如何从mysql数据库检索图像文件路径并使用php显示图像

Question

I'm trying to retrieve an image file path from mysql database using php and display it to a form that will then be used to send the images to our android app. I have search around everywhere and i found one that looks like my problem but it has not been solved so I'm trying to find another help.

here is my php form for uploading images: 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>

this one works just right!

so here is my php form for the displaying of image; 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));
}

?>

this time, this one shows the image path and not the image, what i wanted to do is display the images itself. I just don't have any idea what the right thing to do here. any help will do. Thanks in advance!

Answer 1

Just echo an image tag and put the image variable as it source attribute:

<?php
//I hard coded the array but pull the images id from the db
$rows  = array();
$rows = array("070415togepi.jpg",
              "1384904_681730675185076_523601772_n.jpg",
              "1388517_681730668518410_508160046_n.jpg",
              "1394889_681730678518409_1155435015_n.jpg", 
              "385-jirachi-g.jpg"); 

foreach ($rows as $value){
 $rows['img'] = 'http://www.stagconnect.com/StagConnect/admin/pictures/'.$value;
 echo "<img src='".$rows['img']."' />";
}

?>

Answer 2

Without testing it, I'd say that one of your issues may be that you need to enclose the location of the image:

$rows['image'] = 'url(http://www.stagconnect.com/StagConnect/admin/pictures?image_id=' . $rows['image_id'] . ')';

Give that a go.