使用一個 MySQL 查詢的結果作為另一個 MySQL 查詢中的變量

Question

我正在嘗試根據團隊名稱顯示結果 HTML 表

我能夠回顯正確的團隊名稱，但無法在我的第二個查詢中將其用於變量中 我無法找出我在這里做錯了什么，為什么查詢結果不可見。 我需要更改代碼中的某些內容嗎？

<?php
include_once("connection.php");

$sql = "SELECT TeamName FROM `superuser` WHERE id = '303016'";
$queryRecords2 = mysqli_query($conn, $sql) or die("error to fetch employees data");

while ($row2 = $queryRecords2->fetch_assoc()) {
    echo $row2['TeamName']."<br>";
}

if(isset($_POST['search']))
{
    $valueToSearch = $row2['TeamName'];
    $valueToSearch2 = $_POST['valueToSearch2'];
    $valueToSearch3 = $_POST['valueToSearch3'];
    $sql = "SELECT * FROM `dailydata` WHERE TeamName = '".$valueToSearch."' and Date BETWEEN '".$valueToSearch2."' AND '".$valueToSearch3."'";
    $queryRecords = mysqli_query($conn, $sql) or die("error to fetch employees data");
}
else {
    $sql = "SELECT * FROM `dailydata` WHERE TeamName = ''";
    $queryRecords = mysqli_query($conn, $sql) or die("error to fetch employees data");
}

?>          

Answer 1

出於安全目的，您應該使用准備好的語句，或者更好的是您應該使用 PDO 而不是 mysqli。

要解決此問題，只需在循環內分配 $valueToSearch 變量的值。

<?php
include_once("connection.php");

$sql = "SELECT TeamName FROM `superuser` WHERE id = '303016'";
$queryRecords2 = mysqli_query($conn, $sql) or die("error to fetch employees data");

$valueToSearch;
while ($row2 = $queryRecords2->fetch_assoc()) {
    $valueToSearch = $row2['TeamName'];
}

if(isset($_POST['search']))
{
    $valueToSearch2 = $_POST['valueToSearch2'];
    $valueToSearch3 = $_POST['valueToSearch3'];
    $sql = "SELECT * FROM `dailydata` WHERE TeamName = '".$valueToSearch."' and Date BETWEEN '".$valueToSearch2."' AND '".$valueToSearch3."'";
    $queryRecords = mysqli_query($conn, $sql) or die("error to fetch employees data");
}
else {
    $sql = "SELECT * FROM `dailydata` WHERE TeamName = ''";
    $queryRecords = mysqli_query($conn, $sql) or die("error to fetch employees data");
}

?>