4

我一直在尝试根据在 html 下拉菜单上选择的初始文本找出从数据库中提取信息的正确方法。

这是我的代码：

<html>
<head>
</head>
<script src="testjs.js"></script>
<?php
    $host = "";
    $username = ""; 
    $password = "";
    $database = "";
    mysql_connect($host, $username, $password);
    mysql_select_db($database);
?>
<body>

<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<?php
    $Query = mysql_query("SELECT * FROM population");
    while ($Rows = mysql_fetch_array($Query))
    {
        $ID = $Rows['ID'];
        $Pop = $Rows['Pop'];
        $UniqueID = $Rows['uid'];
        echo "<option value=\"$UniqueID\">$Pop</option>";
    }
?>
</select>
</form>
<br>
<p>DB ID <input type="text" id="ids" name="ID" ></p>
<p>Population <input type="text" id="content" name="contet" ></p>
<p>Unique ID <input type="text" id="uid" name="uid" ></p>
<div id="GetInformation"><b>Person info will be listed here.</b></div>

</body>
</html>

test.js 包含：

function showUser(str)
{
if (str=="")
  {
  document.getElementById("GetInformation").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("GetInformation").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}

获取用户包含：

<?php
$q=$_GET["q"];

$con = mysql_connect('', '', '');
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("DropDown", $con);

$sql="SELECT * FROM population WHERE uid = '".$q."'";

$result = mysql_query($sql);

while($row = mysql_fetch_array($result))
  {
    $ID = $row['ID'];
    $Pop = $row['Pop'];
    $UID = $row['uid'];
?>
<script type="text/javascript">
var ids = '<?php echo json_encode($ID); ?>';
var content = '<?php echo json_encode($Pop); ?>';
var uid = '<?php echo json_encode($UID); ?>';
</script>
 <?php }
mysql_close($con);
?>
4

1 回答 1

1

尝试改变

while($row = mysql_fetch_array($result))
  {
    $ID = $row['ID'];
    $Pop = $row['Pop'];
    $UID = $row['uid'];
?>
<script type="text/javascript">
var ids = '<?php echo json_encode($ID); ?>';
var content = '<?php echo json_encode($Pop); ?>';
var uid = '<?php echo json_encode($UID); ?>';
</script>
 <?php }

至

while($row = mysql_fetch_array($result))
{
    $ID = $row['ID'];
    $Pop = $row['Pop'];
    $UID = $row['uid'];
    echo $ID . ' - ' . $Pop . ' - ' . $UID;
}

这应该有效。但是有更好的方法，因为它们稍后会在您的客户端为您提供更多访问权限。例如发回一个 JSON 对象，一个简单的例子是：

$info = array();
while($row = mysql_fetch_array($result))
{
    $ID = $row['ID'];
    $Pop = $row['Pop'];
    $UID = $row['uid'];

    $info[] = array( 'id' => $ID, 'pop' => $Pop, 'uid' => $UID );
}
echo json_encode($info);

你的JS会是这样的：

if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
    var data = JSON.parse(xmlhttp.responseText);
    for(var i=0;i<data.length;i++) 
    {
      document.getElementById("GetInformation").innerHTML += data[i].id + ' - ' + data[i].pop + ' - ' + data[i].uid;
    }
}

注意：如果您使用的浏览器不包含 JSON 库，则需要加载http://www.json.org/js.html。如果您想使用 jQuery，您的 AJAX/DOM 更改也会变得更加简单

于 2012-12-09T08:46:00.887 回答