How to use GROUP BY method in php-mysql - php

Question

I’m trying to use "group by" instead of "DISTINCT" in my php file to select some rows that all of them have an specific column value and it’s "idchat".
And I want to get more than one columns
please help me!
I’ve checked every pages but I didn’t understand enything
<?php
$connection = mysqli_connect("localhost","---","pass","---");
$id = $_GET["id"];
$mobile = $_GET["mobile"];
$idchat = $_GET["idchat"];
if (strpos($mobile, '9') !== false) {
$query = "SELECT DISTINCT a,b,c,d,idchat FROM database where mobile = '$mobile' ORDER BY id DESC";
$result = mysqli_query($connection,$query);
while ($row = mysqli_fetch_assoc($result)) {
$array[] = $row;
}
header('Content-Type:Application/json');
echo json_encode($array);
}
mysqli_close($connection);?>
and this code gives me this error:
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in /home/---/test.php on line 12

Answer 1

See next output:
mysql> SELECT * FROM database;
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'database' at line 1
mysql> SELECT * FROM mytable;;
ERROR 1146 (42S02): Table 'test.mytable' doesn't exist
ERROR:
No query specified
mysql>
The first query is not understood because database is a reserved word. You should not use that to name a table. A work around is to add backquotes around the name:
mysql> SELECT * FROM `database`;
ERROR 1146 (42S02): Table 'test.database' doesn't exist
mysql>

Answer 2

"Could you correct my codes please?": Ok, but untested:
<?php
$connection = mysqli_connect("localhost","---","pass","---");
$id = $_GET["id"];
$mobile = $_GET["mobile"];
$idchat = $_GET["idchat"];
if (strpos($mobile, '9') !== false) {
$query = "SELECT DISTINCT a,b,c,d,idchat,id FROM `database` where mobile = '$mobile' ORDER BY id DESC";
$result = mysqli_query($connection,$query);
if ($result) {
while ($row = mysqli_fetch_assoc($result)) {
$array[] = $row;
}
header('Content-Type:Application/json');
echo json_encode($array);
}
else {
echo mysqli_error($connection);
}
mysqli_close($connection);?>
}
I also did add id to the query because of comment from #Raymond