我正在使用 AJAX 和 PHP 从数据库中检索数据。在数据库中,其中一列包含图像文件夹的路径。我将路径的值保存在名为 $folder 的 PHP 变量中。这可以在 getuser.php 代码中看到。
我希望此变量在 one.php 中可见/可用,以便可以填充使用此变量的图像。我该怎么做。我也尝试过包含php,但没有用。
获取用户.php
<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'san', '123');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("holidayNet", $con);
$sql="SELECT * FROM image WHERE id = '".$q."'";
$result = mysql_query($sql);
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Picture</th>
</tr>";
while($row = mysql_fetch_array($result))
{
$folder = $row['FirstName'];
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td>" . $row['Hometown'] . "</td>";
/*echo "<td>" . $row['Job'] . "</td>";*/
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
一个.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<script type="text/javascript">
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","getuser.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<option value="1">Sn Qb</option>
<option value="2">Lois Griffin</option>
<option value="3">Glenn Quagmire</option>
<option value="4">Joseph Swanson</option>
</select>
</form>
<br />
<div id="txtHint"><b>Person info will be listed here.</b></div><br />
<img src="<?php echo $folder;?>/pic1.jpg" />
<img src="<?php echo $folder;?>/pic2.jpg" />
<img src="<?php echo $folder;?>/pic3.jpg" />
<img src="<?php echo $folder;?>/pic4.jpg" />
</body>
</html>