I would like to know how can I count the results in case I have two loop ! :
$select = "SELECT X FROM Y"
$result = mysql_query($select,$link);
$total = mysql_num_rows($result);
while($row = mysql_fetch_array($result)) {
$Name = $row['X'];
$select2 = "SELECT id FROM `".$Name."` ";
$result2 = mysql_query($select2,$link) or die ('Erreur : '.mysql_error() );
$total2 = mysql_num_rows($result2);
while($row2 = mysql_fetch_array($result2)) { echo $row2['id']} }
I would like to count how many id I have at the end . Thanks for your help
just give this
$select = "SELECT X FROM Y"
$result = mysql_query($select,$link);
$total = mysql_num_rows($result);
$finalTotal=0;
while($row = mysql_fetch_array($result)) {
$Name = $row['X'];
$select2 = "SELECT id FROM `".$Name."` ";
$result2 = mysql_query($select2,$link) or die ('Erreur : '.mysql_error() );
$total2 = mysql_num_rows($result2);
$finalTotal = $finalTotal + $total2;
while($row2 = mysql_fetch_array($result2)) { echo $row2['id'];} }
echo "total id is ".$finalTotal;
$finalTotal will show you total number of id's
Related
How to add the result of the while loop which is stored on the variable $profit? results are 5 and 70
the code
<?php
$sql2 = "SELECT * from `products`";
$result2 = $link->query($sql2);
while($row2 = $result2->fetch_assoc())
{
$sql3 = "SELECT * from `orders` where product_id = '".$row2['prod_id']."'";
$result3 = $link->query($sql3);
$row3 = $result3->fetch_assoc();
$sql = "SELECT SUM(product_qty) from orders where product_id = '".$row2['prod_id']."'";
$result = $link->query($sql);
$row = mysqli_fetch_array($result);
$res = bcmul($row2['prod_price'], $row[0]);
$profit = $res - $row2['prod_cost'];
if($row[0] == null){
$row[0] = 0;
}
}?>
Try this code
<?php
$sql2 = "SELECT * from `products`";
$result2 = $link->query($sql2);
$totalProfit = 0;
while($row2 = $result2->fetch_assoc())
{
$sql3 = "SELECT * from `orders` where product_id = '".$row2['prod_id']."'";
$result3 = $link->query($sql3);
$row3 = $result3->fetch_assoc();
$sql = "SELECT SUM(product_qty) from orders where product_id = '".$row2['prod_id']."'";
$result = $link->query($sql);
$row = mysqli_fetch_array($result);
$res = bcmul($row2['prod_price'], $row[0]);
$profit = $res - $row2['prod_cost'];
$totalProfit = $totalProfit + $profit
if($row[0] == null){
$row[0] = 0;
}
}?>
now you can use $totalProfit variable value as total.
I tried to count() the data from database with php but it don't show me the total data but it show the datas.
this is how I count
$query = "SELECT
mitra.*,
user.username,
user.privilege,
user.name
FROM
mitra
INNER JOIN
user ON mitra.id_user = user.id "
$result = $connection->query($query);
if ($result->num_rows > 0) {
foreach ($result as $row) :
$id = "" . $row["id"] . "";
$total = "" . $row["total_puas"] . "";
$privilege = "" . $row["privilege"] . "";
if ($privilege == 2) :
$calculate = $total / count($id);
var_dump(count($id));
endif;
endforeach;
}
===================
= id = total =
= 1 = 45.84 =
= 2 = 75.45 =
= 3 = 34.54 =
===================
when I var_dumb it it shows int(1)int(1)int(1) not int(3) that what I wanted.
actually I want to count $calculate with the data in $total that should be there is float and the amount from $total that I want to divided with count id
is there any solution that how to count the amount from $total and can be devided with count $id that should be 3?. please help
what I really trying to do from that table example is like 45.84 + 75.45 + 34.54 / 3
Sounds like you want a COUNT() with GROUP BY in your query. Doing count($id) in PHP will always yield one, as its not an array of values.
