Assign PHP variable to each row - php

The following mysql query is getting the last 4 records from the 'residential' table but I'm trying to assign a php variable ($postcode) to each row (the 3rd [3] column in particular) for use in on another page. The following doesn't seem to be split the rows out correctly to assign to each variable?
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM residential ORDER BY My_id DESC LIMIT 4");
$postcodes = array();
while ($row = mysqli_fetch_array($result))
{
$postcodes[] = $row[3];
}
echo "location one " . $postcodes[0];
echo "<br>";
echo "location two " . $postcodes[1];
echo "<br>";
echo "location three " . $postcodes[2];
mysqli_close($con);
?>

fetch calls fetch a single ROW of data. You're just assigning column #3 of each row, over and over again.
You probably want:
$postcodes = array();
while ($row = ...) {
$postcodes[] = $row[3];
}
var_dump($postcodes);

Related

mysqli_fetch_row() not displaying after foreach() in PHP

when a foreach is followed by a mysqli_fetch_row displaying the same query results, the mysqli_fetch_row() doesn't display. However if the mysqli_fetch_row is executed before the foreach() both are displayed. Why is this?
QUERY:
<?php
$connection = mysqli_connect("localhost", "root", "", "login_app");
$query = "SELECT *
FROM users";
$result = mysqli_query($connection, $query);
if(!$result){
die("db connection failed " . mysqli_error());
}
?>
ONLY DISPLAYS THE RESULTS OF FOREACH:
<?php
foreach($result as $results){
echo "User ID: " . $results['userID'] . "<br>";
echo "Username: " . $results['userName'] . "<br>";
echo "Password: " . $results['password'] . "<br>";
};
$row = mysqli_fetch_row($result);
print_r($row);
?>
DISPLAYS RESULTS OF BOTH FOREACH AND MYSQLI_FETCH_ROW:
<?php
$row = mysqli_fetch_row($result);
print_r($row);
foreach($result as $results){
echo "User ID: " . $results['userID'] . "<br>";
echo "Username: " . $results['userName'] . "<br>";
echo "Password: " . $results['password'] . "<br>";
};
?>
Executing a SELECT query with mysqli_query returns an mysqli_result object. That object implements Traversable, which is why you're able to use a foreach loop to display the results.
foreach($result as $results){ ...
is basically a handier way of doing
while ($results = mysqli_fetch_row($result)) { ...
(See Example #3 here.)
Either way you do it, you're fetching all the results, and when you get to the end of the result set $results will hold the last value fetched (null).
From the manual:
Fetches one row of data from the result set and returns it as an enumerated array, where each column is stored in an array offset starting from 0 (zero). Each subsequent call to this function will return the next row within the result set, or NULL if there are no more rows.
By iterating the result set with your foreach loop, you are effectively exhausting all of the row which is why no data is available on any subsequent row fetches after the loop.

Make PHP array from numerical data in MySQL table and multiply it

Good day,
I'm a noob at PHP and MySQl. Looking to be pointed in the right direction.
I have a MySql table with 5 columns. Each column represents a specific set of data. All numerical.
I want to write PHP code which takes each value in a single column and puts it into an array that I can then modify.
I don't know how to get each column as a separate array. Should I get an array of all the rows and then do something extra to separate each of the 5 numbers in each row into 5 separate arrays?
**
require_once 'login.php';
echo $db;
$conn = mysqli_connect($hn,$un,$pw,$db);
if (!$conn) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
$query = "SELECT xox FROM tablex WHERE id = '1'";
$result = $conn->query($query);
if(!$result) die ("Database access failed: " . $conn->error);
$rows = mysqli_fetch_row($result);
while($rows){
echo $rows['index'];
}
echo $rows[0];
?>
$statement = **yourDatabaseConnection**->query('SELECT column1 FROM table');
$datas = $statement->fetch();
$datas will be an array of your column1 datas
Try reviewing the php documentation, here is a good place to start...php:mysql_fetch_row
Take a look at the example below, if is not what you looking for, please edit your question and provide the approach you are taking (your code) so that we have a better understanding where you are having issues.
<?php
$result = mysql_query("SELECT id,email FROM people WHERE id = '42'");
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result);
echo $row[0]; // 42
echo $row[1]; // the email value
?>

