PHP image display from mysql from different tables [closed] - php

Question

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 7 years ago.
Improve this question
I am creating a used car website for a class. I have a mysql database that has two tables. One for the car details and one table for car photos. I am trying to display 1 single photo of the car next to some details (like you see on cars.com) but I am struggling. The car id is one one table and the "ext_1" in on another table.
<?php
$page_title = "List cars";
require_once ('includes/header.php');
require_once('includes/database.php');
//select statement
$sql = "SELECT * FROM inventory_tbl, inventory_photos_tbl";
//execute the query
$query = $conn->query($sql);
//Handle selection errors
if (!$query) {
$errno = $conn->errno;
$errmsg = $conn->error;
echo "Selection failed with: ($errno) $errmsg<br/>\n";
$conn->close();
//require_once ('includes/footer.php');
exit;
}
//display results in a table
?>
<h2>Inventory</h2>
<table border="1px solid" border="collapse" align ="center" height='300' width='750' >
<tr><center>
<th></th>
<th>Year</th>
<th>Make</th>
<th>Model</th>
<th>Mileage</th>
<th>Price</td>
<th>Details</th>
</center>
</tr>
<?php
//insert a row into the table for each row of data
while (($row = $query->fetch_assoc()) !==NULL){
echo "<tr>";
echo "<td><img src =" .$row['id']. $row['ext_1']. " width = 100 height = 100></td>";
echo "<td>", $row['year'], "</td>";
echo "<td>", $row['make'], "</td>";
echo "<td>", $row['model'], "</td>";
echo "<td>", number_format($row['mileage']), "</td>";
echo "<td>","$", number_format($row['price']), "</td>";
echo "<td><a href='cardetails.php?id=", $row['id'],"'>View Details</td>";
echo "</tr>";
}
?>
<?php
// clean up resultsets when we're done with them!
$query->close();
// close the connection.
$conn->close();
//include the footer
require_once ('includes/footer.php');

Answer 1

I'm moving my answer out of the comments, so future readers are able to benefit from the solution.
First, the query needs to qualify the join condition, so change the query to:
$sql = "SELECT * FROM inventory_tbl, inventory_photos_tbl WHERE inventory_tbl.id = inventory_photos_tbl.ext_1"
Second, the img tag should only refer to the photo source/column, so it should be something like:
echo "<td><img src =" .$row['myPhotoColumn']. " width = 100 height = 100></td>";

Answer 2

What you're wanting to do is a left join. (Join data from the right table into the left table if the data in the right table exists).
Your SQL statement would look like this.
$sql = "SELECT * FROM inventory_tbl
LEFT JOIN inventory_photos_tbl as images
ON images.ext_1 = car_id";
Images serves as an alias to your photos table.
You match up images.ext_1 column to the main table's car_id column which is what the ON statement signifies. Think of it as a where clause for the JOIN
Now you can reference any of the data in the inventory_photos_table you want in your PHP using your $row variable such as $row['image_column_name'].
You'll probably want to wrap the image data in an if statement to show a placeholder image in case there isn't an image associated with the vehicle yet.

Answer 3

As i don't know your table structure. I assumed the relational column name as 'id' in inventory_photos_tbl table.
<?php
$sql = "SELECT * FROM inventory_tbl";
$query = $conn->query($sql);
while (($row = $query->fetch_assoc()) !==NULL){
$carID=$row['id'];
// In this query, i'm assuming the relational column as 'id'. (If you are having different column name, please change it here itself)
$imgQuery = $conn->query("SELECT * FROM inventory_photos_tbl WHERE id=$carID LIMIT 0,1");
while(($rowImg = $imgQuery->fetch_assoc()))
{
$carIMG=$rowImg['ext_1'];
}
echo "<tr>";
echo "<td><img src ="$carID.$carIMG" width = 100 height = 100></td>";
echo "<td>", $row['year'], "</td>";
echo "<td>", $row['make'], "</td>";
echo "<td>", $row['model'], "</td>";
echo "<td>", number_format($row['mileage']), "</td>";
echo "<td>","$", number_format($row['price']), "</td>";
echo "<td><a href='cardetails.php?id=", $row['id'],"'>View Details</td>";
echo "</tr>";
}?>