SQL查詢PHP中的SQL查詢

Question

我正在嘗試為我的網站建立一個列表視圖。 我試圖做到這一點，所以我有更多的概述，然后只有一長串所有數據。

$conn = new mysqli($db_servername, $db_username, $db_password, $db_dbname);
if ($conn->connect_error) { } 

$sql = "SELECT * FROM Home_Rooms";
$result = $conn->query($sql);

if ($result->num_rows > 0) {

    echo "
<div class=\"container\"><hr></div>
<div class=\"container-fluid\">
  <div class=\"row\">
    <div class=\"col-sm-1\">
    </div>
    <div class=\"col-sm-10\">
      <div class=\"panel-group\" id=\"accordion\">";

    while($row = $result->fetch_assoc()) {
        $room = $row["name"];
        echo "
        <div class=\"panel panel-info\">
          <div class=\"panel-heading\">
            <h4 class=\"panel-title\"><a data-toggle=\"collapse\" data-parent=\"#accordion\" href=\"#collapse" .$row["id"]."\">" .$row["name"]."</a></h4>
          </div>
          <div id=\"collapse" .$row["id"]."\" class=\"panel-collapse collapse\">
          <div class=\"panel-body\">";

$conn = new mysqli($db_servername, $db_username, $db_password, $db_dbname);
if ($conn->connect_error) { } 

$sql = "SELECT * FROM Home_Devices WHERE room='$room'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while($row_room = $result->fetch_assoc()) {
        echo "
      <div class=\"media\">
        <div class=\"media-left\">
          <img src=\"/image/pictures/" .$row_room["type"].".png\" class=\"media-object\" style=\"width:60px\">
        </div>
        <div class=\"media-body\">
          <h4 class=\"media-heading\"><a href=\"/dashboard/" .$row_room["sensor_id"]."\">" .$row_room["room"]." - " .$row_room["name"]."</a></h4>
          <p>Sensor ID: " .$row_room["sensor_id"]."</p>
        </div>
      </div>";
    }
} else {
    echo "<p>No devices found, you can add devices <a href=\"/settings\">here</a>.</p>";
}

      echo"</div>
          </div>
        </div>";
    }

    echo "
      </div>
    </div>
  </div>
</div>";

} else {
    echo "<p>No rooms found, you can add rooms <a href=\"/settings\">here</a>.</p>";
}
$conn->close();

但這並沒有加載所有房間，只有第一個房間。 有人可以幫我解決這個問題嗎？ 這是否有可能或者我做錯了什么？

Answer 1

您不必在while語句中定義$ conn。 它已經在代碼開頭定義。 另外，為什么要為房間數據使用與Home_Rooms相同的變量$ result？ 第二次嘗試使用其他變量，因為它可能會使while語句搞砸。

Answer 2

您的Result變量為Overwrite

更改該變量。

$sql1 = "SELECT * FROM Home_Devices WHERE room='$room'";
$result1 = $conn->query($sql1);

並更改各個變量。

謝謝

Answer 3

我有工作

我不得不更改所有不同的數組。

$conn = new mysqli($db_servername, $db_username, $db_password, $db_dbname);
if ($conn->connect_error) { } 

$sql = "SELECT * FROM Home_Rooms";
$result = $conn->query($sql);

if ($result->num_rows > 0) {

    echo "
<div class=\"container\"><hr></div>
<div class=\"container-fluid\">
  <div class=\"row\">
    <div class=\"col-sm-1\">
    </div>
    <div class=\"col-sm-10\">
      <div class=\"panel-group\" id=\"accordion\">";

    while($row = $result->fetch_assoc()) {
        $room = $row["name"];
        echo "
        <div class=\"panel panel-info\">
          <div class=\"panel-heading\">
            <h4 class=\"panel-title\"><a data-toggle=\"collapse\" data-parent=\"#accordion\" href=\"#collapse" .$row["id"]."\">" .$row["name"]."</a></h4>
          </div>
          <div id=\"collapse" .$row["id"]."\" class=\"panel-collapse collapse\">
          <div class=\"panel-body\">";

$conn_2 = new mysqli($db_servername, $db_username, $db_password, $db_dbname);
if ($conn_2->connect_error) { } 

$sql_2 = "SELECT * FROM Home_Devices WHERE room='$room'";
$result_2 = $conn_2->query($sql_2);

if ($result_2->num_rows > 0) {
    while($row_room = $result_2->fetch_assoc()) {
        echo "
      <div class=\"media\">
        <div class=\"media-left\">
          <img src=\"/image/pictures/" .$row_room["type"].".png\" class=\"media-object\" style=\"width:60px\">
        </div>
        <div class=\"media-body\">
          <h4 class=\"media-heading\"><a href=\"/dashboard/" .$row_room["sensor_id"]."\">" .$row_room["room"]." - " .$row_room["name"]."</a></h4>
          <p>Sensor ID: " .$row_room["sensor_id"]."</p>
        </div>
      </div>";
    }
} else {
    echo "<p>No devices found, you can add devices <a href=\"/settings\">here</a>.</p>";
}

      echo"</div>
          </div>
        </div>";
    }

    echo "
      </div>
    </div>
  </div>
</div>";

} else {
    echo "<p>No rooms found, you can add rooms <a href=\"/settings\">here</a>.</p>";
}
$conn->close();