PHP、SQL根據下拉選擇菜單從數據庫填充信息

Question

我設法將 SQL 數據庫中的數據拉入下拉選擇菜單並添加了一個提交按鈕。 但是，當用戶單擊提交時，我正在努力從其他（甚至相同）表中填充來自 DB（例如，ID）的更多數據。

我希望用戶能夠從下拉菜單中選擇 customerName 並提交。 單擊“提交”按鈕時，應從數據庫中填充並顯示鏈接到特定 customerName（例如-Rob）的其他（例如年齡、ID）詳細信息。 請幫忙。 我是 php 的新手，花了幾個小時在這個愚蠢的事情上。 我只被允許使用 php、sql 和 html。

非常感謝。

dropdown.php - 原始版本（無效）

<?php

include "db.php"; // Database connection 


$sql="SELECT customerName FROM customers"; 


echo "<select name=customers value=''>Customer Name</option>"; 

foreach ($conn->query($sql) as $row){

echo "<option value=$row[customerName]>$row[customerName]</option>"; 



}

echo "</select>";

?>

<form id= method="post" action = "iNeedHelpWithThisFile.php">
<input type="submit" value="SUBMIT"/>
</form>
</body>

dropdown.php - 更新版本

<?php
?>


<form id= method="post" action = "iNeedHelpWithThisFile.php">
<select name="customerName">Customer Name</option> 
<?php
include "db.php"; 
$sql="SELECT `customerName` FROM `customers`"; 
foreach ($conn->query($sql) as $row){
  echo "<option value=\"".$row['customerName']."\">".$row['customerName']."</option>"; 
}
?>

</select>
<input type="submit" value="SUBMIT"/>
</form>

數據庫文件

<?php
// to connect database 
define('DB_NAME', 'clients');
    define('DB_USER', 'root');
    define('DB_PASSWORD', '');
    define('DB_HOST', 'localhost');
    $conn = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or
        die ("Cannot connect");
        echo  "Connected to Clients";
?>  

iNeedHelpWithThisFile.php
echo 語句-年齡、名稱、id 沒有被執行

<?php

echo "<pre>";
print_r($_REQUEST);
echo "</pre>";

include('db.php');
echo "<br />";

$customerName = $_POST['submit'];


if(array_key_exists('customerName', $_REQUEST ))
{

    // not sure if there's an issue with intval..?
    $customerName = intval($_REQUEST['customerName']);

    echo("Customer: $customerName <br />");

if ($customerName)
{

    $sql = "SELECT * FROM customers where customerName= ?";  //
    // $conn -defined in db.php
    if($stmt = mysqli_prepare($conn, $sql))
    {

        mysqli_stmt_bind_param($stmt, "i", $customerName);
        mysqli_stmt_execute($stmt);
        $queryresult = mysqli_stmt_get_result($stmt);
    }
    else {
          echo "Could not fetch results!";
           }


    if ($queryresult) 
    {
        if ($currentrow = mysqli_fetch_assoc($queryresult)) {

            $name = $currentrow['customerName'];
            $ID =  $currentrow['customerID'];
            $age  = $currentrow['age'];


           // these statements are not being executed
            echo "Customer Name: $name <br />";
            echo "Customer ID: $ID <br />";
            echo "Age: $age <br />";
    }
}

    else {
        echo "<p>Customer not in the database</p>";
        }
}
}
?>

Answer 1

總而言之 ：

<form id= method="post" action = "iNeedHelpWithThisFile.php">
<select name="customers">Customer Name</option> 
<?php
include "db.php"; 
$sql="SELECT `customerName` FROM `customers`"; 
foreach ($conn->query($sql) as $row){
    echo "<option value=\"".$row['customerName']."\">".$row['customerName']."</option>"; 
}
?>

</select>
<input type="submit" value="SUBMIT"/>
</form>

或者，如果您喜歡簡短的 php 標簽：

<php
include "db.php"; 
$sql="SELECT `customerName` FROM `customers`"; 
?>

<form id="main" method="post" action = "iNeedHelpWithThisFile.php">
<select name="customers">Customer Name</option>

<?php foreach ($conn->query($sql) as $row){ ?>
  <option value="<?=$row['customerName'];?>"><?=$row['customerName'];?></option>

<?php } ?>

</select>
<input type="submit" value="SUBMIT"/>
</form>

如果您還想同時提交年齡和ID以及客戶姓名，您可以像這樣：

<php
include "db.php"; 
$sql="SELECT `age`, `ID`, `customerName` FROM `customers`"; 
?>

<form id= method="post" action = "iNeedHelpWithThisFile.php">
<select name="customers">Customer Name</option>

<?php foreach ($conn->query($sql) as $row){
 $conc=base64_encode($row['age']."|".$row['ID']."|".$row['customerName']); ?>
  <option value="<?=$conc;?>"><?=$row['customerName'];?></option>
<?php } ?>

</select>
<input type="submit" value="SUBMIT"/>
</form>

在iNeedHelpWithThisFile.php 中，你可以這樣

$c = strip_tags(htmlspecialchars($_POST['customers']);
$c = base64_decode($c);

$name = explode("|",$c)[2];
$ID =  explode("|",$c)[1];
$age  = explode("|",$c)[0];

// these statements are not being executed
echo "Customer Name: $name <br />";
echo "Customer ID: $ID <br />";
echo "Age: $age <br />";