PHP:Retrieving images from a database query issues - php

I have read many posts but i cannot find my answer.
I am developing a food order/delivery website, which has many food cuisine categories, African, Alcohol, American... Each category is meant to have a different header image. So if the admin creates a new restaurant, when they select the restaurant cuisine, the correct header image will automatically display on the main websites products page for that said restaurant.
I have manually inputted the images into the database already, now i am trying to retrieve the database, my or die statement prints that it is not, but i have no error messages, which is confusing me.
mysqli_report(MYSQLI_REPORT_INDEX);
if (isset($_GET['rest_id'])) {
$Rest = $_GET['rest_id'];
$get_cat_img = "SELECT Cuisine_category
FROM Rest_Category,Category_img
INNER JOIN Rest_Details
ON Rest_Category.Cat_ID = Rest_Details.Cat_ID
WHERE Rest_Details.Cat_ID='$Rest'";
$results = mysqli_query($dbc, $get_cat_img) or die("query is not working");
$row=mysqli_fetch_array($results) or die ("q not working");
$img=$row['Category_img'];
echo $row['Category_img'];
echo '<img src="'.$img.'" alt="background" style="width:100%;height:300px">';
}
mysqli_close($dbc);

I think you may have just put a column name in the wrong place in your query.
If Category_img is a column name in the Rest_Category table, this is what you want to do
$get_cat_img = "SELECT Cuisine_category,Category_img
FROM Rest_Category
INNER JOIN Rest_Details ON Rest_Category.Cat_ID = Rest_Details.Cat_ID
WHERE Rest_Details.Cat_ID='$Rest'";
You can also shorten things a bit by using Alias's, it often makes the SQL code easier to read when it get past the very simple query.
$get_cat_img = "SELECT rc.Cuisine_category,rc.Category_img
FROM Rest_Category rc
INNER JOIN Rest_Details rd ON rc.Cat_ID = rd.Cat_ID
WHERE rd.Cat_ID='$Rest'";
Also modify your error reporting to actually report the real MYSQL error, it is much more usful that any message you can some up with
Like so
$results = mysqli_query($dbc, $get_cat_img);
if ( $result === false ) {
echo 'query is not working: ' . mysqli_error($dbc);
exit;
}

Related

The above query will not pull the information from the database

$read = "SELECT * FROM elmtree
WHERE id ='$getid' AND
INNER JOIN elmtree_users ON elmtree.userid = elmtree_users.id";
The above query will not pull the information from the database to publish to the website.
Im trying to pull the item from the database with the $getid but also join the item id with the userid who uploaded it. Then using a while loop to print out the item to screen.
Any help would be greatly appreciated.
The SQL syntax you show isn't correct. A join clause describes which tables will be available for the rest of the query; all tables you mention need to be part of the ‘FROM’ clause, so that is where the ‘JOIN’ also belongs.
SELECT *
FROM elmtree
INNER JOIN elmtree_users ON elmtree.userid = elmtree_users.id
WHERE id ='$getid'
Your SQL query was not correctly written. The AND is not used in joining tables and the WHERE statement should be after the JOIN.
Try the following:
$read = "SELECT * FROM elmtree INNER JOIN elmtree_users ON(elmtree.userid = elmtree_users.id) WHERE elmtree.id ='$getid'";
UPDATE
Could you try the following in your code (replacing $this->database with your database variable) and post the result:
if ($result = $conn->query($read)) {
while ($row = $result->fetch_assoc()) {
...
}
} else {
throw new Exception("Database Error [{$this->database->errno}] {$this->database->error}");
}
$read = "SELECT * FROM elmtree INNER JOIN elmtree_users ON(elmtree.userid = elmtree_users.id) WHERE elmtree.itemid ='$getid'
Big thanks to Omari Celestine to finding the answer to the problem!

