Question

我在每次点击下拉列表时创建了一个下拉列表，另一个下拉列表显示在[u] drop1 [/ u]数据库中。以下是相同的代码：

dropdown1.php

  <html>
    <head>
    <script>
    function showUser(str)
    {
    if (str=="")
      {
      document.getElementById("txtHint").innerHTML="";
      return;
      } 
    if (window.XMLHttpRequest)
      {// code for IE7+, Firefox, Chrome, Opera, Safari
      xmlhttp=new XMLHttpRequest();
      }
    else
      {// code for IE6, IE5
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
    xmlhttp.onreadystatechange=function()
      {
      if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
        document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
        }
      }
    xmlhttp.open("GET","drop2.php?q="+str,true);
    xmlhttp.send();
    }
    </script>
    </head>
    <body>

    <form>
    <select name="users" onchange="showUser(this.value)">
    <option value="">Select a plan:</option>
    <option value="1">Dedicated</option>
    <option value="2">VPS</option>

    </select>
    </form>
    <br>
    <div id="txtHint"><b>Person info will be listed here.</b></div>

    </body>
    </html>

**drop2.php:**

<?php
$q = intval($_GET['q']);

$con = mysqli_connect('localhost','root','','test');
if (!$con)
  {
  die('Could not connect: ' . mysqli_error($con));
  }

mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM drop WHERE id = '".$q."'";

$result = mysqli_query($con,$sql);

/*echo "<table border='1'>
<tr>

<th>plan1</th>
<th>plan2</th>

</tr>";*/
  echo "Plan";
  echo "<select>";

while($row = mysqli_fetch_array($result))
  {

   echo "<option>" . $row['plan1'] . "</option>";
   echo "<option>" . $row['plan2'] . "</option>";

  }
//echo "</table>";
echo "</select>";
mysqli_close($con);
?>

上面的代码非常适合显示正确的下拉列表。现在我想在另一个数据库中插入第二个下拉列表值。我完全陷入了困境。：（

Answer 1

与插入相同的数据库没有什么不同，除了您将连接如下：

$con2 = mysqli_connect('localhostOrServerName','root','','otherDatabaseName');

然后通过$ con2：

引用第二个数据库

$result2 = mysqli_query($con2,$insert_sql);

Answer 2

你可以这样做：

 $needInsert = false;
 $query = 'INSERT INTO table(plan1, plan2) VALUES ';
 while($row = mysqli_fetch_array($result))
 {
     $needInsert = true;
     $query .= "('{$row['plan1']}','{$row['plan2']}'),";

     echo "<option>" . $row['plan1'] . "</option>";
     echo "<option>" . $row['plan2'] . "</option>";
 }

 if ($needInsert)
 {
     $query = substr($query, 0, -1);
     mysqli_select_db($con,"another_db");
     mysqli_query($con,$query);
 }

我建议您使用PDO扩展而不是mysqli，因为它会在插入数据库之前自动清理数据。否则，您必须像这样清除$row['plan1']和$row['plan2']：

 mysql_real_escape_string($row['plan1']);
 mysql_real_escape_string($row['plan2']);

使用数据库中的数据填充另一个下拉列表并插入另一个数据库

2 个答案: