I'm quite new to working with PHP and databases. So, like all beginners, I'm having problems with the mysqli extension. I was trying to avoid using mysqli_results, as I didn't want to deal with an array and a loop every time I want a simple piece of data. But that might not be possible.
I need to echo $user_count, but nothing seems to be stored there. My code seems to be okay according to the API, but maybe I'm just trying to use the wrong functions altogether. How do I put the result I need into $user_name?
mysqli_query($conn, "SELECT * FROM `wp_users` WHERE 1");
$result_user_count = mysqli_query($conn, "SELECT COUNT(`ID`) FROM `wp_users`");
$user_count = mysqli_fetch_field_direct($result_user_count, 0);
echo $user_count . ' users found on ' . $dbname . ':</br>';
Based on your provided code, you need to change it like below:-
$query = mysqli_query($conn, "SELECT COUNT(ID) AS COUNT FROM wp_users") or die(mysqli_error($conn));
while ($row = mysqli_fetch_assoc($query)) {
echo "Total Users Are:- ". $row["COUNT"];
}
Note:- try to add mysqli error reporting code so that you will what problem actually exist in your query code, and you can easily resolve them. thanks
Try this. comment if not worked.
$query = mysqli_query($conn, "SELECT COUNT(ID) AS COUNT FROM wp_users");
if ($u_count = $mysqli->query($query)) {
while ($res = $u_count->fetch_assoc()) {
echo "Total Users: ". $res["COUNT"];
}
}
Related
I cannot get "LIKE %M" to work on XAMPP with MariaDB.
$sql = "SELECT * FROM mytable WHERE names LIKE 'm%' ORDER by names ASC";
$result = mysqli_query($connection, $sql) or die("Error in selecting " .
mysqli_error($connection));
$emparray = array();
while($row = mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
this returns a blank screen, no errors or anything.
but if i switch to something like this
$sql = "SELECT * FROM mytable WHERE names = 'SomeName' ORDER by names ASC";
it runs just fine, so I know its not a problem connecting to the DB.
And I have researched through this site for the answer first, so I am pretty confident that my code is correct.
just not sure if this is a problem with XAMPP and/or MariaDB, or if I just missed something stupid.
Thanks in advance, any advice would be appreciated.
figured it out....
one of my names had an umlaut over the O (Ö).
I didn't realize it was there, but once I noticed it and removed it.. it worked just fine.
I am a novice when it comes to PHP but I don't understand if my syntax is wrong in this statement, or how would I grab an int from my MySQL server.
I know that my server credentials are working fine. How would I fix this statement to give me a returned integer of the number of reviews in the userinfo table?
$numberofpreviousreviews = mysql_query("SELECT `number_of_reviews` FROM `userinfo`") or die(mysql_error()); //Check to see how many reviews user has previously created
$amountofreviews = $numberofpreviousreviews + 1;
$query2 = mysql_query("ALTER TABLE userinfo ADD `amountofreviews` VARCHAR(10000)") or die(mysql_error()); //Make another column in database for the new review
You need to fetch your results after you run your query. There are several ways to do this but using mysql_fetch_assoc() will work for you.
$numberofpreviousreviews = mysql_query("SELECT `number_of_reviews` FROM `userinfo`") or die(mysql_error()); //Check to see how many reviews user has previously created
$row = mysql_fetch_assoc($numberofpreviousreviews);
$amountofreviews = $row['number_of_reviews'] + 1;
FYI, you shouldn't be using mysql_* functions anymore. They are deprecated and going away. You should use mysqli or PDO.
Assume you have a table userinfo which has the following structure and data :
Scenario #1 :
If you want to retrieve the all number_of_reviews, then do like this,
$query = "SELECT `number_of_reviews` FROM `userinfo`";
$result = mysqli_query($db,$query);
while ($row = mysqli_fetch_assoc($result)) {
echo "Number of reviews : " . $row['number_of_reviews'] . "<br/>";
}
It will give you,
Number of reviews : 20
Number of reviews : 40
Since, the result has many rows, it will display like above.
Scenario #2:
If you want to retrieve only the specific number_of_reviews for some user id (which is unique). I take id as 1 as a example here. Then do like,
$query2 = "SELECT `number_of_reviews` FROM `userinfo` WHERE `id` = 1";
$result2 = mysqli_query($db,$query2);
while ($row2 = mysqli_fetch_assoc($result2)) {
echo $row2['number_of_reviews'] . "<br/>";
}
This will print,
20.
Because, number_of_reviews is 20 for id 1.
I am having problems trying to get these queries with a WHERE clause to work. I have two tables which look like this :
What I am trying to do is return the genre that each film has. At the moment no data is returning at all from what I can see. Here are the two queries:
$film_id = $row_movie_list['film_id'];
mysql_select_db($database_fot , $fot);
$query_get_genre = "SELECT * FROM film_genre WHERE `id_film` ='". $film_id. "'";
$get_genre = mysql_query($query_get_genre, $fot) or die(mysql_error());
$row_get_genre = mysql_fetch_assoc($get_genre);
$totalRows_get_genre = mysql_num_rows($get_genre);
$genre_id = $row_get_genre['id_genre'];
mysql_select_db($database_fot , $fot);
$query_genre = "SELECT * FROM genre WHERE `id_genre` ='". $genre_id. "'";
$genre= mysql_query($query_genre, $fot) or die(mysql_error());
$row__genre = mysql_fetch_assoc($genre);
$totalRows_genre = mysql_num_rows($genre);
PHP with content area. I fairly new to PHP so any help would be appreciated.
