I executed 4 count queries from one table. but I am getting the same output from all the queries. but the actual value is different in the table.
Here is my table.
ID || notify_type || status
__________________________________________
1 || resume_uploaded || 1
Here are my queries:
$notify_query1 = "select count(*) from notify where status = 1 and notify_type = 'resume_uploaded'";
$row1 = mysqli_query($db_manager->connection,$notify_query1);
$rcount = mysqli_num_rows($row1);
$notify_query2 = "select count(*) from notify where status = 1 and notify_type = 'detail_filled'";
$row2 = mysqli_query($db_manager->connection,$notify_query2);
$dcount = mysqli_num_rows($row2);
$notify_query3 = "select count(*) from notify where status = 1 and notify_type = 'job_detailed'";
$row3 = mysqli_query($db_manager->connection,$notify_query3);
$jcount = mysqli_num_rows($row3);
$notify_query4 = "select count(*) from notify where status = 1 and notify_type = 'msg_sent'";
$row4 = mysqli_query($db_manager->connection,$notify_query4);
$mcount = mysqli_num_rows($row4);
I am getting output 1 from all the four queries:
Please help me.
Use fetch_row() instead mysqli_num_rows().
$result = $db->query("select count(*) from notify where status = 1 and notify_type = 'resume_uploaded'");
$row = $result->fetch_row();
echo 'No of count: '. $row[0];
Related
This question already has answers here:
Select Name instead OF ID in table with ID-Ref Column SQL
(2 answers)
Show Name Instead of ID from Different Table
(2 answers)
Closed 1 year ago.
I have 2 tables
rankID | name
1 | new
2 | learner
3 | experienced
4 | pro
And another with all the user info and passwords and stuff
id | username | rankID
1 | hello | 3
2 | hey | 3
I have come so far so I can display their rank number, but I want to display the rank name. How can I do that? I have tried a lot of things but I'm not so good at sql and the php part of it.
This is the code I use to display the rank number
//Get rankID
$query = "SELECT rankID FROM users WHERE id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rankID'];
And to display the rank number:
Rank: <?php echo $rank; ?>
Simple JOIN query :-
"SELECT rank.name as rank_name,users.rankID as rankID from users LEFT JOIN rank ON rank.rankID = users.rankID WHERE id = '$userId'"
And then After:-
$query = "SELECT rank.name as rank_name,users.rankID as rankID from users LEFT JOIN rank ON rank.rankID = users.rankID WHERE id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
Do:-
$rank = $row['rankID'];
$rank_name = $row['rank_name'];
Rank: <?php echo $rank; ?>
RankName: <?php echo $rank_name; ?>
Or
$rank_data = $row;
Rank: <?php echo $rank_data['rankID']; ?>
RankName: <?php echo $rank_data['rank_name']; ?>
Not:- lot of other possible ways are there which are listed by other programmers in comment and answer as well.
//Get datas
$query = "SELECT rankID, name FROM users WHERE id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rankID'];
$rank = $row['name'];
And to display the datas:
Rank: <?php echo $rank; ?>
Name: <?php echo $name; ?>
Hope so this should make a trick for you.
$query = "SELECT rankID FROM users WHERE id = '".$userId."'";
$result = $conn->query($query);
$count = $result->num_rows;
if($count==0)
{
return false;
}
else
{
$rows=[];
while($row = $result->fetch_assoc())
{
$rows[] = $row;
}
return $rows;
}
Please use below code
$query = "SELECT name FROM users as u JOIN rank as r ON r.rankID = u.rankID WHERE u.id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$name = $row['name'];
Name: <?php echo $name; ?>
When you want to get data from two different table.You need join query.
Here is your query which will solve your proble definitely :
$q="select a.name,b.rankID from rankname as a INNER JOIN user as b
ON a.rankID = b.rankID";
For more know about How to join two tables see this:http://www.tutorialspoint.com/sql/sql-using-joins.htm
Hope this will help you better.
Please try this
//Get rankID
$query = "SELECT r.name as rank_name FROM rank as r inner join users as u on r.rankID = u.rankID WHERE u.id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rank_name'];
echo 'Rank: '. $rank;
try this:
//Get rankID
$query = "SELECT rankID, rank.name AS rank_name FROM rank, users WHERE id = '$userId' and users.rankid = rank.rankid";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rank_name'];
echo $rank;
I am doing sum with the below query but it is not giving result properly.
If there are four items and it is showing the result like:
1.000 2.000 3.000 4.000 and it should be like 10.000
I don't know where I am mistaken please help.
<?php
$order_temp = mysql_query("select * from temp_cart
where item_id = '".$mitem_idC."' and ses_mem=113 order by id");
while ($torder = mysql_fetch_array($order_temp)) {
$prITTC = $torder['item_id'];
$qtyT = $torder['qty'];
$chTP = mysql_query("
select * from temp_choices
where item_id = '".$prITTC."'
