We can connect a single PHP page with more than one databases. But what about executing multiple dependent queries that belong to different databases?
mysqli_multi_query(connection,query);
does not seems helpful as it asks for a single shot while my second query uses result of the first. I tried using two connection strings and using them with two queries but this does not seems to be helpful. My code uptil now is here:
<div class="image fit">
<?php
//Connection with DB
$user = "root";
$pass = "";
$db1 = "raw_images";
$db2 = "labelled_images";
$conn1 = mysqli_connect('localhost',$user,$pass,$db1,true);
$conn2 = mysqli_connect('localhost',$user,$pass,$db2,true);
if(!$conn1||!$conn2)
{
die(mysqli_error());
}
*/
//query to display image
$query1 = "SELECT * FROM raw_images.tbl_raw_image WHERE id IN
(SELECT id FROM (SELECT id FROM raw_images.tbl_raw_image ORDER BY RAND() LIMIT 1) t)";
$sth = $conn1->query($query1);
$result=mysqli_fetch_array($sth);
$id=$result['id'];
$image=$result['image'];
$query2="INSERT INTO `labelled_image`.`tbl_labelled_image` (`id`,`image`, `label`, `id_in_raw_image_table`) VALUES (NULL, '$image', '0', '$id')";
$rs = mysqli_query($conn2, $query2);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $image ).'"/>'; ?>
</div>
When using such approach, first query executes successfully but not the second one. While executing it simply does not inserts values which I retrieved from the first table but only the values which are given directly to the query like label which I pass zero hence it results in the following outcome:
id_in_raw_image_tablecolumn. What is the result you get from the first query? Do records in thetbl_raw_imagetable withid0 and 404 (these ids are shown in the screenshot) have values in theimagecolumn?image? If it isBLOB, you have escaping troubles.INSERT ... '$image' ...is very likely to make a mess. Think of what happens if$imagecontains a'!