Hi I'm trying to display all car names,their images ,rental rate all from different tables in a page. How do I get id for each car in select statement?
for instance:
while($row_showall = mysql_fetch_array($result_showall))
{
$car_id[] = $row_showall['carName_id'];
}
foreach($car_id as $id)
{
$id;
echo $id.'<br/>';
//result is
27773
27774
27778
27779
}
when I place $id outside foreach loop obviously it doesn't loop and the result is
2779.
so when I use select statement here like this:
$query_showdays_1="SELECT ........WHERE car_name.carName_id='$id'";
how do I make it select each and every of the car id instead of only one id?
I'm guessing that your table name is cars, because you haven't provided that info. If it's not then change the cars with your table name.
$query_showdays_1 = "SELECT * FROM cars";
while ($row_showall = mysql_fetch_array($query_showdays_1))
{
echo $row_showall['carName_id'] . '<br />';
}
Related
I have two different sql statements. $sql grabs all the items whose title matches a certain search text. $cat_sql grabs all the category_items that are in a certain category. An item has an ID. A category_item has a field called item_id which is a foreign key to IDs in the items table
...
mysqli setup code
...
$title = $_POST["title"];
$cat_id = $_POST["cat_id"];
$cat_sql = "SELECT * FROM category_items WHERE category_id = '".$cat_id."'";
$sql = "SELECT * FROM items where title LIKE '%". $title ."%' Limit 70";
if (!$result_cat = $mysqli->query($cat_sql)) {
// The query failed.
echo "<h2 >ERROR</h2>";
exit;
}
if (!$result = $mysqli->query($sql)) {
// The query failed.
echo "<h2 >ERROR</h2>";
exit;
}
Then I display all items:
while ($item = $result->fetch_assoc()) {
include 'item_card.php';
}
Currently this just displays all items fetched in the $sql query. Is there some way to remove all items from $result that do not have their ID represented as an item_id in $result_cat?
NOTE:
I would strongly prefer not to do just combine both SELECT statements into a table join because the actual $sql and $cat_sql are not nearly as simple as I have represented here. Also, they vary depending on which if statement they are in.
My question is: given $result and $result_cat, can I remove items from $result?
EDIT 1
As suggested by comments I am making an array if item_ids then doing an in_array query. Progress thus far:
$result_cat_ids = [];
while ($cat_item = $result_cat->fetch_assoc()) {
$result_cat_ids[] = $cat_item['item_id'];
}
EDIT 2 Here is the working code following the suggestions in the comments
if (in_array($item['id'], $result_cat_ids)) {
include 'item_card.php';
}
You may also use 'INTERSECT' sql clause.
$sql = "SELECT * FROM items WHERE id IN (SELECT item_id FROM category_items WHERE category_id = '".$cat_id."' INTERSECT SELECT id FROM items where title LIKE '%". $title ."%')";
This way, you can query for items that accomplish both conditions.
Note: I'm not using "limit 70" but you may add it as well.
I have a search results page which displays items from a MySQL database (table1). The code I am using to the display the results is:
if (!empty($data)) {
foreach ($data as $item) {
echo '<div class="item">';
if (strlen($item['item_desc']) > 10) {
if (strlen($item['item_link']) > 10) {
echo '<a href="/item.php?id='.$item['item_id'].'">';
} else {
echo 'No Results Found';
}
}
}
}
The images for each search result are stored in a separate table (table2). I am trying to use the code below to count the number of images in table2 for each result and display the number against each result on the search results page, but it returns a 0 value?
$result=mysql_query("SELECT count(*) as total from table2 where id = '" .$item['item_id']. "'");
$query=mysql_fetch_assoc($result);
echo $query['total'];
I am missing some data about your database structure but I can imagine you store your image id as an integer and you are requesting the count with the id being a string.
So, if I'm true this may help. Remove the quotes around the php variable with the item_id
"SELECT count(*) as total from table2 where id = " .$item['item_id'];
you can try this..
$str=$item['item_id'];
mysql_query("SELECT count(*) as total from table2 where id = '$str'");
There is a mysql function to count rows outputted from your mysql query, it is mysql_num_rows(). From what I can understand about your database your code could be this:
$item_id = $item['item_id'];
$result = mysql_query("SELECT * FROM table2 where id = '$item_id'");
$count = mysql_num_rows($result);
echo $count;
This will output how many records match your query.
Thanks in advance for any time you spend on my question.
I am trying to display data in a way that will display the manufacturer as a name instead of a number.
Basically when they store the data they choose a manufacturer from a drop down which is generated from a table.. IE Trogues = 1 so products stores the #1 so I know that any beer is associated with trogues is 1. Now I want to display the data but instead of having a 1 I would like to have Trogues be displayed. Where you see manufacturer in the echo code below..
