How to display photo from database using PHP?

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Hi,

I am having a hard time figuring out how to display photos from the database using php. So, far I can successfully add text and photos into the database, in addition to displaying the text, however I am just not able to display my photos from the database. I have a feeling that it has to do with the fact that I am not able to add any of the uploaded photos into my images folder or the path to this folder is not correct. Please see my code below, and it would be great if I could get some help:

Here is my uploadtest.php file:

<?php

if(isset($_POST['upload'])){//if upload button is pressed

//the path to store uploaded images
$target = "images/" .basename($_FILES['image']['name']);

//connect to database
$conn = mysqli_connect('127.0.0.1', 'root', '', 'photos');
// get all the submitted data from the form
$image=$_FILES['image']['name'];
$text =$_POST['text'];

$sql ="INSERT INTO images (image, text) VALUES ('$image', '$text')";
mysqli_query($conn, $sql); //stores data into database table images

//lets move uploaded image into folder
if(move_uploaded_file($_FILES['image']['tmp_name']. $target, "images/")){
$msg="Image uploaded successfully.";
}else{
$msg="There was a problem uploading the image";
}

}

?>

<!DOCTYPE html>

<title>Image upload

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">

<?php
$conn = mysqli_connect('127.0.0.1', 'root', '', 'photos');
$sql="SELECT * FROM images";
$result = mysqli_query($conn,$sql);

while($row = mysqli_fetch_array($result)){
echo "

";
echo "<img src='images/'".$row['image']."'>";
echo "

".$row['text']."

";
echo "

";

}
?>

And here is my css file:

#content{
width:50%;
margin:20px auto;
border:1px solid #cbcbcb;
}
form{
width:50%;
margin:20px auto;
}
form div{
margin-top:5px;
}
#img_div{
width:80%;
padding:5px;
margin:15px auto;
border:1px solid #cbcbcb;
}
#img_div:after{
content:"";
display:block;
clear:both;
}
img{
float:left;
margin:5px;
width:300px;
height:140px
}

Thanks again!

What I have tried:

I have been watching many youtube videos but am still struggling with this problem.

Posted 2-Aug-17 17:33pm

Member 13342809

Updated 2-Aug-17 19:37pm

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GrpSMK · Answer 1 · 2017-08-02T19:37:00

Solution 1

To get image from database you have to try like this.

<?php

  $id = $_GET['id'];
  // do some validation here to ensure id is safe

  $link = mysql_connect("localhost", "root", "");
  mysql_select_db("dvddb");
  $sql = "SELECT dvdimage FROM dvd WHERE id=$id";
  $result = mysql_query("$sql");
  $row = mysql_fetch_assoc($result);
  mysql_close($link);

  header("Content-type: image/jpeg");
  echo $row['dvdimage'];
?>

Posted 2-Aug-17 19:37pm

GrpSMK

Comments

Member 13342809 3-Aug-17 20:59pm

Thank you for the information, however I am now getting this error message:

The image "http://localhost/Email/uploadtest.php" can't be displayed because it contains errors.

Mohibur Rashid 4-Aug-17 6:02am

Change content type to text and read the error

Member 13342809 4-Aug-17 9:55am

Thank you for the response. Here are the error messages that I am getting:

Notice: Undefined index: id in C:\xampp\htdocs\Email\images\uploadtest.php on line 68

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Email\images\uploadtest.php on line 75

Notice: Undefined variable: link in C:\xampp\htdocs\Email\images\uploadtest.php on line 76

Warning: mysql_close() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Email\images\uploadtest.php on line 76

This is my code so far:

<?php

if(isset($_POST['upload'])){//if upload button is pressed

//$ext_type = array('gif','jpg','jpe','jpeg','png');
//the path to store uploaded images
$target = "images/" .basename($_FILES['image']['name']);

//$ext_type = array('gif','jpg','jpe','jpeg','png');//checks to see if file is one of these

//connect to database
$conn = mysqli_connect('127.0.0.1', 'root', '', 'photos');
//get all the submitted data from the form
$image=$_FILES['image']['name'];
$text =$_POST['text'];

$sql ="INSERT INTO images (image, text) VALUES ('$image', '$text')";
mysqli_query($conn, $sql); //stores data into database table images

//lets move uploaded image into folder
if(move_uploaded_file($_FILES['image']['tmp_name']. $target, "images/")){
$msg="Image uploaded successfully.";
}else{
$msg="There was a problem uploading the image";
}

//$sql="SELECT * FROM images";
//$result = mysqli_query($conn, $sql);
//while($row= mysqli_fetch_array($result)){
//echo "<img src = 'Image/".$row['image'].'" />';
//echo ' ';
// $res= mysql("SELECT * FROM images");
// $res = mysqli_query($conn, sql);
//echo "";
// echo "";

//while($row = mysqli_fetch_array($res)){
// echo "";
// echo '< height='100' width='100'/>;

}

//if(isset($_POST['sumbit2'])){
// $res = mysql_query("SELECT * FROM image");
//echo "";
//echo "";
//while($row = mysql_fetch_array($res)){
//echo "";
//echo '<img scr = data:image/jpeg; base_64'.base64_decode($row['image']).'" height="100" width="100"/>';
//}
//echo "";
//}
?>

<!DOCTYPE html>

<title>Image upload

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">

<?php
$conn = mysqli_connect('127.0.0.1', 'root', '', 'photos');

$id = $_GET['id'];
// do some validation here to ensure id is safe

// $link = mysql_connect("localhost", "root", "");
// mysql_select_db("photos");
$sql = "SELECT image FROM images WHERE id=$id";
$result = mysql_query("$sql");
$row = mysql_fetch_assoc($result);
mysql_close($link);

header("text: image/jpeg");
echo $row['image'];
/*
$sql="SELECT * FROM images";
$result = mysqli_query($conn,$sql);

while($row = mysqli_fetch_array($result)){
echo "";
echo "<img src='images/'".$row['image']."'>";
echo "".$row['text']."";
echo "";

}
?>*/
?>