$query = "SELECT COUNT(id) as cnt, id, total_puas
FROM table
GROUP BY id";
$result = $connection->query($query);
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$calculate = $row["total_puas"] / $row['cnt'];
echo $row['id']." has a count of ".$row['cnt'].", a total of ".$row['total_puas']." and calculated to ".$calculate."<br />\n";
}
}
From your updated question, the output becomes a bit more clear. Based on the output data and the result you desire, you want to calculate the sum of all total_paus, divided by the number of rows. You can do this directly in one query, like this
$query = "SELECT SUM(total_puas) / COUNT(u.id) as total
FROM mitra AS m
INNER JOIN user AS u
ON mitra.id_user = user.id";
$result = $connection->query($query);
$row = $result->fetch_assoc();
$total = $row['total'];
echo $total;
You can try this code:
<?php
$i = 0;
$total =0.00;
if ($result->num_rows > 0) {
while ($row = $result->fetch_array()):
if ($row["privilege"] == 2) :
$total = $total + $row["total"];
$i++;
endif;
endwhile;
echo $total."<br>";
echo $i."<br>";
echo $calculate = $total / $i;
}
?>
output
=====================================
$total = 155.83;
$i = 3;
$calculate = $total/$i;
$ans = 51.943333333333;
=====================================
You can try this code:
$query = "SELECT * FROM table"
$result = $connection->query($query);
if ($result->num_rows > 0) {
$total = 0.0;
foreach ($result as $row) :
$id = "" . $row["id"] . "";
$total = $total + $row['total'];
endforeach;
$calculate = $total / $result->num_rows;
echo $total.<br>;
echo $calclulate;
}
<?php
include"../database_conn.php";
$con=mysqli_connect("localhost","admin","123456","ayurveadic");
$query_pag_data = "SELECT id from diseases";
$result_pag_data = mysql_query($query_pag_data) or die('MySql Error' . mysql_error());
while ($row = mysql_fetch_array($result_pag_data)) {
$diseases_id = $row['id'];
$result = mysqli_query($con,"SELECT $diseases_id FROM treatment WHERE gender_id = '10' AND diseases_id = '$diseases_id'");
$row_gid = mysqli_fetch_array($result);
if ($row_gid == TRUE){
$sl_dise = mysqli_query($con,"SELECT Diseases_type, id FROM diseases WHERE id = '$diseases_id'");
$rowss = array();
while($r = mysqli_fetch_assoc($sl_dise)) {
$rowss[] = $r;
}
print json_encode($rowss);
}
}
Output is:
[{"Diseases_type":"fever","id":"114"}][{"Diseases_type":"rhrh","id":"123"}]
How can i get this output:
[{"Diseases_type":"fever","id":"114"},{"Diseases_type":"rhrh","id":"123"}]
You need to initialize the JSON array before the outer loop, and print it at the very end, not each time through.
<?php
include"../database_conn.php";
$con=mysqli_connect("localhost","admin","123456","ayurveadic");
$query_pag_data = "SELECT id from diseases";
$result_pag_data = mysql_query($query_pag_data) or die('MySql Error' . mysql_error());
$rowss = array();
while ($row = mysql_fetch_array($result_pag_data)) {
$diseases_id = $row['id'];
$result = mysqli_query($con,"SELECT $diseases_id FROM treatment WHERE gender_id = '10' AND diseases_id = '$diseases_id'");
$row_gid = mysqli_fetch_array($result);
if ($row_gid == TRUE){
$sl_dise = mysqli_query($con,"SELECT Diseases_type, id FROM diseases WHERE id = '$diseases_id'");
while($r = mysqli_fetch_assoc($sl_dise)) {
$rowss[] = $r;
}
}
}
print json_encode($rowss);
Here is you optimised code
<?php
include"../database_conn.php";
$con=mysqli_connect("localhost","admin","123456","ayurveadic");
$query_pag_data = "select diseases.Diseases_type,diseases.id from diseases inner join treatment on treatment.diseases_id = diseases.id where treatment.gender_id = '10' ";
$result_pag_data = mysql_query($query_pag_data) or die('MySql Error' . mysql_error());
$rowss = array();
while ($row = mysqli_fetch_assoc($result_pag_data)) {
$rowss[] = $row;
}
print json_encode($rowss);
?>
Put your
print json_encode($rowss);
after end of thiswhile ($row = mysql_fetch_array($result_pag_data)) {
Also remove this $rowss = array(); array initialization
This is the entire code the error must be in query 3 or 4. As you can see query 3 just gets info to build query 4 which should return the results.
