中选择行。这将仅选择第二个表具有isik的匹配值的行。

SELECT `data`.* FROM `data` INNER JOIN `test` ON `data`.`isik` = `test`.`kood`

Answer 2

SELECT *
FROM First a
WHERE EXISTS (SELECT 1
          FROM Second b
          WHERE a.isik = b.isik);

如何在另一个表中选择具有匹配行的行

2 个答案: