i created a form which store clients passports in my database. i want to retrive the images but its has not been giving me the images on the screen. instead , it write long alphabeth of different symbols .
below is my codes , pls help me out. thanks
$sql = "SELECT * FROM `file` WHERE id = 8";
$mq = mysqli_query($dbconnect, $sql) or die ("not working query");
$row = mysqli_fetch_array($mq) or die("line 44 not working");
$s=$row['data'];
echo $row['data'];
echo '<img src="'.$s.'" alt="HTML5 Icon" style="width:128px;height:128px">';
what is wrong with my codes please.
thanks
Because you say you see lots of characters instead of the filename, I assume that you don't store a path to that image (then you could use src as you tried to use), but the image content itself and that it is jpg (you can change to png or whatever type you have there). If so, use data URI syntax for src:
$src = "data:image/jpg;base64,".base64_encode($row['data']);
echo '<img src="'.$src.'" alt="HTML5 Icon" style="width:128px;height:128px">';
It is called data URI - you can read more about it for example here: https://en.wikipedia.org/wiki/Data_URI_scheme
Related
I am making a social media and I'm echoing profile pictures this way (code beneath), but I need to rescale the images that the users upload before echoing them. Either that or resize them before they get uploaded to the server.
Does anyone have some suggestions?
$stmt = $conn->prepare("SELECT ID, fname, profilePicture from users");
$stmt->execute();
$out = "";
while($row = $stmt->fetch()){
$picture = "<img src='profilePictures/$row[2]'>";
$out .="<a href='profile.php?user=$row[0]'> $picture $row[1] </a> <br/>";
}
echo $out;
The most common PHP library for this type of thing is imagemagick https://www.imagemagick.org/script/index.php. It should be fairly easy to do the manipulations with that.
When the users upload the photos you can save a smaller version as a thumbnail or profile pic (as well as the original if you'll need it later).
There's a nice library called Intervention Image, which can both utilize PHP's ImageMagick- and GD-extensions as drivers. Here's a very basic example:
Image::make('path/to/image.jpg')
->resize(320, 240)
->save('path/to/ouput/image.jpg');
I'm very new to development, and I'm trying to retrieve an image from an SQL server over an odbc connection using the below:
<?php
require ('connect.inc.php');
$sql = "SELECt image FROM test.dbo.clients WHERE id = 1";
$sql_exec = odbc_exec($con, $sql);
if($sql_array = odbc_fetch_array($sql_exec)){
$image = base64_encode($sql_array['image']);
header('Content-type: image/jpeg');
echo "<img src=".$image."/>";
}
?>
The issue is that I'm getting an icon showing a broken image.
The query is correct as when I change it to the code below, it returns this string instead of the image:
Q2hyeXNhbnRoZW11bS5qcGc=
<?php
require ('connect.inc.php');
$sql = "SELECt image FROM test.dbo.clients WHERE id = 1";
$sql_exec = odbc_exec($con, $sql);
if($sql_array = odbc_fetch_array($sql_exec)){
$image = base64_encode($sql_array['image']);
echo $image;
}
?>
I know the code is not correct and might be vulenrable to SQL injection, however I'd appreciate if you can help me retrieve the image.
Many thanks in advance,
J
If you decode the base64 string you get, then you'll see that the decoded data is Chrysanthemum.jpg. This is just the filename of the image, not the image data.
You need to either store the image in the database (not the filename) or add some code to read the image from the filesystem.
BTW, Content-Type image/jpeg requires that the data is the raw image, but your content (<img src=...> ...</img>) is an HTML fragment.
This seems like such a simple thing to do but for some reason, I can't get this to work.
I have a database with vehicle data stored in it. (Used Cars (I'm developing a car dealership website)).
I successfully display results from the database without images.
The images for each record aren't stored in the database, instead they're dumped on the server in a directory and those images are only referenced in the database.
If I echo the image name out it works fine, and if I echo the actual image out, the path is correct if you look at the image info. But in the info it states that the image is of text. i don't know how to change this.
Please find some of the code below.
<?php
$dbName = "F:/Domains/autodeal/autodeal.co.za/wwwroot/newsite/db/savvyautoweb.mdb";
// Throws an error if the database cannot be found
if (!file_exists($dbName)) {
die("Could not find database file.");
}
// Connects to the database
// Assumes there is no username or password
$conn = odbc_connect("Driver={Microsoft Access Driver (*.mdb)};Dbq=$dbName", '', '');
//this is selecting individual records
$selected_id = intval($_GET['Id']);
//this is the query
$sql = "SELECT Id, Make, Model, Year, Price, SpecialPrice, Branch, StockNO, MainPic FROM Vehicle WHERE Id = $selected_id";
// Runs the query above in the table
$rs = odbc_exec($conn, $sql);
$id = odbc_result($rs, Id);
$make = odbc_result($rs, Make);
$model = odbc_result($rs, Model);
$mainPic = odbc_result($rs, MainPic);
//this is a failsafe for when there are no images for a specific record
$no_image = "<a href='db/Nopic.gif' data-lightbox='nopic' title='No Image Available'><img src='db/Nopic.gif' /></a>";
//This successfully displays the name of the image referenced in the database
$main_pic = "<img src='db/vehicleImages/'" .$mainPic. "/>";
//this is supposed to display the actual
echo $main_pic . "<br />";
echo $no_image;
odbc_free_result($rs);
odbc_close($conn);
// This message is displayed if the query has an error in it
if (!$rs) {
exit("There is an error in the SQL!");
}
?>
Any help in the regard would be greatly appreciated.