I've got query 1 & 2 working ok which have the correct data showing.
<?php
session_start();
//printf('<pre>%s</pre>', print_r($_SESSION, true));
require('includes/config.inc.php');
require('includes/session.php');
//WORKING
DB_Connect();
$query = "SELECT * FROM food_delivery_orders_items WHERE food_delivery_orders_items.type = 'product' AND food_delivery_orders_items.order_id=".$_SESSION['pdf_quote']['id']."";
$result = mysql_query( $query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$hash = $row['hash'];
$foreignid = $row['foreign_id'];
$qty = $row['cnt'];
}
$query2 = "SELECT * FROM food_delivery_products WHERE food_delivery_products.id = ".$foreignid."";
$result2 = mysql_query( $query2) or die(mysql_error());
while($row = mysql_fetch_array($result2))
{
$name = $row['name'];
$description = $row['description'];
}
//---------------------------------------------------------------------------------------------------------------
$query3 = "SELECT * FROM food_delivery_orders_items WHERE food_delivery_orders_items.type = 'extra' AND food_delivery_orders_items.order_id=".$_SESSION['pdf_quote']['id']."";
$result3 = mysql_query( $query3)or die(mysql_error()) ;
while($row = mysql_fetch_array($result3))
{
$foreignidextra = $row['foreign_id'];
$qtyextra = $row['cnt'];
}
$query4 = "SELECT * FROM food_delivery_extras WHERE food_delivery_extras.id = ".$foreignidextra."";
$result4 = mysql_query( $query4) or die(mysql_error());
while($row = mysql_fetch_array($result4))
{
$nameextra = $row['name'];
}
echo $nameextra;
echo $qtyextra;
DB_Disconnect();
//$products = implode(", ",$_SESSION['pdf_quote']['product_arr']);;
//print $products
?>
print $row; is only valid inside the loop if you want to print all of them. Otherwise it will print only the last $row. Make it
while($row = mysql_fetch_array($result3))
{
$foreignidextra = $row['foreign_id'];
$qtyextra = $row['cnt'];
print $row;
}
its because you are printing the $row after the while loop gets executed
So it only prints the final result set
Use something like this instead:
$query3 = "SELECT * FROM food_delivery_orders_items WHERE food_delivery_orders_items.type = 'extra' AND food_delivery_orders_items.order_id=".$_SESSION['pdf_quote']['id']."";
$result3 = mysql_query( $query3) ;
while($row = mysql_fetch_array($result3))
{
$foreignidextra = $row['foreign_id'];
$qtyextra = $row['cnt'];
print_r($row);
}
you can also use the var_dump($row); instead of print_r($row);
this will output all the values stored in the $row array each time the loop iterates
I have a table in PHP which displays the result of the MySQL query.
The database table item_em has the price details of the items that are available in hypermarkets available in the database table hypermarket_em.There are some items that aren't available in the hypermarket. Hence when i display the prices in the table I want those cells with no price as N/A.
So my problem is I tried a code myself to display message when database table field has no value.But it doesn't work and I don't get any error as well. Any ideas much appreciated. Thank you.
