Hi guys i need help with mysqli
I want to order by fiend with specific value, here is 1 field
$results = mysqli_query($con,"SELECT * FROM `data` ORDER BY field(mu,'Legija Stranaca Elite','Legija Stranaca') ASC, level Desc");
how do apply a 2nd ORDER BY using the same
You need to simply use another field() condition in the order by part of your statement e.g.
$results = mysqli_query($con,"SELECT * FROM `data` ORDER BY field(mu,'Legija Stranaca Elite','Legija Stranaca') ASC, field(anotherfield,'Something Else','Another thing') ASC, level Desc");
Related
I want to show data from database if status is enable and package type is domestic and order by id desc. Here is my code:
$sql = "select *
from holidays
where pkg_status = 'Enable'
and pkg_type = 'Domestic'
limit 6";
I want to show this data as order by id desc. When I am trying to add query like this, it is not working.
$sql = "select *
from holidays
where pkg_status = 'Enable'
and pkg_type = 'Domestic'
limit 6
order by id desc";
Please help me.
limit is always written after group clause in SQL
Here you go, just copy and paste, you will be fine :
SELECT * FROM holidays
WHERE pkg_status='Enable'
AND pkg_type = 'Domestic'
ORDER BY id DESC LIMIT 6;
I thought this would be simple but I'm having a tough time figuring out why this won't populate the the data array.
This simple query works fine:
$queryPrice = "SELECT price FROM price_chart ORDER BY id ASC LIMIT 50";
$resultPrice = mysqli_query($conn, $queryPrice);
$data = array();
while ($row = mysqli_fetch_array($resultPrice)) {
$data[] = $row[0];
}
But instead I want it to choose the last 10 results in Ascending order. I found on other SO questions to use a subquery but every example I try gives no output and no error ??
Tried the below, DOESN'T WORK:
$queryPrice = "SELECT * FROM (SELECT price FROM price_chart ORDER BY id DESC LIMIT 10) ORDER BY id ASC";
$resultPrice = mysqli_query($conn, $queryPrice);
$data = array();
while ($row = mysqli_fetch_array($resultPrice)) {
$data[] = $row[0];
}
I also tried specifying the table name again and using the IN, also doesn't work:
$queryPrice = "SELECT price FROM price_chart IN (SELECT price FROM price_chart ORDER BY id DESC LIMIT 10) ORDER BY id";
$resultPrice = mysqli_query($conn, $queryPrice);
$data = array();
while ($row = mysqli_fetch_array($resultPrice)) {
$data[] = $row[0];
}
In both examples my array is blank instead of returning the last 10 results and there are no errors, so I must be doing the subquery wrong and it is returning 0 rows.
The subquery doesn't select the id column, so you can't order by it in the outer query. Also, MySQL requires that you assign an alias when you use a subquery in a FROM or JOIN clause.
$queryPrice = "SELECT *
FROM (SELECT id, price
FROM price_chart
ORDER BY id DESC LIMIT 10
) x ORDER BY id ASC";
$resultPrice = mysqli_query($conn, $queryPrice) or die (mysqli_error($conn));
$data = array();
while ($row = mysqli_fetch_assoc($resultPrice)) {
$data[] = $row['price'];
}
You would have been notified of these errors if you called mysqli_error() when the query fails.
Your second query is the closest. However you need a table alias. (You would have seen this if you were kicking out errors in your sql. Note you will need to add any field that you wish to order by in your subquery. In this case it is id.
Try this:
SELECT * FROM (SELECT price, id
FROM price_chart ORDER BY id DESC LIMIT 10) as prices
ORDER BY id ASC
You must have errors, because your SQL queries are in fact incorrect.
First, how to tell you have errors:
$resultPrice = mysqli_query (whatever);
if ( !$resultprice ) echo mysqli_error($conn);
Second: subqueries in MySQL need aliases. So you need this:
SELECT * FROM (
SELECT id, price
FROM price_chart
ORDER BY id DESC LIMIT 10
) AS a
ORDER BY id ASC";
See the ) AS a? That's the table alias.
I'm trying to return the row from my database, with the highest UID, where the URL column matches http://urltocheck.com.
I've tried all manner of things I can think of, and this is the closest I can get, but I'm getting an SQL syntax error.
