how do include multiple pieces of data on one page? - php

I cant figure out how to get multiple entries of data from my database. You can see below that I have newsID = 1, which allows me to get that to show on my database, but how should I change the code to be able to access other entries like newsID = 2 as well?
<?php
$queryAffinity = "SELECT * FROM affnews WHERE newsID = 1";
$stmt = $pdo->query($queryAffinity);
$row = $stmt->fetchObject();
?>
<?php if($row->newsID == 1) echo "<p>{$row->newsDescription}</p>"; ?>
this is the code im trying to use to echo the id 1

Edit your WHERE-Clause and put it in a loop. For example:
$queryAffinity = "SELECT * FROM affnews WHERE newsID = 1 OR newsID = 2";
$stmt = $pdo->query($queryAffinity);
while($row = $stmt->fetchObject())
{
//do something with $row
}

You Can use the following code:
$queryAffinity = "SELECT * FROM affnews WHERE newsID IN (1, 2)";
$stmt = $pdo->query($queryAffinity);
while($row = $stmt->fetchObject())
{
echo $row->id;
}

Related

My for loop only allows one post to be displayed, over and over again

/* To sort the id and limit the post by 40 */
$sql = "SELECT * FROM requests";
$result = $conn->query($sql);
$sqlall= "SELECT * FROM requests ";
$resultall = $conn->query($sqlall);
$i = 0;
if ($result->num_rows > 0) {
// Output data of each row
$idarray= array();
while($row = $result->fetch_assoc()) {
echo "<br>";
// Create an array to store the
// id of the blogs
array_push($idarray,$row['id']);
}
}
else {
echo "0 results";
}
?>
<?php
for($x = 1; $x < 40; $x++) {
// This is the loop to display all the stored blog posts
if(isset($x)) {
$query = mysqli_query(
$conn,"SELECT * FROM `requests`");
$res = mysqli_fetch_array($query);
$email1 = $res['email1'];
$msg1= $res['msg1'];
$subject1 = $res['subject1'];
$name1 = $res['name1'];
$id = $res['id'];
the output is 40 cards reading data from the first row in my database. can anyone help?
I'm using xampp.
This code is to show the loop, but if anyone wants the full code is here
You are storing all the IDs in the array $idarray, but then you don't really use them properly. You loop over them, but you just run SELECT * FROM requests` 40 more times, and always extract the same first row. You never use the ID to change the query.
But it really makes no sense to run lots of separate queries anyway. If you just want the first 40 rows then use MySQL's LIMIT keyword. It usually works best when combined with ORDER BY as well. Something like this:
$sql = "SELECT * FROM requests ORDER BY id LIMIT 40";
$result = $conn->query($sql);
while ($res = $result->fetch_assoc()) {
$email1 = $res['email1'];
$msg1 = $res['msg1'];
$subject1 = $res['subject1'];
$name1 = $res['name1'];
$id = $res['id'];
//example output, just for demo:
echo $email1." ".$msg1." ".$subject1." ".$name1." ".$id;
}
Documentation: https://dev.mysql.com/doc/refman/8.0/en/limit-optimization.html

Output is array, how to fix this?

Right now I am selecting some data from Mysql which I want to echo out, but I don't know how..
Code
<?php
// Database connection
require_once($_SERVER["DOCUMENT_ROOT"] . "/includes/config.php");
require_once($_SERVER["DOCUMENT_ROOT"] . "/includes/opendb.php");
// News 1
$searchroutenews1 = "SELECT newsid FROM NewsHomepage WHERE id = '1'";
$handlenews1 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenews1);
$news1 = mysqli_fetch_row($handlenews1);
$searchroutenewsimg1 = "SELECT newsimg FROM NewsHomepage WHERE id = '1'";
$handlenewsimg1 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenewsimg1);
$NewsImg1 = mysqli_fetch_row($handlenewsimg1);
// News 2
$searchroutenews2 = "SELECT newsid FROM NewsHomepage WHERE id = '2'";
$handlenews2 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenews2);
$news2 = mysqli_fetch_row($handlenews2);
$searchroutenewsimg2 = "SELECT newsimg FROM NewsHomepage WHERE id = '2'";
$handlenewsimg2 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenewsimg2);
$NewsImg2 = mysqli_fetch_row($handlenewsimg2);
// News 3
$searchroutenews3 = "SELECT newsid FROM NewsHomepage WHERE id = '3'";
$handlenews3 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenews3);
$news3 = mysqli_fetch_row($handlenews3);
$searchroutenewsimg3 = "SELECT newsimg FROM NewsHomepage WHERE id = '3'";
$handlenewsimg3 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenewsimg3);
$NewsImg3 = mysqli_fetch_row($handlenewsimg3);
?>
After this I require_once this in an other file, and then I echo the variables $news1, $news2, $news3, $NewsImg1, $NewsImg2 and $NewsImg3. But if I echo this variables out now it says: array.
You can fetch all this information via a single query, instead of the 6 you currently are running. Then it's a matter of putting mysqli_fetch_*() as the argument of a while, as you'll then fetch all the rows, until that function returns null - at which point you've fetched all the rows returned by the query.
$result = mysqli_query($GLOBALS["___mysqli_ston"], "SELECT newsid, newsimg FROM NewsHomepage WHERE id IN (1, 2, 3)");
while ($row = mysqli_fetch_assoc($result)) {
echo $row['newsid']." ".$row['newsimg'];
}
Change however you need it to be displayed inside the while loop, and use the two variables as they are inside.
Alternatively, you can use WHERE id BETWEEN 1 AND 3 instead, but using IN (1, 2, 3) can more easily be changed to the exact ids you need.
http://php.net/mysqli-result.fetch-assoc
First you should read http://php.net/manual/en/mysqli-result.fetch-row.php
you can find there mysqli_result::fetch_row -- mysqli_fetch_row — Get a result row as an enumerated array mysqli_fetch_row always return array so now you can't echo array thats why it gives you array.
You can try foreach loop or for loop or while loop to display your data. there are also various methods to get array value.
Below is an example you can use.
while ($news1 = mysqli_fetch_row($handlenews1)) {
echo $news1[0];
}