Creating An Array from MYSQLI query

I am updating all my code to Mysqli, before I did I had this code and it worked:
while($row = mysql_fetch_assoc($WildHorses)){
$InWild[] = $row['id'];
}
$RandomWild = array_rand($InWild);
$RandomHorse = $InWild[$RandomWild];
This is my SELECT statement:
$sql = "SELECT Horse.id, Horse.Name, Horse.Age, Horse.Image_name, Horse.Owner, Horse.Barn, Images.Image_path, Images.Image_name FROM Horse, Images WHERE Horse.Owner = '$colname_WildHorseList' AND Images.Image_name = Horse.Image_name";
$result = mysqli_query($con,$sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"]. " Name: " . $row["Name"]. " ImageName: " . $row["Image_name"]. "<br>";
}
} else {
echo "0 results";
}
The SELECT statement ends up echoing all of the correct information, but I want to make an array of only the Id's so that I can pick one at random each time a button is clicked.
I have tried multiple different copies and pastes of code to try and get what I want, but nothing seems to come out right.
Can someone point me in the right direction or explain what I'm doing wrong?
In your while loop you can simply do this :-
$i=0;
$animals=array();
$animals[$i]=$row["id"]; //puts id in array
And then you can create a random number by "rand()" between the length of 0-$i
and can get the job done.

How do I dynamically display the results of my PHP results in separate divs according to their ID?

I'm trying to create a simple e-commerce system. First thing I did was select the rows with the same Order ID from mysql. My table looks like this:
Now, I'd like to know how I can group them into separate divs, it should look like this:
Here's my code:
$con = mysqli_connect("localhost","root","","test");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result2 = mysqli_query($con, "SELECT DISTINCT request_date, department_id FROM tbl_requests WHERE request_id=".$_GET['request_id']) ;
while($row2 = mysqli_fetch_array($result2)) {
echo "" . $row2['request_date'] . "<br/>";
echo "Order from Department " . $row2['department_id'] . "<br/>";
echo "<br/>";
echo "<hr/>";
echo "<br/>";
}
$result = mysqli_query($con,"SELECT * FROM tbl_requests WHERE request_id=".$_GET['request_id']);
while($row = mysqli_fetch_array($result)) {
echo "" . $row['request_details'] . "";
echo "<br/>";
}
I'm sorry if ever this question is incomplete, please feel free to ask me any more questions. Thank you in advance :)
You can check for every product in array using Javascript or jQuery(much easier).
This way you can check if your page does contain any existing div with that manufacture id (ie. #m_1055, #m_1040) or not.
If div does exist, append product in that div (in jQuery.append())
If not, then first append the div with that manufacture id to the document and then append the product to it.

Select one value from database

I have the code bellow, it's ok but I want to be able to use for example the 4th value extracted from the database, use it alone, not put all of them in a list, I want to be able to use the values from database individually. How do I echo them?
Edit: I was thinking to simplify things, to be able to add the values from database into one array and then extract the value I need from the array (for example the 4th - ordered by "order_id"). But how?
Right now I can only create a list with all the values one after the other..
(Sorry, I am new to this). Thank you for all your help..
<?php
include '../../h.inc.php';
$con = mysql_connect($db['host'],$db['user'],$db['passwd']);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("database", $con);
$result = mysql_query("SELECT * FROM options WHERE Name LIKE 'x_swift%' ORDER BY order_id");
echo "<table border='1'>
<tr>
<th>Values</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
// echo "<td>" . $row['Name'] . "</td>";
echo "<td>" . $row['VALUE'] . "</td>";
echo "</tr>";
$array = array(mysql_fetch_array($strict));
}
echo "</table>";
mysql_close($con);
?>
To select the value in the value column of the row where order_id is 4, use this SQL:
$query = 'select value from options where order_id = 4';
Then you can access this result in many ways. One is to get the entire result row (which in this case is just one cell) as an associative array:
if ($result = mysql_query($query)) {
$row = mysql_fetch_assoc($result);
echo 'value = ' . $row['value'];
}
You can also get the value directly:
if ($result = mysql_query($query)) {
echo 'value = ' . mysql_result($result, 'value');
}
It would just be a query like...
$result = mysql_query("SELECT * FROM options WHERE ID = 3");
mysql_fetch_row($result);
Unless Im misunderstanding you....let me know
But you really should use PDO, instead of deprecated mysql_* functions
http://php.net/manual/en/book.pdo.php

Categories