如何从mysql数据库检索图像文件路径并使用php显示图像

我正在尝试使用php从mysql数据库检索图像文件路径，并将其显示为一种表单，该表单随后将用于将图像发送到我们的android应用程序。我到处都有搜索，发现一个看起来像我的问题，但尚未解决，因此我试图寻找另一种帮助。

这是我上传图片的php表格：inSTAGram.php

<?php

require('admin.config.inc.php');

if(isset($_POST['upload'])){
$image_name = $_FILES['image']['name'];
$image_type = $_FILES['image']['type'];
$image_size = $_FILES['image']['size'];
$image_tmp_name = $_FILES['image']['tmp_name'];

$path = "/home/stagcon2/public_html/StagConnect/admin/pictures/$image_name";

if($image_name==''){
echo "Don't just click! select an image please .";
exit();
}
else{
move_uploaded_file($image_tmp_name, $path);
$mysql_path = $path."/".$image_name;
$query = "INSERT INTO `inSTAGram`(`image_name`,`path`) VALUES ('$image_name','$mysql_path')";

$query_params = array(
    ':image_name' => $image_name,
    ':mysql_path' => $path,
    );

//execute query
try {
    $stmt   = $db->prepare($query);
    $result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
    // For testing, you could use a die and message. 
    //die("Failed to run query: " . $ex->getMessage());

    //or just use this use this one:
    $response["success"] = 0;
    $response["message"] = "Database Error. Couldn't Upload Image!";
    die(json_encode($response));
}

$response["success"] = 1;
$response["message"] = "Image Uploaded Succesfully!";
echo json_encode($response);
}
}   
?>

<form action="inSTAGram.php" method="post" enctype="multipart/form-data">
<input type="file" name="image" >
<input type="submit" name="upload" value="Upload" >
</form>

这个工作正好！

所以这是我的用于显示图像的php表单；inSTAGramDisplay.php

<?php
require("admin.config.inc.php");

//initial query
$query = "Select * FROM inSTAGram";

try {
$stmt   = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error!";
die(json_encode($response));
}

// Finally, we can retrieve all of the found rows into an array using fetchAll 
$rows = $stmt->fetchAll();


if ($rows) {
$response["success"] = 1;
$response["message"] = "Image Available!";
$response["images"]   = array();

foreach ($rows as $row) {
    $rows             = array();
    $rows['image'] = 'http://www.stagconnect.com/StagConnect/admin/pictures?image_id=' . $rows['image_id'];
    $rows[] = $rows;        

    //update our repsonse JSON data
    array_push($response["images"], $image);
}

// echoing JSON response
echo json_encode($rows);


} else {
$response["success"] = 0;
$response["message"] = "No Image Available!";
die(json_encode($response));
}

?>

这次，这只显示图像路径而不是图像，我要做的是显示图像本身。我只是不知道该怎么做。任何帮助都可以。提前致谢！