Echo contents of JOIN SQL tables with MySQLi

I'm working on a system, and this module is supposed to echo the contents of the database.
It worked perfectly until I added some JOIN statements to it.
I've checked and tested the SQL code, and it works perfectly. What's not working is that part where I echo the content of the JOINed table.
My code looks like this:
$query = "SELECT reg_students.*, courses.*
FROM reg_students
JOIN courses ON reg_students.course_id = courses.course_id
WHERE reg_students.user_id = '".$user_id."'";
$result = mysqli_query($conn, $query);
if (mysqli_fetch_array($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
echo $row["course_name"];
echo $row["course_id"];
The course_name and course_id neither echo nor give any error messages.
UPDATE: I actually need to increase the query complexity by JOINing more tables and changing the selected columns. I need to JOIN these tables:
tutors which has columns: tutor_id, t_fname, t_othernames, email, phone number
faculty which has columns: faculty_id, faculty_name, faculty_code
courses which has columns: course_id, course_code, course_name, tutor_id, faculty_id
I want to JOIN these tables to the reg_students table in my original query so that I can filter by $user_id and I want to display: course_name, t_fname, t_othernames, email, faculty_name
I can't imagine that the user_info table is of any benefit to JOIN in, so I'm removing it as a reasonable guess. I am also assuming that your desired columns are all coming from the courses table, so I am nominating the table name with the column names in the SELECT.
For reader clarity, I like to use INNER JOIN instead of JOIN. (they are the same beast)
Casting $user_id as an integer is just a best practices that I am throwing in, just in case that variable is being fed by user-supplied/untrusted input.
You count the number of rows in the result set with mysqli_num_rows().
If you only want to access the result set data using the associative keys, generate a result set with mysqli_fetch_assoc().
When writing a query with JOINs it is often helpful to declare aliases for each table. This largely reduces code bloat and reader-strain.
Untested Code:
$query = "SELECT c.course_name, t.t_fname, t.t_othernames, t.email, f.faculty_name
FROM reg_students r
INNER JOIN courses c ON r.course_id = c.course_id
INNER JOIN faculty f ON c.faculty_id = f.faculty_id
INNER JOIN tutors t ON c.tutor_id = t.tutor_id
WHERE r.user_id = " . (int)$user_id;
if (!$result = mysqli_query($conn, $query)) {
echo "Syntax Error";
} elseif (!mysqli_num_rows($result)) {
echo "No Qualifying Rows";
} else {
while ($row = mysqli_fetch_assoc($result)) {
echo "{$row["course_name"]}<br>";
echo "{$row["t_fname"]}<br>";
echo "{$row["t_othernames"]}<br>";
echo "{$row["email"]}<br>";
echo "{$row["faculty_name"]}<br><br>";
}
}

MYSQL query so it also returns values if has no value in the other table

Hi guys in the code below you can see what my JSON returns.
{"lifehacks":[{
"id":"2",
"URLtoImage":"http:\/\/images.visitcanberra.com.au\/images\/canberra_hero_image.jpg",
"title":"dit is nog een test",
"author":"1232123",
"score":"2",
"steps":"fdaddaadadafdaaddadaaddaadaaaaaaaaaaa","category":"Category_2"}]}
What the JSON returns is fine. The only problem is it is only displaying lifehacks if it has one like or more. So what should I change about my Query so it would display lifehacks without likes aswell.
//Select the Database
mysql_select_db("admin_nakeitez",$db);
//Replace * in the query with the column names.
$result = mysql_query("select idLifehack, urlToImage, title, Lifehack.Users_fbId, idLifehack, steps, Categorie, count(Lifehack_idLifehack) as likes from Lifehack, Likes where idLifehack = Lifehack_idLifehack AND idLifehack > " . $_GET["id"]. " group by idLifehack;", $db);
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$row_array['id'] = $row['idLifehack'];
$row_array['URLtoImage'] = $row['urlToImage'];
$row_array['title'] = $row['title'];
$row_array['author'] = $row['Users_fbId'];
$row_array['score'] = $row['likes'];
$row_array['steps'] = $row['steps'];
$row_array['category'] = $row['Categorie'];
//push the values in the array
array_push($json_response,$row_array);
}
echo "{\"lifehacks\":";
echo json_encode($json_response);
echo "}";
//Close the database connection
fclose($db);
I hope my problem is clear like this. Thank you in advance I can't figure it out myself.
You need a LEFT JOIN here. Your query has an INNER JOIN.
select
idLifehack,
urlToImage,
title,
Lifehack.Users_fbId,
idLifehack,
steps,
Categorie,
count(Lifehack_idLifehack) as likes
from Lifehack
left join Likes on idLifehack = Lifehack_idLifehack
where idLifehack > whatever
group by idLifehack;
There's an excellent explanation of the different join types here.
A couple additional points...
Use prepared statements in your PHP. Your code is wide-open to SQL Injection, which has ruined careers and led to millions of innocent people having their personal information stolen. There are plenty of web sites showing how to do this so I won't go into it here, though I'll say my favorite is bobby-tables.
Avoid the implicit join anti-pattern in your queries. This is an implicit join:
FROM a, b
WHERE a.id = b.id
Use explicit joins instead; they separate your join logic from your filtering (WHERE) logic:
FROM a
INNER JOIN b ON a.id = b.id