My Table is called Adam, and I have the columns... UID (unique), URL (plus loads more). I'm trying to access the MySQL databse via PHP.
$query = "SELECT * FROM `Adam`
WHERE URL='http://urltocheck.com'
ORDER BY `UID` ASC;
LIMIT 1;";
Can anyone help please?
You shoul use order DESC and remove the ";" after ASC
$query = "SELECT * FROM `Adam`
WHERE URL='http://urltocheck.com'
ORDER BY `UID` DESC
LIMIT 1";
Try like this. Also, remove ; at this line ORDER BY UID ASC; (didn't noticed that earlier) because of which limit 1 not coming to picture.
SELECT * FROM `Adam`
WHERE URL='http://urltocheck.com'
and `UID` = (select max(`uid`) from `Adam`)
with the highest UID
You should order by UID desc and limit to 1.
You can also ORDER BY MAX ID.
<?php
$query = "SELECT * FROM `Adam`
WHERE URL='http://urltocheck.com'
ORDER BY MAX(`UID`) DESC;";
This is executed faster.
$query = "SELECT * FROM `Adam`
WHERE URL='http://urltocheck.com'
ORDER BY MAX(`UID`);";
?>
I have a database with rated items in it. When I want to display all the items from a specific category and sort the results by rate than by number of likes, it's easy and it's working:
$query = "SELECT * FROM infos WHERE category = '".$categories."'";
$query .= "ORDER BY `rate` DESC, `like` DESC";
The problem is when the results have the same value, they show up in alphabetical order. So, I would like to randomize the database before I sort it by rate and like. I just want to let the same chance to all items and those that have the same value not to be advantaged by alphabetical order.
I tried this, but it didn't work:
$query = "SELECT * FROM infos WHERE category = '".$categories."'";
$query .= "ORDER BY RAND(), ORDER BY `rate` DESC, `like` DESC";
Can somebody help me? I'm stuck.
You should turn it around:
SELECT * FROM infos WHERE category=...
ORDER BY rate DESC, like DESC, RAND();
That way it sorts by rate, then like and finally random if rate and like are equal.
Also your original ORDER BY with RAND() doesn't work because you use two ORDER BY clauses.
ok so i coded in a news section and everytime i insert new news,its shows below the old one.
i want to make it ORDER BY id but make it start like backwards.i dont know how to explain but yeh
i want them to be ordered by the newest added by the id so if the 1st row that was inserted's id is 1 then i want it to show below the next id.
so row with id = 2 will be here
so row with id = 1 will be here
thats how i want it to be, instead of it being like this
so row with id = 1 will be here
so row with id = 2 will be here
.
Sorry for my bad explanation i hope this is understandable
heres my code so far
<?php
require("include/config.php");
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id");
while($row = mysql_fetch_array($sqlnews)) {
$dbdate = $row['date'];
$dbnews = $row['news'];
echo "<h1><strong>$dbdate</strong></h1>";
echo "<div class='content'>$dbnews</div><br><br>";
}
?>
add DESC in your ORDER BY clause
SELECT * FROM news ORDER BY id DESC
by default, it is in ASC mode.
SELECT * FROM news ORDER BY id DESC
DESC is the descending keyword ASC is ascending
If you specify neither then default behaviour is ascending
Just use DESC keyword in your sql query.
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id DESC");
However, it isn't really such a good idea to use id, because semantically, there is nothing preventing somebody from changing the sequence which counts up automatically assigning the ID.
Therefore, you should add a column created_at. Everytime you insert a row, you can use the SQL function NOW().
The advantage is that you can say:
SELECT * FROM news WHERE created_at <= NOW() ORDER BY created_at DESC
This means that you can schedule news items ahead of time, and it will automatically display when the date/time arrives!
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id DESC");
try this:
just have to add
order by id DESC
Just replace your code by this code:
<?php
require("include/config.php");
$sqlnews = mysql_query("SELECT * FROM news ORDER BY id DESC");
while($row = mysql_fetch_array($sqlnews)) {
$dbdate = $row['date'];
$dbnews = $row['news'];
echo "<h1><strong>$dbdate</strong></h1>";
echo "<div class='content'>$dbnews</div><br><br>";
}
?>