Need Some Help Regarding Fetching Data from Mysql using explode function

In the below script I want to fetch data from mysql using a explode function and also a variable within an explode function.
Here's how I want to get
<?php
include ('config.php');
$track = "1,2,3";
$i = 1
$trackcount = explode(",",$track);
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>
This is the code
$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";
I want sql to fetch data from tracks table where id = $trackcount[$i]
Whatever the value of $trackcount[$i] mysql should fetch but it shows a blank screen.
If I put this
$sql = "SELECT * FROM tracks WHERE id='$trackcount[1]'";
It works perfectly
save your $trackcount[$i] in one variable and then pass it in the query as given below
<?php
include ('config.php');
$track = "1,2,3";
$i = 1;
$trackcount = explode(",",$track);
$id=$trackcount[$i];
$sql = "SELECT * FROM tracks WHERE id='$id'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>
and one more thing check your previous code with echo of your query and see what is passing ok.
echo $sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//like this
problem is with your query
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//change
to
$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";
Generally you would want to use the IN operator with this type of query, so for you this would be:-
$sql="SELECT * FROM `tracks` WHERE `id` in (".$track.");";
or, if the $ids are in an array,
$sql="SELECT * FROM `tracks` WHERE `id` in (".implode( ',', $array ) .");";

how correctly get country and other info from sql for specific id

I am trying to get country and other info from sql for specific id, only "id" displays correctly in both cases.
<?php
$sql = "SELECT * FROM `list` ORDER BY category ASC";
$result = mysql_query($sql);
$rows = mysql_fetch_assoc($result);
$id = $_GET['id'];
$country = $_GET['country'];
echo $id;
echo $country;
?>
and
<?php
$sql = "SELECT * FROM `list` ORDER BY category ASC";
$result = mysql_query($sql);
$id = $_GET['id'];
$country = $_GET['country'];
if (mysql_num_rows($result) > 0) {
while($rows = mysql_fetch_assoc($result)) {
echo $id;
echo $country;
}
}
?>
I think you have a column named id and another named country from which you want to collect the data in the country row of a certain record.
I have similar ideas to some of the comments and would like to highlight my thoughts in this answer.
$_GET superglobal is used to extract data form a URL parameter. If it is used to get the data from the record, you might want to use mysql_fetch_assoc($result)["country"]. For more information, please take a look at https://www.php.net/manual/en/function.mysql-fetch-assoc.php
and https://www.w3schools.com/php/php_superglobals_get.asp
To filter certain records according to some input id, you would want to use the WHERE clause. Link for more information:- https://www.w3schools.com/sql/sql_where.asp
The possible code might be:-
<?php
$id = $_GET["id"]
$sql = "SELECT * FROM `list` WHERE id='$id' ORDER BY category ASC";
$result = mysql_query($sql);
if (mysql_num_rows($result) > 0) {
while($rows = mysql_fetch_assoc($result)) {
echo $rows['id'];
echo $rows['country'];
}
}
?>
PS: I have tried my best to answer the asked question provided the less information given. Hope it helps

PHP MySQL single column/row display?

I'm trying to get a single result from my database, just one name.
I tried using;
$row = mysql_fetch_array(mysql_query("SELECT * FROM persons WHERE id = '$id'"));
echo $row['name'];
But that din't work, any other way to simply show only one result?
Thanks in advance!
[EDIT:]
(I'm using PHP 5.3)
<?php
include("connection.php");
$id = $_GET['deletid'];
$result = mysql_query("SELECT * FROM persons WHERE id = '$id' LIMIT 1");
if(!$result){
echo mysql_error();
}
if ($row = mysql_fetch_array($result)){
echo $row['name'];
}
echo "<p>id:$id</p>";
?>
If you need just the name and you need just one result you should rewrite your query as follow:
$row = mysql_fetch_array(mysql_query("SELECT name FROM persons WHERE id = '". (int) $id ."' LIMIT 1"));
Now to get the result you should just get it with a
$row['name'];
EDIT
Now that you posted your entire code i got what's wrong: You are deleting that result before getting its name. Basically you delete that user and then you attempt to get its name.
EDIT
<?php
include("connection.php");
if (empty($_GET['deleteid'])) {
exit('"deleteid" is empty');
}
$id = mysql_real_escape_string($_GET['deletid']);
$result = mysql_query("SELECT name FROM persons WHERE id = '". (int) $id ."' LIMIT 1");
if(!$result){
echo mysql_error();
}
$row = mysql_fetch_assoc($result); // for just one result you don't need of any loop
echo $row['name'];
echo "<p>id:". htmlspecialchars($id) ."</p>";
?>
try
$row = mysql_fetch_array(mysql_query("SELECT name FROM persons WHERE id = ". (int) $id));
echo $row['name'];

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