Display BLOB data from SQLITE database using PHP - php

Question

Hope You all have a Great Day, I new with SQLITE DB and so I'm little confused about handling BLOB data.I tried lot of things ,but nothing get in to favor of me.
Here is My Code
INSERTING DATA
insert.php
<form method="POST" enctype="multipart/form-data">
model no:<input type="text" name="model_no"\><span style="color: red"><?php echo $messages["model_no"]; ?></span>
image:<input type="file" name="image" id="image"\><span style="color: red">
<input type="submit" value="save" id="save" name="save"/>
<input type="reset" value="Clear"/>
</form>
include.php
<?php
$flag = 0;
class MyDB extends SQLite3
{
function __construct()
{
$this->open('../database.db');
}
}
//checking save button is clicked
if(isset($_POST["save"])){
$model_no = $_POST['model_no'];
$image = $_FILES['image']['name'];
if($model_no != '' && $image != ''){
$flag = 1;
}
if($flag == '1'){
$db = new MyDB();
if(!$db){
echo $db->lastErrorMsg();
}
else{
$dbb = new MyDB();
$sql = 'SELECT COUNT(*) as count FROM fisfis WHERE model_no = "'.$model_no.'"';
//echo $sql;
$rows = $dbb->query($sql);
$row = $rows->fetchArray();
$numRows = $row['count'];
//echo $numRows;
if($numRows != 0){
$flag = 0;
echo "error:this model no is already taken";
}else{
$sqlp = "INSERT INTO fisfis(model_no, image) values('$model_no','$image')";
if($dbb->exec($sqlp)){
echo "data inserted successfully\n";
}
else{
echo"error:\n";
echo "data not inserted \n";
echo"contact admin for details\n";
}
}
}
}
else{
echo"error:\n";
echo "data not inserted : data fields cannot be empty\n";
}
}
?>
this is very successfully inserting data in to my table (table name:fisfis)
DISPLAYING DATA
display.php
<form method="POST" enctype="multipart/form-data">
<p>Enter Barcode : <input type="text" name="search_model" id="search_model" /></p>
<!--<p><img src='image.php?id=<?php //echo $row['model_no'];?>'/></p>-->
<p><input type="submit" value="search" name="search" id="search"/></p>
</form>
display_include.php
<?php
$flag2 = 0;
$flag3 = 0;
class MyDB extends SQLite3
{
function __construct()
{
$this->open('../database.db');
}
}
$db = new MyDB();
if(isset($_POST["search"])){
$model_no = '';
$model_no = $_POST['search_model'];
if($model_no != ''){
$flag2 = 1;
}
if($flag2 == '1'){
$db = new MyDB();
if(!$db){
echo $db->lastErrorMsg();
}
else{
$dbb = new MyDB();
$sql = 'SELECT COUNT(*) as count FROM fisfis WHERE model_no = "'.$model_no.'"';
$rows = $dbb->query($sql);
$row = $rows->fetchArray();
$numRows = $row['count'];
echo $numRows;
if($numRows == 0){
echo 'sorry this model no is not exists';
echo '<br/>';
echo 'create new';
$flag2 = 1;
}else{
echo "this is exisists";
$flag3 = 1;
}
if($flag3 == 1){
$sqla = 'SELECT * FROM fisfis WHERE model_no = "'.$model_no.'"';
$result = $dbb->query($sqla);
while($row = $result->fetchArray(SQLITE3_ASSOC) ) {
echo "MODEL NO = ". $row['model_no'] ."\n";
$img = '\'<img src="'. $row['image'] .'" width="100" height="100"/>\'';
echo $img;
}
}
}
}
}
?>
the display part displaying all data other than blob ,what i have tried
1.try to echo the image inside the php script
$img = '\'<img src="'. $row['image'] .'" width="100" height="100"/>\'';
try to display image inside html tag
but this is not working,please help me to fix this,thanks in advance

Answer 1

My explaining is not good, so here is a show-and-tell. I made a repro so I wouldn't forget any necessary bits.
The database table is defined:
CREATE TABLE `imgrepro` (
`id` INTEGER NOT NULL,
`photo` BLOB,
PRIMARY KEY(`id`)
);
It is seeded with one row, id = 1, photo is NULL.
The POST block will insert the image into the BLOB column in the db. The rest of the script will fetch the image and the html will display it.
<?php
$db = new SQLite3("***********\stacktest.db");
if ($_SERVER['REQUEST_METHOD'] == "POST") {
// insert photo
$photo = file_get_contents($_FILES['fname']['tmp_name']);
$query = $db->prepare("UPDATE imgrepro set photo = :photo where id = 1");
$query->bindValue(':photo',$photo,SQLITE3_BLOB);
$result = $query->execute();
}
// get photo if there is one
$rows = $db->query("SELECT * from imgrepro")->fetchArray(SQLITE3_ASSOC);
$showphoto="";
if (count($rows) > 0) {
$showphoto=base64_encode($rows['photo']);
}
$db->close();
?>
<!DOCTYPE html>
<head>
<title>Say Cheese!</title>
</head>
<!-- show the photo -->
<img alt="no photo yet" src="data:image/jpeg;base64,<?= $showphoto ?>" >
<form action="imgrepro.php" method="post" enctype="multipart/form-data">
<input type="file" name="fname" accept=".jpg">
<input type="submit">
</form>