使用一个 MySQL 查询的结果作为另一个 MySQL 查询中的变量

Question

我正在尝试根据团队名称显示结果 HTML 表

我能够回显正确的团队名称，但无法在我的第二个查询中将其用于变量中 我无法找出我在这里做错了什么，为什么查询结果不可见。 我需要更改代码中的某些内容吗？

<?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

出于安全目的，您应该使用准备好的语句，或者更好的是您应该使用 PDO 而不是 mysqli。

要解决此问题，只需在循环内分配 $valueToSearch 变量的值。

<?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");
}

?>