由于我如何在表格中獲取信息，我需要建立一個(gè)聯(lián)合來顯示交叉引用。Tablebrand1 是表的名稱。2 和 3Tablebrand1CODE    CROSSREFERENCEA15     SAMSUNG A10A16     SAMSUNG A20A63     SAMSUNG A30Tablebrand2CODE    CROSSREFERENCEX63     SAMSUNG A10X64     SAMSUNG A20X65     SAMSUNG A30Tablebrand3CODE    CROSSREFERENCETOP99   SAMSUNG A10TOP98   SAMSUNG A20TOP97   SAMSUNG A30SQL查詢(SELECT "Tablebrand1" AS `brand`, CROSSREFERENCE as `code`    FROM Tablebrand1 WHERE CROSSREFERENCE = 'SAMSUNG A10')UNION (SELECT "Tablebrand2" AS `brand`, CROSSREFERENCE as `code`    FROM Tablebrand2 WHERE CROSSREFERENCE = 'SAMSUNG A10')UNION(SELECT "Tablebrand3" AS `brand`, CROSSREFERENCE as `code`    FROM Tablebrand3 WHERE CROSSREFERENCE = 'SAMSUNG A10');查詢正確地顯示了它，但是當(dāng)嘗試在 php 中顯示它時(shí)，我收到一個(gè)來自查詢的錯誤，因?yàn)槿绻倚薷乃赃M(jìn)行更簡單的查詢，它會正確地顯示數(shù)據(jù)。<?phpecho "<table style='border: solid 1px black;'>";echo "<tr><th>brand</th><th>code</th></tr>";class TableRows extends RecursiveIteratorIterator {  function __construct($it) {    parent::__construct($it, self::LEAVES_ONLY);  }  function current() {    return "<td style='width:150px;border:1px solid black;'>" . parent::current(). "</td>";  }  function beginChildren() {    echo "<tr>";  }  function endChildren() {    echo "</tr>" . "\n";  }}$servername = "localhost";$username = "user";$password = "pass";$dbname = "dbname";try {  $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);  $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);  $stmt = $conn->prepare("((SELECT "Tablebrand1" AS `brand`, CROSSREFERENCE as `code`    FROM Tablebrand1 WHERE CROSSREFERENCE = 'SAMSUNG A10')UNION (SELECT "Tablebrand2" AS `brand`, CROSSREFERENCE as `code`    FROM Tablebrand2 WHERE CROSSREFERENCE = 'SAMSUNG A10')UNION我的代碼如下，我對 php 和 union 的了解很少。