<?php do { echo $genre['genre']; } while($row_get_genre = mysql_fetch_assoc($get_genre)); ?>
Update: I am now able to get first genre but not second it just echos the first one twice and I have tried but still no luck:
do {do { echo $row_genre['genre']; } while($row_genre = mysql_fetch_assoc($genre));} while($row_get_genre = mysql_fetch_assoc($get_genre)); ?>
Avoiding the fact that you're using a deprecated way to establish connection and interact with MySQL, what you're doing is getting a single relation genre-film and then getting the row of the genre that matches. You should surround part of your code with a while that executes while it's still genres of the film with id. Something like:
$film_id = $row_movie_list['film_id'];
mysql_select_db($database_fot , $fot);
$query_get_genre = "SELECT * FROM film_genre WHERE `id_film` ='". $film_id. "'";
$get_genre = mysql_query($query_get_genre, $fot) or die(mysql_error());
while($row_get_genre = mysql_fetch_assoc($get_genre)){
$genre_id = $row_get_genre['id_genre'];
$query_genre = "SELECT * FROM genre WHERE `id_genre` ='". $genre_id. "'";
$genre= mysql_query($query_genre, $fot) or die(mysql_error());
$row__genre = mysql_fetch_assoc($genre);
// You should do whatever you want to do with $row__genre here. Otherwise it will be cleared.
}
I must insist this is a deprecated and insecure way of communication with a MySQL Database. I recommend you read about MySQLi or PDO extensions.
MySQLi: http://www.php.net/manual/en/book.mysqli.php
PDO: http://www.php.net/manual/en/book.pdo.php
Was hoping someone could give me some help. I have been trying to learn pagination but have run into an issue. when I run this to get my total rows count:
$sql = "SELECT COUNT (*) FROM item WHERE fid='17'";
$query = mysqli_query($con, $sql);
$row = mysqli_fetch_row($query);
$rows = $row[0];
$rows comes back with no value, I am under the impression that $rows should contain the total number of records from item with the fid of 17, when I run SELECT COUNT (*) FROM item WHERE fid='17' in phpmyadmin it returns 98 which is the correct count. Directly before the above code I use this code to connect to the db, which I use again later to display the records which works fine.
$con=mysqli_connect("$host","$username","$password","$dbname");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
This statement displays records later in the script
$sql = mysqli_query($con,"SELECT * FROM item WHERE fid='17' ORDER BY id DESC $limit ");
So there is data and the info is correct.
I have been following this tutorial on the subject http://www.developphp.com/view.php?tid=1349 and it works like a charm on his example in the video, all I have changed is the database particulars to mine. Can't figure this one out, been stuck for days so now I am bothering you fine folks.
Update: Changed
$query = mysqli_query($con, $sql);
$row = mysqli_fetch_row($query);
$rows = $row[0];
to
if ($result=mysqli_query($con,$sql))
{
// Return the number of rows in result set
$rows=mysqli_num_rows($result);
// Free result set
mysqli_free_result($result);
}
And everything is working now. Still wish I knew what was wrong with my original code but I will be moving on. Thanks Phil Perry for all your help this morning!
Well, you could SELECT * FROM... and then call mysqli_num_rows($query). Probably slower than SELECT count(*). BTW, I presume that item is not a reserved word in SQL. You can always wrap table and field names in backticks (not quotes ' or "). You could also try SELECT count(*) AS myCount....
try this,
$sql = "SELECT COUNT(*) FROM item WHERE fid='17'";
$query = mysqli_query($sql);
$row = mysqli_fetch_array($query);
$rows = $row[0];
I am trying to print out some topic information, but it is not going so well. This is my query:
SELECT * FROM topics WHERE id='$read'
This doesn't work. I've echo'ed the $read variable, it says 1. So then if I do like this:
SELECT * FROM topics WHERE id='1'
It works perfectly. I don't get what is the problem. There's no hidden characters in $read or anything else like that.
Try like this:
$query = "SELECT * FROM topics WHERE id='" . $read . "'"
ID is normally a numeric field, it should be
$id = 1;
$query = "SELECT * FROM topics1 WHERE id = {id}"
If you are using strings for some reason, fire a query like
$id = '1';
$query = "SELECT * FROM topics1 WHERE id = '{$id}'"
SELECT * FROM topics WHERE id=$read
it consider it as string if you put i single quotes
I wonder why all the participants didn't read the question that clearly says that query with quotes
SELECT * FROM topics WHERE id='1'
works all right.
As for the question itself, it's likely some typo. Probably in some other code, not directly connected to $read variable
try
$query = sprintf("SELECT * FROM topics WHERE id='%s';",$read);
Also remember to escape the variable if needed.
Looks like you might have an issue with the query generation as everyone else is pointing to as well. As Akash pointed out it's always good to build your query in to a string first and then feed that string to the MySQL API. This gives you easy access to handy debugging techniques. If you are still having problems try this.
$id = 1;
$query = "SELECT * FROM `topics1` WHERE `id`={$id}";
echo ": Attempting Query -> {$query}<br />";
$res = mysql_query($query, $dblink);
if($res <= 0)
die("The query failed!<br />" . mysql_error($dblink) . "<br />");
$cnt = mysql_num_rows($res);
if($cnt <= 0)
{
$query = "SELECT `id` FROM `topics1`";
echo "No records where found? Make sure this id exists...<br />{$query}<br /><br />";
$res = mysql_query($query, $dblink);
if($res <= 0)
die("The id listing query failed!<br />" . mysql_error($dblink) . "<br />");
while($row = mysql_fetch_assoc($res))
echo "ID: " . $row['id'] . "<br />";
}
This will at least let you monitor between calls, see what your query actually looks like, what mysql says about it and if all else fails make sure that the ID you are looking for actually exists.
try with this : SELECT * FROM topics WHERE id=$read