Foreach looping too many times

I'm using this to display information from a queried db in Wordpress. It displays the correct information but it loops it too many times. It is set to display from a SELECT query and depending on the last entry to the db seems to be whether or not it prints double or triple each entry.
foreach ($result as $row) {
echo '<h5><i>'.$row->company.'</i> can perform your window installation for <i>$'.$row->cost.'</i><br>';
echo 'This price includes using<i> '.$row->material.'</i> as your material(s)<br>';
echo '<hr></h5>';
}
Does anyone know what could be producing this error?
Thanks
The query powering that script is:
$result = $wpdb->get_results( "SELECT bp.*, b.company
FROM `windows_brands_products` bp
LEFT JOIN `windows_brands` b
ON bp.brand_id = b.id
JOIN Windows_last_submissions ls
JOIN windows_materials wm
JOIN Windows_submissions ws
WHERE ws.username = '$current_user->user_login'
AND bp.width = ROUND(ls.width)
AND bp.height = ROUND(ls.height)
AND bp.material IN (wm.name)
AND bp.type = ls.type
AND IF (ls.minimumbid != '0.00',bp.cost BETWEEN ls.minimumbid AND ls.maximumbid,bp.cost <= ls.maximumbid)
ORDER BY b.company ASC");
I can't seem to see the duplicate but I agree it must be there.
EDIT-- when I replace the WHERE clause to WHERE ws.username = 'password' , it still repeats. It it displaying a result for each time a result has username='password' , and displaying that set twice as well.
I think you want the following, if you're using MySQLi:
while ($row = $result->fetch_object()) {
echo '<h5><i>'.$row->company.'</i> can perform your window installation for <i>$'.$row->cost.'</i><br>';
echo 'This price includes using<i> '.$row->material.'</i> as your material(s)<br>';
echo '<hr></h5>';
}
Redundant JOIN clauses in my query which was pretty much pulling the same results from two tables (one of which was just a VIEW of the other).

SQL Inner Join Issue (Could be easy fix)

So I am trying to populate a drop down by grabbing my playersID from the players table whilst sharing the same playersID with my Payments table.
When I put my query as below, it works fine.
$sql = "SELECT playersID FROM players";
But when I want to join the two tables together, I get an empty drop down.
<?php
include "dbconnect.php";
$sql = "SELECT players.playersID FROM players INNER JOIN Payments ON Payments.playersID = players.playersID";
if (!$result = mysql_query($sql, $conn))
{
die('Error in querying the database' . mysql_error());
}
echo "<br><select name = 'listbox' id = 'listbox' onclick = 'populate()'>";
while ($row = mysql_fetch_array($result))
{
$playersID = $row['playersID'];
$allText = "$playersID,";
echo "<option value = '$allText'> $playersID</option>";
}
echo "</select>";
mysql_close($conn); ?>
Note I have never done an INNER JOIN before, so it may be a simple issue. The aim of the code is to enter payment information into a form and save the data to the data table. I want to show the playerID in the payments table to show that particular player paid for this product.
Cheers!
It looks like you want to LEFT OUTER JOIN with the payments table instead of inner join, because the payment table does not have a payment for all players yet or for any at all.

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