问题
I am a newbie to PHP and I am stuck at a certain point. I tried looking up a solution for it however, I didn't find exactly what I need.
My goal is to create a leaderboard, in which the values are displayed in descending order plus the rank and score are displayed. Furthermore, it should also display whether or not a tie is present.
The database should look like this:
+---------+------+----------------+-------+------+
| user_id | name | email | score | tied |
+---------+------+----------------+-------+------+
| 1 | SB | sb@gmail.com | 1 | 0 |
+---------+------+----------------+-------+------+
| 2 | AS | as@web.de | 2 | 0 |
+---------+------+----------------+-------+------+
| 3 | BR | br@yahoo.com | 5 | 1 |
+---------+------+----------------+-------+------+
| 4 | PJ | pj@gmail.com | 5 | 1 |
+---------+------+----------------+-------+------+
And the outputted table should look something like this:
+------+-------------+-------+------+
| rank | participant | score | tied |
+------+-------------+-------+------+
| 1 | BR | 5 | Yes |
+------+-------------+-------+------+
| 2 | PJ | 5 | Yes |
+------+-------------+-------+------+
| 3 | AS | 2 | No |
+------+-------------+-------+------+
| 4 | SB | 1 | No |
+------+-------------+-------+------+
I managed to display the rank, participant and the score in the right order. However, I can't bring the tied column to work in the way I want it to. It should change the value, whenever two rows (don't) have the same value.
The table is constructed by creating the <table> and the <thead> in usual html but the <tbody> is created by requiring a php file that creates the table content dynamically.
As one can see in the createTable code I tried to solve this problem by comparing the current row to the previous one. However, this approach only ended in me getting a syntax error. My thought on that would be that I cannot use a php variable in a SQL Query, moreover my knowledge doesn't exceed far enough to fix the problem myself. I didn't find a solution for that by researching as well.
My other concern with that approach would be that it doesn't check all values against all values. It only checks one to the previous one, so it doesn't compare the first one with the third one for example.
My question would be how I could accomplish the task with my approach or, if my approach was completely wrong, how I could come to a solution on another route.
index.php
<table class="table table-hover" id="test">
<thead>
<tr>
<th>Rank</th>
<th>Participant</th>
<th>Score</th>
<th>Tied</th>
</tr>
</thead>
<tbody>
<?php
require("./php/createTable.php");
?>
</tbody>
</table>
createTable.php
<?php
// Connection
$conn = new mysqli('localhost', 'root', '', 'ax');
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// SQL Query
$sql = "SELECT * FROM names ORDER BY score DESC";
$result = $conn->query("$sql");
// Initalizing of variables
$count = 1;
$previous = '';
while($row = mysqli_fetch_array($result)) {
$current = $row['score'];
$index = $result['user_id']
if ($current == $previous) {
$update = "UPDATE names SET tied=0 WHERE user_id=$index";
$conn->query($update);
}
$previous = $current;
?>
<tr>
<td>
<?php
echo $count;
$count++;
?>
</td>
<td><?php echo $row['name'];?></td>
<td><?php echo $row['score'];?></td>
<td>
<?php
if ($row['tied'] == 0) {
echo 'No';
} else{
echo 'Yes';
}
?>
</td>
</tr>
<?php
}
?>
回答1:
I think the problem is here
$index = $result['user_id'];
it should be
$index = $row['user_id'];
after updating tied you should retrieve it again from database
回答2:
instead of the update code you've got use something simular
$query = "select score, count(*) as c from names group by score having c > 1";
then you will have the scores which have a tie, update the records with these scores and your done. Make sure to set tie to 0 at first for all rows and then run this solution
UPDATE for an even faster solution sql based:
First reset the database:
$update = "UPDATE names SET tied=0";
$conn->query($update);
All records have a tied = 0 value now. Next update all the records which have a tie
$update = "update docs set tied = 1 where score IN (
select score from docs
group by score having count(*) > 1)";
$conn->query($update);
All records with a tie now have tied = 1 as we select all scores which have two or more records and update all the records with those scores.
回答3:
So I solved my question by myself, by coming up with a different approach.
First of all I deleted this part:
$current = $row['score'];
$index = $result['user_id']
if ($current == $previous) {
$update = "UPDATE names SET tied=0 WHERE user_id=$index";
$conn->query($update);
}
$previous = $current;
and the previous variable.
My new approach saves the whole table in a new array, gets the duplicate values with the array_count_values() method, proceeds to get the keys with the array_keys() method and updates the database via a SQL Query.
This is the code for the changed part:
// SQL Query
$sql = "SELECT * FROM names ORDER BY score DESC";
$result = $conn->query("$sql");
$query = "SELECT * FROM names ORDER BY score DESC";
$sol = $conn->query("$query");
// initalizing of variables
$count = 1;
$data = array();
// inputs table into an array
while($rows = mysqli_fetch_array($sol)) {
$data[$rows['user_id']] = $rows['score'];
}
// -- Tied Column Sort --
// counts duplicates
$cnt_array = array_count_values($data);
// sets true (1) or false (0) in helper-array ($dup)
$dup = array();
foreach($cnt_array as $key=>$val){
if($val == 1){
$dup[$key] = 0;
}
else{
$dup[$key] = 1;
}
}
// gets keys of duplicates (array_keys()) and updates database accordingly ($update query)
foreach($dup as $key => $val){
if ($val == 1) {
$temp = array_keys($data, $key);
foreach($temp as $k => $v){
$update = "UPDATE names SET tied=1 WHERE user_id=$v";
$conn->query($update);
}
} else{
$temp = array_keys($data, $k);
foreach($temp as $k => $v){
$update = "UPDATE names SET tied=0 WHERE user_id=$v";
$conn->query($update);
}
}
}
Thank you all for answering and helping me get to the solution.
来源：https://stackoverflow.com/questions/46019325/php-compare-column-values-and-edit-database-accordingly