I am not understanding the process logic here..
error_reporting(E_ALL);
ini_set('display_errors', 1);
$sql = "SELECT * FROM products
LEFT JOIN manufacturer
ON product.manufacturer = manufacturer.id
ORDER BY manufacturer.id, product.id";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)) {
echo "
<div class=reportclientproduct>".$row['manufacturer']." - <a href=".$row['website']." target=_blank>".$row['product']."</a></div>";
}
Have you tried the query like this:
$sql = "SELECT man.id AS manufac, products.product AS prod FROM products
LEFT JOIN manufacturer as man
ON product.manufacturer = manufacturer.id
ORDER BY manufacturer.id, product.id";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)) {
echo "
".$row['manufac']." - ".$row['prod']."
";
}
Assuming that the table products had a column named manufacturer which holds the ID of the manufacturer, and that both tables have columns name ID which hold the ID of the table item.
Also the JOIN functions may vary based on the database you use. But the aforementioned method is for mysql.
I have the following query which I run via PHP:
select
{$tableProducts}.*,
{$tableImages}.*
from {$tableProducts}
left join {$tableImages}
on {$tableImages}.product_id = {$tableProducts}.product_id
group by {$tableProducts}.product_id;
Each product (from product table) can have several images (in images table). I loop through the results with a simple while statement:
while($row = $results->fetch_object()) {
echo $row->product_name; // Product table
echo $row->image_src; // Image table
}
Problem: Only the first image of each product is printed, but I want to display all of them. All images are printed if I remove the "order by" part, but then the product_name is printed once for each image (so if one product has three images, the product_name would be printed three times as well).
How do I best solve this?
That's how GROUP BY works.
If you want to get all images for all products, you can solve that (at least) 3 ways:
1: Do not use GROUP BY, handle it in the loop, like:
$last_product = null;
while($row = $results->fetch_object()) {
if ($last_product !== $row->product_id) {
// new product starts here
$last_product = $row->product_id;
echo $row->product_name; // Product table
}
echo $row->image_src; // Image table
}
2: Use GROUP BY & query all images with different statements within the loop.
$products = <query products>;
while($row = $products->fetch_object()) {
echo $row->product_name; // Product table
$images = <query images for product in $row>;
while($row = $images->fetch_object()) {
echo $row->image_src; // Image table
}
}
3: Use aggregate string functions to get all images for a product. This is only works in special cases, f.ex. here, as URL cannot consist new lines, for example.
In MySQL:
select
{$tableProducts}.*,
group_concat({$tableImages}.image_src SEPARATOR '\n') as image_srcs
from {$tableProducts}
left join {$tableImages}
on {$tableImages}.product_id = {$tableProducts}.product_id
group by {$tableProducts}.product_id;
In PostgreSQL:
select
{$tableProducts}.*,
string_agg({$tableImages}.image_src, '\n') as image_srcs
from {$tableProducts}
left join {$tableImages}
on {$tableImages}.product_id = {$tableProducts}.product_id
group by {$tableProducts}.product_id;
In the loop:
while($row = $products->fetch_object()) {
echo $row->product_name; // Product table
foreach (explode("\n", $row->image_srcs) as $image_src) {
echo $image_src; // Image table
}
}
I am working on a php project with MySQL as database. Here is the table structure
product_id (PK)
shop_id (FK)
product_name
product_desc
Product_id is the primary key and shop_id is the foreign key (shows the product belongs to which shop). Now I want to select these product and display them in a <div>. For each shop products, I want to display a different <div>. e.g if there are three products say p1,p2 and p3, and they have same shop_id (say) 1, i want them to show in a separate <div>. there will be separate <div> for each shop_id, and all products belong to that shop_id will be displayed in that <div>. How can I do this with My_SQL query or using PHP. I have used a simple SELECT query like this but with this i could not find a way to initiate different <div> for separate shop.
$q = mysql_query("SELECT * FROM table");
echo "<div>";
while($product = mysql_fetch_assoc($q)){
echo $product['product_name']."<br/>";
}
echo "</div>";
It just throw all products in a single dive, I want separate div for separate shop (which contains all products of that shop). Thanks for reading..
Simplest way is to group the products by id in a array and loop through them
$all_products = array();
while($product = mysql_fetch_assoc($q)){
$all_products[$product['shop_id']][] = $product;
}
foreach($all_products as $shop_id => $product_array){
echo "<div id='shop_".$shop_id."'>";
foreach($product_array as $product){
echo $product['product_name']."<br/>";
}
echo "</div>";
}
OR call product data fro each shop id
$q = mysql_query("SELECT * FROM shops");
while($shops = mysql_fetch_assoc($q)){
$product_q = mysql_query("SELECT * FROM product_table WHERE shop_id=$shops['id']");
echo "<div>";
while($product = mysql_fetch_assoc($product_q)){
echo $product['product_name']."<br/>";
}
echo "</div>";
}