I want to show the count of users which have the status 1 (see code) within PHP MySQL.
<?php
// something like this
$result = mysqli_query($mysqli, "SELECT COUNT(*) FROM users WHERE status = '1'");
echo "$result";
?>
Try this:
$query = "SELECT COUNT(*) as countvar FROM users where status = '1'";
$result = mysqli_query($con,$query);
$row = mysqli_fetch_array($result);
$count= $row['countvar '];
I am using the following to fetch the latest articles from the database:
//Latest Article
$title_query1 = "SELECT title FROM articles ORDER BY id DESC LIMIT 1";
$description_query1 = "SELECT description FROM articles ORDER BY id DESC LIMIT 1";
$content_query1 = "SELECT content FROM articles ORDER BY id DESC LIMIT 1";
$image_query1 = "SELECT image FROM articles ORDER BY id DESC LIMIT 1";
$title_result1 = mysqli_query($con, $title_query1) or die(mysqli_error($con));
$description_result1 = mysqli_query($con, $description_query1) or die(mysqli_error($con));
$content_result1 = mysqli_query($con, $content_query1) or die(mysqli_error($con));
$image_result1 = mysqli_query($con, $image_query1) or die(mysqli_error($con));
//Second Latest Article
$title_query2 = "SELECT title FROM articles ORDER BY id DESC LIMIT 2,1";
$description_query2 = "SELECT description FROM articles ORDER BY id DESC LIMIT 2,1";
$content_query2 = "SELECT content FROM articles ORDER BY id DESC LIMIT 2,1";
$image_query2 = "SELECT image FROM articles ORDER BY id DESC LIMIT 2,1";
$title_result2 = mysqli_query($con, $title_query2) or die(mysqli_error($con));
$description_result2 = mysqli_query($con, $description_query2) or die(mysqli_error($con));
$content_result2 = mysqli_query($con, $content_query2) or die(mysqli_error($con));
$image_result2 = mysqli_query($con, $image_query2) or die(mysqli_error($con));
However, i'm not sure how I can then do something like this:
<h1>Here is the first article: <?php $title_result1 ?><h1>
<h2>Here is the first article description: <?php $description_result1 ?>
<h1>Here is the second article: <?php $title_result2 ?><h1>
<h2>Here is the second article description: <?php $description_result2 ?>
Also, is this method not good? If I am going to do this for 100+ articles, will it cause the web page to load slowly?
Thanks
You do not need to do a single query for each column. You can get all columns for a row by doing select * Also, as mentioned, you can fetch as many rows as you want, and loop through them.
I prefer the while method.. Example, show last 100 articles
// fetch latest 100
$sql = "SELECT * FROM articles ORDER BY id DESC LIMIT 100";
if ($result = mysqli_query($con, $sql)){
// got results, convert result object to array called $row
while ($row = mysqli_fetch_array($result)) {
// echo out $row array elements with
//the column names from the database as array index
echo '<h2>'. $row['title'] .'</h2>'; // example wrap results with HTML
echo '<b>' .$row['description'] .'</b>';
echo $row['content'];
echo <img src="'. $row['image'] .'" title="'.$row['description'].'">';
echo '<br>'; //so next result is on new line
} // end of while loop
} else {
//no result, error
}
i have a question which is my limit statement didn't work i want the content select from database and limit the show content in 40 only but it didn't work
here is my SQL statement with php code
$chatroomID=$_GET['chatroomID'];
$userID = $_SESSION['id'];
$sql="SELECT * FROM chatroom_chat WHERE chatroom_id ='$chatroomID'";
$result1 = mysqli_query($connection, $sql) or die(mysqli_error($connection));
while ($row = mysqli_fetch_array($result1)) {
$chat = $row['chat_id'];
$sql3 ="SELECT * FROM (
SELECT * FROM chat WHERE id = '$chat' ORDER BY id DESC LIMIT 0,40
) sub
ORDER BY id ASC ";
$getChatData = mysqli_query($connection,$sql3) or die(mysqli_error($connection));
/*here have statement to get username*/
while($row3 = mysqli_fetch_array($getChatData)) {
echo "<div>all content</div>";
}
}
does my code have any syntax error? i no sure why it didn't work
SELECT * FROM (
SELECT * FROM chat WHERE id = '$chat' ORDER BY id DESC LIMIT 40
) sub
ORDER BY id ASC
I have a query that gets 5 lines of data like this example below
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
}
I want to run a query inside each results like this below
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
but for some reason it's only display the first line when I add the query inside it. I can seem to make it run all 5 queries.
