Question

我正在尝试从数据库中创建一个下拉菜单。我现在什么也没回来。对于菜单，我需要选择一个对象，然后对其运行SQL查询。此查询将填充表格并将是动态的。这是代码，请帮忙。

<html>
<?php
include('header.php');
?>

  <h1>Chemicals Search</h1>

  <br/ >
  <br/ >

  </head>


  <h1>
    <center>Chemical Search</center>
  </h1>
  <form action="chemicals.php" method="post">
    <label>Search By Product:</label>
    <?php
    //making the chemical array
    $con = mysqli_connect("...");

    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $p = mysqli_query($con, "SELECT product_name FROM product");

    $products = mysqli_fetch_array($p);

    echo '<select name="product">';
    foreach ($product as $key => $value) {
      echo "<option value=\"$key\"> $value</option>\n";
    }
    echo '</select>';
    echo "<br>";
    ?>
  </form>

Answer 1

您的查询仅选择产品名称对其进行一些更改

 $p = mysqli_query($con,"SELECT product_id,product_name FROM product");
 echo'<select name="product">';
 if(mysqli_num_rows($p) > 0)
 {
  while($products =mysqli_fetch_array($p))
  {
    echo"<option value='".$products['product_id']."'>".$products['product_name']."</option>\n";       
  }
 }
echo '</select>';

echo"<br>";

我相信这会有效，每次都适合我

我试图用PHP中的数据库列填充下拉菜单

1 个答案: