使用搜索结果对html表进行多个sql查询

lokaqttq  于 2021-06-25  发布在  Mysql
关注(0)|答案(1)|浏览(318)

我发现了类似的主题,但是许多结果使用mysqli\u multi\u查询,这是我希望避免的,因为将来会实现用户生成的查询。
目前,我正在使用下面的php实现我想要的结果,但是我有一种感觉,我缺少一种更有效的方法。我还担心并发查询和这样一个复杂的进程占用连接的可能性。尽管任何时候的userload都不应该大于20。
这里是代码,任何批评或见解欣赏(我仍然在学习php,因此我的代码在这方面可能是狗屎!):
表模式(这需要很好地工作—从本质上说,我需要开发最佳结构):

CREATE TABLE `EAM`(
    `EAM_ID` INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
    `EAM_IPADDR` VARCHAR(15) NOT NULL, 
    `EAM_PORT` INT(5) NOT NULL,
    `EAM_STATE` VARCHAR(6) NOT NULL);

CREATE TABLE `ACTIVE`(
    `EAM_ID` INT NOT NULL,
    `ACTIVE_STATUS` VARCHAR(25) NOT NULL,
    `ACTIVE_TIME` TIME DEFAULT NULL,
    FOREIGN KEY (EAM_ID) REFERENCES EAM(EAM_ID));

CREATE TABLE `MAP_IMG` (
  `MAP_IMG_BLDG` varchar(25) NOT NULL,
  `MAP_IMG_ROOM` varchar(25) NOT NULL,
  `MAP_IMG_X` int(8) DEFAULT NULL,
  `MAP_IMG_Y` int(8) DEFAULT NULL,
  `MAP_IMG_ROOM_STATUS` varchar(10) NOT NULL DEFAULT 'NOTCLEAR');

CREATE TABLE `LOCATION`(
    `LOCATION_ROOM` VARCHAR(25) NOT NULL PRIMARY KEY,
    `LOCATION_BLDG` VARCHAR(25) NOT NULL,
    `EAM_ID` INT NOT NULL,
    `LOCATION_COMMENT` VARCHAR(250) DEFAULT NULL,
    FOREIGN KEY (EAM_ID) REFERENCES EAM(EAM_ID));
$dbQuery = "SELECT EAM_ID FROM LOCATION WHERE LOCATION_BLDG = 'LQ1'"; //TODO: Eventually make this 'LQ1' a variable for page selected.
$dbQueryResult = mysqli_query($dbConnection, $dbQuery) OR DIE("Bad Query: $dbQuery");

    echo "<table class='table table-striped'><thead>"; 
    echo "<tr><th>BUILDING</th>";
    echo "<th>ROOM NUMBER</th>";
    echo "<th>ROOM STATUS</th>";
    echo "<th>LAST UPDATE</th>";
    echo "<th>EAM IP ADDRESS</th>";
    echo "<th>EAM PORT</th></tr></thead>";

  while($row = mysqli_fetch_assoc($dbQueryResult)) {

      $eamID = $row['EAM_ID']; //Assign EAM_ID from initial query to $eamID for use throughout subsequent queries.

      $dbq2 = "SELECT LOCATION_BLDG, LOCATION_ROOM FROM LOCATION WHERE EAM_ID = $eamID";
      $dbqr2 = mysqli_query($dbConnection, $dbq2) OR DIE("Bad Query: $dbq2");

      while($r2 = mysqli_fetch_assoc($dbqr2)) {

        echo "<tr><td>{$r2['LOCATION_BLDG']}</td>";
        echo "<td>{$r2['LOCATION_ROOM']}</td>";

      }

      $dbq2 = "SELECT ACTIVE_STATUS, ACTIVE_TIME FROM ACTIVE WHERE EAM_ID = $eamID";
      $dbqr2 = mysqli_query($dbConnection, $dbq2) OR DIE("Bad Query: $dbq2");

       while($r2 = mysqli_fetch_assoc($dbqr2)) {

         echo "<td>{$r2['ACTIVE_STATUS']}</td>";
         echo "<td>{$r2['ACTIVE_TIME']}</td>";

       }

      $dbq2 = "SELECT EAM_IPADDR, EAM_PORT FROM EAM WHERE EAM_ID = $eamID";
      $dbqr2 = mysqli_query($dbConnection, $dbq2) OR DIE("Bad Query: $dbq2");

       while($r2 = mysqli_fetch_assoc($dbqr2)) {

        echo "<td>{$r2['EAM_IPADDR']}</td>";
        echo "<td>{$r2['EAM_PORT']}</td></tr>";

       }
  }
9bfwbjaz

9bfwbjaz1#

不要在循环中执行查询来获取相关数据,而是在顶部使用一个查询来返回数据。

SELECT 
    a.`EAM_ID`,
    a.`LOCATION_BLDG`,
    a.`LOCATION_ROOM`,
    b.`ACTIVE_STATUS`,
    b.`ACTIVE_TIME`
FROM `LOCATION` a
LEFT JOIN `ACTIVE` b
ON a.`EAM_ID` = b.`EAM_ID`
LEFT JOIN `EAM` c
ON a.`EAM_ID` = c.`EAM_ID`
WHERE a.`LOCATION_BLDG` = 'LQ1'

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