这里是我的代码我试图使用ajax从数据库中获取图像，但它无法正常工作，请将我解雇。当我尝试使用锚标记获取图像时图像上传正常工作，但它在另一页上显示我的图像。但我想在窗口加载时在同一页面上。
if(filter_has_var(INPUT_GET, "image_id") !== false && filter_input(INPUT_GET, 'image_id', FILTER_VALIDATE_INT) !== false)
{
$image_id = filter_input(INPUT_GET, "image_id", FILTER_SANITIZE_NUMBER_INT);
try {
$dbh = new PDO("mysql:host=localhost;dbname=flipbook", 'root', '');
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "SELECT image, image_type FROM testblob WHERE user_id=$image_id";
/*** prepare the sql ***/
$stmt = $dbh->prepare($sql);
$stmt->execute();
/*** set the fetch mode to associative array ***/
$stmt->setFetchMode(PDO::FETCH_ASSOC);
$array = $stmt->fetch();
if(sizeof($array) == 2)
{
header("Content-type: ".$array['image_type']);
echo $array['image'];
}
else
{
throw new Exception("Out of bounds Error");
}
}
catch(PDOException $e)
{
echo $e->getMessage();
}
catch(Exception $e)
{
echo $e->getMessage();
}
}
else
{
echo 'Please use a real id number';
}
}
<body onload="showUserProfilePic(<?php echo $_SESSION['current_user_id'];?>)">
<div id="txtHint">Child Picture will be listed here.</div>
<script>
function showUserProfilePic(str) {
//alert(str);
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
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","getImage.php?image_id="+str,true);
xmlhttp.send();
}
}
答案 0 :(得分：0)
尝试使用
$sql = "SELECT image, image_type FROM testblob WHERE user_id='" . $image_id . "';
你的标签在哪里？