SELECT
t1.ref,
t1.user,
t1.id,
t2.domain,
t2.title
FROM
table AS t1
LEFT JOIN anothertable AS t2 ON
t2.domain = t1.ref
LIMIT
0, 5
The problem is that inside the while-cycle you use the same variable $result, which then gets overridden. Use another variable name for the $result in the while cycle.
You change the value of your $query in your while loop.
Change the variable name to something different.
Ex:
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$qry = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$rslt = mysql_query($qry) or die(mysql_error());
if (mysql_num_rows($rslt) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
Use the following :
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query_domain = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result_domain = mysql_query($query_domain) or die(mysql_error());
if (mysql_num_rows($result_domain) )
{
$row_domain = mysql_fetch_row($result_domain);
$title = $row_domain['title'];
} else {
$title = "No Title";
}
echo "$ref - $title";
}
This is a logical problem. It happens that way, because you are same variable names outside and inside the loop.
Explanation:
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
// Now $results hold the result of the first query
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
//Using same $query does not affect that much
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
//But here you are overriding the previous result set of first query with a new result set
$result = mysql_query($query) or die(mysql_error());
//^ Due to this, next time the loop continues, the $result on whose basis it would loop will already be modified
//..............
Solution 1:
Avoid using same variable names for inner result set
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$sub_result = mysql_query($query) or die(mysql_error());
// ^ Change this variable so that it does not overrides previous result set
Solution 2:
Avoid the double query situation. Use joins to get the data in one query call. (Note: You should always try to optimize your query so that you will minimize the number of your queries on the server.)
SELECT
ref,user,id
FROM
table t
INNER JOIN
anothertable t2 on t.ref t2.domain
LIMIT 0, 5
Learn about SQL joins:
SELECT table.ref, table.user, table.id, anothertable.title
FROM table LEFT JOIN anothertable ON anothertable.domain = table.ref
LIMIT 5
You're changing the value of $result in your loop. Change your second query to use a different variable.
it is not give proper result because you have used same name twice, use different name like this edit.
$query = "SELECT ref,user,id FROM table LIMIT 0, 5";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ref = $row['ref'];
$query1 = "SELECT domain,title FROM anothertable WHERE domain = '$ref'";
$result1 = mysql_query($query1) or die(mysql_error());
if (mysql_num_rows($result1) )
{
$title = $row['title'];
} else {
$title = "No Title";
}
echo "$ref - $tile";
}
I am trying to show all of the data in the 'status' column of my table but am having troubles. What am I doing wrong:
<?php
$query1 = "SELECT id, status FROM alerts WHERE customerid='".$_SESSION['customerid']."' ORDER BY id LIMIT $start, $limit ";
$result = mysql_query($query1);
while ($row = mysql_fetch_array($result))
{
echo $row['status'] ;
}
?>
Try this:
$query1 = "SELECT id, `status` FROM alerts WHERE customerid='".$_SESSION['customerid']."' ORDER BY id LIMIT $start, $limit ";
$result = mysql_query($query1) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
echo $row['status'];
}
Also, make sure that:
$_SESSION['customerid'], $start and $limit are not empty. You can test the constructed query with echo $query1;
Note: Addition of mysql_error() in in the mysql_query will allow you to see if there is an error in the query.
I am trying to show all of the data in
the 'status' column of my table
If you want to show all the rows, your query should be:
$query1 = "SELECT id, `status` FROM alerts ORDER BY id";
But if you want to show for a specific customer, your query should be:
$query1 = "SELECT id, `status` FROM alerts WHERE customerid='".$_SESSION['customerid']."' ORDER BY id";