Question

我一直试图找出一种基于初始文本从数据库中提取信息的正确方法在html下拉菜单中选择。

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);
?>

Answer 1

尝试更改

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 }

到

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，因为它们会在您的客户端为您提供更多访问权限。例如发送一个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就像：

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。如果你想使用jQuery

，你的AJAX / DOM更改也会变得简单得多

PHP，AJAX从HTML中自动填充输入字段选择？

1 个答案: