php mysql random rows no duplicates - php

Hi im trying to get 5 random rows from a database and then display them. i currently do this but it does result in duplicates.
i need to change the limit to 5 and then store them in an array but how do i do that? or is there a better way?
function GetPlayer($link){
if (isset($_SESSION['username'])) {
$x = 0;
while($x <= 5) {
$sql = "SELECT * FROM userstats ORDER BY RAND() LIMIT 1; ";
$result = mysqli_query($link,$sql);
$row = mysqli_fetch_assoc($result);
if($row['username'] !== $_SESSION['username']){//add so it dosent put duplicates
echo ("<tr>");
echo ("<th>".$row['username']." </th>");
echo ("<th>Level: ".$row['Level']." </th>");
echo ("<th>Player Stats:".$row['Attack']."/".$row['Defence']." </th>");
echo ("<th>Win Chance: ");
echo CalculateWinChance($link,$row['Defence']);
echo ("<th><input type ='submit' name = 'Attack_Btn' value ='Attack'></th>");
echo ("</tr>");
$x++;
}
}
}
}

Why dont't you try to request 5 results (LIMIT 5) AND loop this? It will no return any duplicates. Four queries less would be a side effect.
$sql = "SELECT * FROM userstats ORDER BY RAND() LIMIT 5; ";
while($row = mysqli_fetch_assoc($result)){
...
}

Instead of calling the query 5 times within the loop, you should have a single query. That's the most optimal approach.
You can use GROUP BY clause to select unique rows.
Your query should be like the following:
SELECT *
FROM userstats
GROUP BY username
ORDER BY RAND()
LIMIT 5

You'll want to sanitize the inputs in the query and/or use a prepared statement, but this should get you pretty close to what you want:
$sql = 'SELECT * FROM userstats WHERE username != ? GROUP BY username ORDER BY RAND() LIMIT 5';
$stmt = $mysqli->prepare($sql);
$stmt->bind_param('s', $_SESSION['username']);

Related

How to mysqli_fetch_row while using mysqli_multi_query

I'm trying to get the latest info about some specific person, and I'm using a query like
SELECT * FROM Table WHERE Name LIKE 'Peter' ORDER BY ID DESC LIMIT 1
and
SELECT * FROM Table WHERE Name LIKE 'Mary' ORDER BY ID DESC LIMIT 1
because in the Table each day will insert new data for different person at the instant of updating it (for record reference), so I would like to print out a few persons latest info by "ORDER BY ID DESC LIMIT 1"
I have tried to print it out with "mysqli_multi_query" and "mysqli_fetch_row"
like
$con=mysqli_connect($localhost,$username,$password,'Table');
$sql = "SELECT * FROM Table WHERE Name LIKE 'Peter' ORDER BY ID
DESC LIMIT 1 ";
$sql .= "SELECT * FROM Table WHERE Name LIKE 'Mary' ORDER BY ID
DESC LIMIT 1";
// Execute multi query
if (mysqli_multi_query($con,$sql))
{
do
{
// Store first result set
if ($result=mysqli_store_result($con)) {
// Fetch one and one row
while ($row=mysqli_fetch_row($result))
{
echo '<tr>'; // printing table row
echo '<td>'.$row[0].'</td>';
echo '<td>'.$row[1].'</td>';
echo '<td>'.$row[2].'</td>';
echo '<td>'.$row[3].'</td>';
echo '<td>'.$row[4].'</td>';
echo '<td>'.$row[5].'</td>';
echo '<td>'.$row[6].'</td>';
echo '<td>'.$row[7].'</td>';
echo '<td>'.$row[8].'</td>';
echo '<td>'.$row[9].'</td>';
echo '<td>'.$row[10].'</td>';
echo '<td>'.$row[11].'</td>';
echo '<td>'.$row[12].'</td>';
echo '<td>'.$row[13].'</td>';
echo '<td>'.$row[14].'</td>';
echo'</tr>'; // closing table row
}
// Free result set
mysqli_free_result($result);
}
}
while (mysqli_next_result($con));
}
mysqli_close($con);
?>
In the result page , it doesn't show any error message , but no results are printed.
The individual queries were tested.
Please advise, much thanks
Is there another way to keep the query simple, so there is no need to use mysqli_multi_query?
Best practice indicates that you should always endeavor to make the fewest number of calls to the database for any task.
For this reason, a JOIN query is appropriate.
SELECT A.* FROM test A INNER JOIN (SELECT name, MAX(id) AS id FROM test GROUP BY name) B ON A.name=B.name AND A.id=B.id WHERE A.name IN ('Peter','Mary')
This will return the desired rows in one query in a single resultset which can then be iterated and displayed.
Here is an sqlfiddle demo: http://sqlfiddle.com/#!9/2ff063/3
P.s. Don't use LIKE when you are searching for non-variable values. I mean, only use it when _ or % are logically required.
This should work for you:
$con = mysqli_connect($localhost,$username,$password,'Table');
// Make a simple function
function personInfo($con, $sql)
{
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_array($result))
{
echo '<table>
<tr>';
for($i=0; $i < count($row); $i++) echo '<td>'.$row[$i].'</td>';
echo '</tr>
</table>';
}
}
}
$sql1 = "SELECT * FROM Table WHERE Name='Peter' ORDER BY ID DESC LIMIT 1 ";
$sql2 = "SELECT * FROM Table WHERE Name='Mary' ORDER BY ID DESC LIMIT 1";
// Simply call the function
personInfo($con, $sql1);
personInfo($con, $sql2);

How to echo result from query

Maybe I am overthinking this, but I have narrowed my query to find a row down to 1 result that I need, and it will not display. Wondering if someone could tell me what I am doing wrong.
$result = mysqli_query($link, "SELECT pageid FROM article ORDER BY id DESC LIMIT 1");
$row = mysqli_use_result($result);
echo $row;
I have it selecting the last row and supplying me with the stored data from the pageid of the last row.
I had to adapt my code. I believe it was because I use mysql. However, this code will work if you use mysqli
$pageid = "SELECT pageid FROM articles ORDER BY id DESC LIMIT 1";
$resultpageid = $link->query($pageid);
if ($resultpageid->num_rows > 0) {
while ($row = $resultpageid->fetch_assoc()) {
$pagenumber = $row["pageid"];
}
} else {
echo "0 results";
}
mysqli doesn't have any function to get a single column from a single row. You need to use one of the fetch methods e.g. fetch_array(). You don't need any loop if you use LIMIT 1.
Just fetch a single row and get the column from the returned array:
$pageid = "SELECT pageid FROM articles ORDER BY id DESC LIMIT 1";
$resultpageid = $link->query($pageid);
$row = $resultpageid->fetch_assoc();
// or
$row = $resultpageid->fetch_array();
if ($row) {
echo $row["pageid"];
} else {
echo "No record found!";
}

shuffle : Display only one row at the same time

How to display only one row at random at the same time from DB. Everything works fine, but all rows are displayed. thanks
<?php
$sql = "SELECT id,name FROM table ";
$rows = array();
$result = $objCon->query($sql);
while($row = $result->fetch_assoc())
{
$rows[] = $row;
}
shuffle($rows);
echo '<ol>';
foreach($rows as $row)
{
echo '<li><h3>'.$row['id'].' = '.$row['name'].'</h3></li>';
}
echo '</ol>';
?>
Change your SQL request:
SELECT id,name FROM table ORDER BY RAND() LIMIT 1;
You can do it using PHP:
....
shuffle($rows);
$randomRow = reset($rows);
....
But the better way is to change your SQL query:
$query = "SELECT id, name FROM table ORDER BY RAND() LIMIT 1;"
<?php
$sql = "
SELECT id, name
FROM table
ORDER BY RAND()
LIMIT 1 ";
$result = mysql_query($sql);
// As you are only return a single row you do you require the while()
$row = mysql_fetch_array($result);
echo '<ol>';
echo '<li><h3>'.$row['id'].' = '.$row['name'].'</h3></li>';
echo '</ol>';
?>
By adding an ORDER BY RAND() in your sql query you are asking MySQL to randomly order the results then at a LIMIT to restrict the number of rows you would like returned.
The example code is written based on selecting a single row. If you would like more, e.g. 5, you will need to add a while loop.

Select Random Value MYSQL But Not Current Value

I have the following code to pull a random value from a database.
$result = mysqli_query($con,"SELECT column1 FROM table1
ORDER BY RAND()
LIMIT 1");
while($row = mysqli_fetch_array($result))
{
echo $row['column1'];
echo "<br>";
}
At the moment I only have about 10 values in total. Sometimes they appear more than once in a row upon refresh.
Can the query be amended to show a random value that is not the current value?
in order to do that you need to somehow remember the last row.
Then do something like this
$result = mysqli_query($con,"SELECT column1 FROM table1
where id_of_your_row != ".mysqli_real_escape_string($id_of_your_last_row)."
ORDER BY RAND()
LIMIT 1");
For remembering the last row, you could use sessions.
Your whole code could look like this.
// sessions need to bestarted
$result = mysqli_query($con,"SELECT column1 FROM table1
where column1 != ".(isset($_SESSION["lastid"]) ? mysqli_real_escape_string($_SESSION["lastid"]) : '')." ORDER BY RAND()
LIMIT 1");
while($row = mysqli_fetch_array($result))
{
echo $row['column1'];
$_SESSION["lastid"] = $row['column1'];
echo "<br>";
}

echo random id numbers from mysql database without repeating numbers?

How do I echo random id numbers from mysql database without repeating numbers?
this is my sample code:
$query = mysql_query("SELECT * FROM store");
$number=mysql_num_rows($query);
for ($count=1; $count<= $number ; $count++)
{
$id = mysql_query ("SELECT id FROM store ORDER BY RAND() LIMIT $number");
$id = mysql_fetch_assoc($id);
$id = $id['id'];
echo $id;
}
It will echo six random numbers but have instances like "1 1 3 2 4 5" where 1 is echoed twice instead of once. thanks in advance
Just order your results by rand and limit their number, your id has to be unique :
SELECT * FROM store ORDER BY RAND() LIMIT 0,6
The Problem is, that you do a SELECT inside of the loop, instead of selecting once and loop over the result.
$query = mysql_query("SELECT * FROM store");
$number=mysql_num_rows($query);
$result = mysql_query ("SELECT id FROM store ORDER BY RAND() LIMIT $number");
while ($row = mysql_fetch_assoc($result)) {
echo $row["id"];
}
BTW: SELECT * to get the number of recordsets is ugly, use SELECT count(id)instead
If you're coming out of php you're probably better off (faster, easier, no locking issues) to randomize your numbers there. And SQL queries inside loops is an antipattern.

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