Group the rows by date, use COUNT(*) to get the count of rows in each group, and then use ORDER BY and LIMIT to get the highest count. SELECT DATE(timestampColumn) AS date, COUNT(*) AS count FROM yourTable GROUP BY date ORDER BY count DESC LIMIT 1 9 solved How to get the count of day … Read more
There’s no such thing as “sort by” in SQL. ORDER BY lets you establish the sort order of your result set. GROUP BY means something entirely different. You should read about that. solved difference between order by and sort by
You don’t need to store image directly into to database. Just store the image name in the database and fetch it when you want to show. For e.g. $target_dir = “uploads/”; $target_file = $target_dir . basename($_FILES[“fileToUpload”][“name”]); $uploadOk = 1; $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION); // Check if image file is a actual image or fake image if(isset($_POST[“submit”])) … Read more
I can’t explain this Let me explain it for you. You are looking for pivoting the comments values for the concern for each student. Unfortunately, MySQL has no pivot table operator. However, you can use the CASE expression to do this. Like so: SELECT student_name, MAX(CASE WHEN concern = ‘Academics’ THEN comments END) AS ‘Accademics’, … Read more
Do not name the field order that is a reserved word in MySQL and it is a hassle to use reserved words as field names. Let’s name it rank. — first add new field. alter table table_name add column rank int(3); — then fill the field with a rank. set @rank = 0; set @ref … Read more
Yes you are using a prepared statement with a parameter. That’s the right thing to do. Parameters are the best way to write safe SQL statements in the majority of cases. There are just a few edge cases where they don’t help (see my answer to how safe are PDO prepared statements) I can suggest … Read more
\DB::table(‘cars’) ->join(‘garage’, ‘garage.id’, ‘=’, ‘cars.garage’) ->where(‘garage.name’, ‘main’) The above solves the garage and cars part, but you never specified what is aliased to p. If p is a different table then you need to add another call to join() and making the following \DB::table(‘cars’) ->join(‘garage’, ‘garage.id’, ‘=’, ‘cars.garage’) //<Other table join goes here for table … Read more
Do you mean this? $data[$your_date_field_index] = date(‘Y-m-d H:i:s’, strtotime($data[$your_date_field_index])); 1 solved Change value in an array [closed]
You should better a try var_dump($row) or print_r($row) first to see if the array contains the datas at right keys. $lang[$row[0]] doesnt work because $row[0] is empty. Your are probably not assigning Hello to $row[0]. So try print_r($row); to see whats stored in the whole array. Add : $lang[$row[0]] will give you $lang[hello]. It should … Read more
$con = mysqli_connect(“localhost”, “root”, “”, “zoo”); if (mysqli_connect_error()) { echo “Failed to connect” . mysqli_connect_error(); } $sel = “select (quantity* bill) as total from animal inner join price on animal.id=price.animal_id group by price.animal_id”; $result = mysqli_query($con, $sel); while ($row = mysqli_fetch_array($result)) { echo “<tr>”; echo”<td>” . $row[‘total’] . “</td>”; } Before copy and paste check … Read more
There are multiple issues with your current code. You fetch the first row with $row = mysqli_fetch_array($select); (line 3), but you don’t do anything with it. This means that the first result is discarded. Your while loop attempts to loop over an incorrect variable ($query is a string, not the result-object), and you’ve quoted it … Read more
Change your code following way- <?php $date2=date(‘Y’, strtotime(‘+1 Years’)); for($i=date(‘Y’); $i<$date2+5;$i++){ echo ‘<option>’.$i.’-‘.($i+1).'</option>’; } ?> solved PHP How to generate School Year?
SELECT cities.cityname, states.statename, countryname FROM cities JOIN states ON cities.states_id = states.id JOIN country ON states.country_id = country.id WHERE cities.id=3; Assuming 3 is the id of the city you are searching for… solved I need a sql query to view data
$values = explode(PHP_EOL, $row); foreach($values AS $data){ echo ‘<li>’.$data.'</li>’; } You should normalize this data, it’s not the right way to store it. 0 solved PHP/MySQL adding before and after every row
This is how to get the rows based on the row number: SET @rownum = 0; Select sub.*, sub.rank as Rank FROM ( Select *, (@rownum:=@rownum+1) as rowNum FROM Categories ) sub WHERE rowNum IN (4, 6) However, Mysql has a built in function ORDER BY LIMIT to do this. 1 solved How to print … Read more