如何从MySQL数据库映像文件路径和使用PHP显示图像 [英] How to retrieve image file paths from mysql database and display the images using php

问题描述

我试图使用PHP来获取MySQL数据库映像文件的路径，并显示给将被用来将图像发送到我们​​的Andr​​oid应用程序的形式。我已搜索周围到处都是，我发现了一个看起来像我的问题，但一直没有得到解决，所以我试图找到另外的帮助。

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.

下面是我的PHP形式上传图片： inSTAGram.php

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>

该一期工程刚刚好！

所以这里是我的PHP形式图像的显示; inSTAGramDisplay.php

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));
}

?>

这个时候，这一个显示图片的路径，而不是形象，我想要做的是显示图像本身。我没有任何想法，什么是正确的事情在这里。任何帮助都行。在此先感谢！

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!

推荐答案

只是呼应的图像标签，并把图像的变量，因为它的源属性：

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']."' />";
}

?>

这篇关于如何从MySQL数据库映像文件路径和使用PHP显示图像的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！