AND ses_mem = 113
AND status = 1
");
while($chGET = mysql_fetch_array($chTP)){
$fID = $chGET['id'];
$field = $chGET['choice_id'];
$order_tempCHP = mysql_query("
select sum(price) as total, id, ename, choice_id, item_id, price
from choice_price
WHERE
id IN('$field')
");
while ($torderCP = mysql_fetch_assoc($order_tempCHP)){
$totalCH = $torderCP['total'];
$tsl = $totalCH+($qtyT*$prIDTC);
$altsl = number_format($tsl, 3, '.', '');
echo $altsl;
} }
}
?>
according to my question above after trying and trying i found the solution and resolved my problem according to below code:
Thanks to #John Kugelman at MySQL query using an array
$order_temp = mysql_query("select * from temp_cart
where item_id = '".$mitem_idC."' and ses_mem=113 order by id");
while ($torder = mysql_fetch_array($order_temp)) {
$prITTD = $torder['id'];
$prITTC = $torder['item_id'];
$chTPaa = mysql_query("
select choice_id
FROM temp_choices
WHERE item_id = '$prITTC'
AND ses_mem = 113
AND status = 1
group by choice_id
");
while ($chGETaa = mysql_fetch_assoc($chTPaa)){
$temp[] = $chGETaa['choice_id'];
}
$thelist = implode(",",$temp);
$order_tempCHP = mysql_query("
select sum(price) as total
from choice_price
WHERE
id IN ($thelist)
AND
item_id = '".$prITTC."'
");
while($torderCP = mysql_fetch_assoc($order_tempCHP)){
$totalCH = $torderCP['total'];
$tsl = $totalCH+($qtyT*$prIDTC);
$altsl = number_format($tsl, 3, '.', '');
echo $altsl;
} }
I have the following code to validate a user login at database.
Running this SELECT query at phpMyadmin, the result is "Showing rows 0 - 0 ( 1 total, Query took 0.0011 sec)".
For this reason the mysqli_num_rows($query) count is null when I echoes it.
What is wrong?
$sql = "SELECT
Id_Usuario,
Nome_Usuario,
Matricula_Usuario,
Email_Usuario,
Senha_Usuario,
Perfil_Usuario,
Status_Usuario
FROM USUARIO
WHERE Email_Usuario = '".$email."' AND Senha_Usuario = '".$senha."'
LIMIT 1";
$query = mysqli_query($conecta, $sql, MYSQLI_STORE_RESULT) or die('Problem running query: ' . mysql_error());
$row = mysqli_num_rows($query);
while ($row = mysqli_fetch_assoc($query)) {
$Id_Usuario = $row['Id_Usuario'];
$Nome_Usuario = $row['Nome_Usuario'];
$Matricula_Usuario = $row['Matricula_Usuario'];
$Email_Usuario = $row['Email_Usuario'];
$Senha_Usuario = $row['Senha_Usuario'];
$Perfil_Usuario = $row['Perfil_Usuario'];
$Status_Usuario = $row['Status_Usuario'];
}
Here is my query, I am trying to get numbers from another table using a number from another table here is my query ...
$order_id = $template_vars['{order_name}'];
// Query to find the product id for the current order and then set it to a variable
$query="SELECT product_id FROM ps_order_detail WHERE id_order = $order_id";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
$Product_id = $row['0'];
// get all the custom part numbers and set them to the variables
$customnumbers ="SELECT API, SWAIM, JOHN_CRANE, SNOW_WELL, MIDAS, QUINN, WILSON, WEATHERFORD, HF, BLACK_GOLD, EDI, SO_CAL_PUMPS, WEST_RIVER
FROM ps_product_part_number WHERE Product_ID = $Product_id";
$secondresult = mysql_query($customnumbers);
$secondrow = mysql_fetch_row($secondresult);
$API = $secondrow['0'];
$SWAIM = $secondrow['1'];
$JOHN_CRANE = $secondrow['2'];
$SNOW_WELL = $secondrow['3'];
$MIDAS = $secondrow['4'];
$QUINN = $secondrow['5'];
$WILSON = $secondrow['6'];
$WEATHERFORD = $secondrow['7'];
$HF = $secondrow['8'];
$BLACK_GOLD = $secondrow['9'];
$EDI = $secondrow['10'];
$SO_CAL_PUMPS = $secondrow['11'];
$WEST_RIVER = $secondrow['12'];
How about doing it in one step with a join?
SELECT ppn.*
FROM ps_product_part_number ppn
join ps_order_detail od on od.product_id = ppn.Product_ID
WHERE od.id_order = $order_id
I am trying to show a count but only where the column deleted is equal to 0
here is what i have tried
$result=mysql_query("SELECT * FROM users where deleted=0")or die('Error ' );
$counter = mysql_query("SELECT COUNT(*) as personID FROM users");
$row = mysql_fetch_array($result);
$personID = $row['personID'];
$personFname = $row['personFname'];
$personSname = $row['personSname'] ;
$llmail = $row['llmail'];
$mainadmin = $row['mainadmin'];
$delete = $row['delete'];
$num = mysql_fetch_array($counter);
$count1 = $num["personID"];
This shows a count of 4 however of the 4, 2 are deleted so it should only show 2 if this makes sense?
SELECT COUNT(*) as personID FROM users WHERE deleted=0