我试图使用php在mysql上创建一个表。我发送一个包含列名和实际数据的数组。第一个位置是列名，其余的是实际数据。我设法创建了表和列。虽然我不能用数据填充表。这是我使用的PHP：

<?php
// Assuming your MySQL database connection details
$DATABASE_HOST = "xxxx";
$DATABASE_USER = "xxxx";
$DATABASE_PASS = "xxxx";
$DATABASE_NAME = "xxxx";
$conn = mysqli_connect($DATABASE_HOST, $DATABASE_USER, $DATABASE_PASS, $DATABASE_NAME);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
$data = json_decode($_POST['data'], true);
$columnName = array_shift($data); // Remove and return the first element
$columnData = $data; // The rest of the array is the data to be inserted
// Build the SQL query dynamically based on the number of data elements
$placeholders = implode("', '", array_fill(0, count($columnData), "?"));
$query = "CREATE TABLE IF NOT EXISTS tmp_transactions (rownumber INT AUTO_INCREMENT PRIMARY KEY);";
$result = $conn->query($query);
if ($result) {
    echo "Table created successfully";
} else {
    echo "Error creating table: " . $conn->error;
    $queryDrop = "DROP TABLE tmp_transactions";
    $result = $conn->query($queryDrop);
}
$queryAlter = "ALTER TABLE tmp_transactions ADD COLUMN $columnName VARCHAR(255);";
$result = $conn->query($queryAlter);
$sqlInsert = "INSERT INTO tmp_transactions ($columnName) VALUES ($placeholders);";
$stmtInsert = $conn->prepare($sqlInsert);
$stmtInsert->bind_param(str_repeat('s', count($columnData)), ...$columnData);
$resultInsert = $stmtInsert->execute();
if ($resultInsert) {
    echo "Data inserted successfully";
} else {
    echo "Error inserting data: " . $stmtInsert->error;
}
$stmtInsert->close();
$conn->close();
?>

字符串
输入

["Amount","-43.85","200.00","4925.00","-210.00","-62.00","1638.00","25435.00","-12.50","-74.00"] 
or 
["transactiondate","2023-12-22","2023-12-22","2023-12-22","2023-12-21","2023-12-21","2023-12-22","2023-12-22","2023-12-18","2023-12-18"]

型
尝试插入原始字符串而不是$placeholder没有成功。