Here is my code
$res = mysql_query("SELECT h_id FROM hypermarket_em") or die(mysql_error());
echo"<tbody><tr>";
while($row = mysql_fetch_array( $res ))
{
$result = mysql_query("SELECT item_name FROM items_em WHERE h_id=".$row['h_id'])or die(mysql_error());
while($drop_4 = mysql_fetch_array( $result ))
{
echo"<th scope=row>".$drop_4['item_name']."</th>";
$rslt = mysql_query("SELECT price FROM items_em WHERE item_name='".$drop_4['item_name']."' and h_id=".$row['h_id'])or die(mysql_error());
if (mysql_num_rows($rslt) == 0)
echo"<td>N/A</td>";
while($drop = mysql_fetch_array( $rslt ))
{
echo"<td>".$drop['price']."</td>";
}
}
}
echo"</tr></tbody>";
Try reversing your mysql_num_rows and check if you have > than 0 rows-
if (mysql_num_rows($rslt) > 0){
while($drop = mysql_fetch_array( $rslt )){
echo"<td>".$drop['price']."</td>";}
}
else{
echo"<td>N/A</td>";}
Edit-
I think what you are wanting to do then is something like this-
while($drop = mysql_fetch_array( $rslt )){
if ($drop['price'] == ''){
echo"<td>N/A</td>";}
else{
echo"<td>".$drop['price']."</td>";}
$data = mysql_fetch_array( $res );
if (!empty($data)) {
while($row = mysql_fetch_array( $res ))
{
$result = mysql_query("SELECT item_name FROM items_em WHERE h_id=".$row['h_id'])or die(mysql_error());
while($drop_4 = mysql_fetch_array( $result ))
{
echo"<th scope=row>".$drop_4['item_name']."</th>";
$rslt = mysql_query("SELECT price FROM items_em WHERE item_name='".$drop_4['item_name']."' and h_id=".$row['h_id'])or die(mysql_error());
if (mysql_num_rows($rslt) == 0)
echo"<td>N/A</td>";
while($drop = mysql_fetch_array( $rslt ))
{
echo"<td>".$drop['price']."</td>";
}
}
}
} else {
echo "<td>No record found</td>";
}
Not really sure what you're trying to do here, but it seems like you might want to simplify your query as well. Try something like:
SELECT
*
FROM
hypermarket_em h
LEFT JOIN items_em i ON ( h.h_id = i.h_id );
Using this query, you can do something like:
while($drop = mysql_fetch_array( $rslt )){
echo"<td>" . ( empty( $drop['price'] ) ? 'N/A' : $drop['price'] ) . "</td>";
}
So here is my answer.
$res = mysql_query("SELECT DISTINCT h_id FROM hypermarket_em") or die(mysql_error());
echo"<tbody><tr>";
while($row = mysql_fetch_array( $res ))
{
$result = mysql_query("SELECT item_name FROM items_em WHERE h_id=".$row['h_id'])or die(mysql_error());
while($drop_4 = mysql_fetch_array( $result ))
{
echo"<th scope=row>".$drop_4['item_name']."</th>";
$rslt = mysql_query("SELECT price FROM items_em WHERE item_name='".$drop_4['item_name']."' and h_id=".$row['h_id'])or die(mysql_error());
while($drop = mysql_fetch_array( $rslt ))
{
echo"<td>".$drop['price']."</td>";
}
}
}
$res = mysql_query("SELECT DISTINCT h_id FROM hypermarket_em") or die(mysql_error());
while($row = mysql_fetch_array( $res ))
{
$result = mysql_query("SELECT item_name FROM items_em WHERE h_id=".$row['h_id'])or die(mysql_error());
if(mysql_num_rows($result)<=0)
echo"<td>N/A</td>";
}
echo"</tr></tbody>";
I had to execute the same query twice as i want the item names and the price values in the order. So I first print the item names and then only the prices or "N/A" message.
Thank you all for your help..:)