Thank you. :)
you should use quotations around the attributes of html objects (it looks like you might be breaking things via your method):
yours:
<a href=db/Nopic.gif data-lightbox=nopic title=No Image Available><img src=db/Nopic.gif /></a>
correct:
<a href='db/Nopic.gif' data-lightbox='nopic' title='No Image Available'><img src='db/Nopic.gif' /></a>
whether or not this fixes your problem will be determined by what you come back with :p
User forward slash in front of your image paths. This way you can be sure the image path always starts from the root folder. The image path is relative to the location of the executed php script.
For example if your script.php file is in the folder /root/some_folder/script.php and you use the image path db/vehicleImages/image.jpg the script is starting the path from the same folder where it is itself which results in a path like this /root/some_folder/db/vehicleImages/image.jpg.
If you use forward slash in front of the path like this /db/vehicleImages/image.jpg it tells the script to start from the root folder like so /root/db/vehicleImages/image.jpg.
I think this is the case with your script - you are giving it the wrong path which results in a file not found.
As good practice, I'm only storing my image's link in my database, the questios are:
How should I store my image's link? (let's say it's on c:)
c://image.jpg?
Which peace of PHP code should I use to display that image?
I'm only displaying the path, what should I do to display the image?
Can I use this:
$query = "SELECT ImageURL from WhateverTable";
$result = mysql_query($query) or die(mysql_error());
$Image = mysql_fetch_row($result);
echo "<img src='$Image[0]' alt='This is an image'>";
Thank you all lads
You only want to store the relative path, not the absolute path, as linking to something like
<img src="/var/www/vhosts/website.com/images/file.jpg">
would return a 404 on any real website. store your files in the database via a relative path (/images/file.jpg) or by only the filename if they are all in the same directory.
alternatively, you can learn MongoDB and it allows you to actually store files in the database itself.
I would strongly suggest that you use PDO instead.
Use relative URLs to your image folder in case you need to move them one day.
Here is an example.
// relative to your public webroot
$publicImageUrl = '/images/in/here';
// Pull up some record, maybe of a product
$select = 'SELECT imageFilename FROM products WHERE id = 2332';
$results = mysql_query($select);
if(!$results) {
// issue with query. deal with it here
} else {
if( mysql_num_rows($result) ) {
// record not found. deal with it here
}
$row = mysql_fetch_array($result);
$imageSrc = $publicImageUrl . '/' . $row['imageFilename'];
}
And then your HTML would be as follows
<img src="<?php echo $imageSrc; ?>" />
use PDO for php <-> mysql connection
post mysql query output
I have a site that currently uses images on a file server. The images appear on a page where the user can drag and drop each as is needed. This is done with jQuery and the images are enclosed in a list. Each image is pretty standard:
<img src='//network_path/image.png' height='80px'>
Now however I need to reference images stored as a BLOB in an Oracle database (no choice on this, so not a merit discussion). I have no problem retrieving the BLOB and displaying on it's own using:
$sql = "SELECT image FROM images WHERE image_id = 123";
$stid = oci_parse($conn, $sql);
oci_execute($stid);
$row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS);
$img = $row['IMAGE']->load();
header("Content-type: image/jpeg");
print $img;
But I need to [efficiently] get that image as the src attribute of the img tag. I tried imagecreatefromstring() but that just returns the image in the browser, ignoring the other html. I looked at data uri, but the IE8 size limit rules that out.
So now I am kind of stuck. My searches keep coming up with using a src attribute that loads another page that contains the image. But I need the image itself to actually show on the page. (Note: I say image, meaning at least one image but as many as eight on a page).
Any help would be greatly appreciated.
Well, you can do a few things. You can either make a page that will render the image
<img src="image.php?id=123" />
That image.php page would have this:
$sql = "SELECT image FROM images WHERE image_id = " . (int) $_GET['id'];
$stid = oci_parse($conn, $sql);
oci_execute($stid);
$row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS);
if (!$row) {
header('Status: 404 Not Found');
} else {
$img = $row['IMAGE']->load();
header("Content-type: image/jpeg");
print $img;
}
Or, you could base64 encode it into the src (note, not all browsers handle this well):
<img src="data:image/jpeg;base64,<?php echo base64_encode($img); ?>" />
But I need to [efficiently] get that image as the src attribute of the img tag
As Byron already says, the accepted and right way is to output the blob in an independent image resource, and to embed that using an img tag. It's the only good way. You can use data: URIs but they
fatten your HTML code
don't work in IE < 8 and are limited to 32 KB in IE 8,
expand the data volume by 33%, and
take away the brower's possibility to cache the image resource.
Almost never a good option.
The normal way to do this is with a <img src=/path/to/script?id=32> field. It will show up on the page not as a link. What is the problem with this? Do you want to embed the image data into HTML?
To make it more efficient, you can implement some type of caching ie, write the image data to a file and do a header(location...) if you find it instead of querying the db again. Also the browser caching headers should be set so the browser doesn't download the image if it has it cached locally.
You may try this :
$img = $row['IMAGE']->load();
print('<img src="data:image/png;base64,'.base64_encode($img).'" />');
<?php
if(isset($_POST['']))//get the id
$roll_no=$_POST[''];
$conn = oci_connect("", "", "");//DB connection
$query = 'SELECT image FROM TABLE where id=:id';
$stmt = oci_parse ($conn, $query);
oci_bind_by_name($stmt, ':id', $id);
oci_execute($stmt);
$arr = oci_fetch_array($stmt, OCI_ASSOC);
$result = $arr['image']->load();
header("Content-type: image/JPEG");
echo $result;
oci_close($conn);
?>