I want to echo the contents of a table (a product list) yet still have control over the individual fields if possible. While loops did not work as you cannot then go to pick how each field is displayed.
What I am after is to create a product list which then has each record as a hyperlink setting a variable to the product id. Something like:
www.domain.com/catalog/product.php?productid=1
If there isnt a way to echo the fields individually within each row, how would I be able to get this outcome without adding the link into the database itself for all the products?
Its fairly easy
<?php
$con = mysql_connect("localhost","username","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_db", $con);
$result = mysql_query("SELECT * FROM Product_List");
while($row = mysql_fetch_array($result))
{
echo "<a href='www.domain.com/catalog /product.php?productid=".$row['productid']."'>".$row['productid']."</a>";
echo "<br />";
}
mysql_close($con);
?>
You'll still need an while loop. Consider this:
<?php
$getRows = mysql_query("SELECT id FROM products");
while($row = mysql_fetch_assoc($getRows)) {
echo 'Product link';
}
Related
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.
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
I'm fairly new to php and mysql and I'm on the home stretch of finishing my page but I've been banging my head on the keyboard all day trying to figure out how to fix this problem. I've set up a php script to run as a cron even every 24 hours. The script assigns a random number between 10 and 30 to each field in my table. That works fine and every time I load the script the values change.
The problem I'm having is when I try to use those values. The result keeps printing as the word Array instead of the number in the table. So I'll give you some snippets of the code I'm running here. This is the cron event.
?php
$con = mysql_connect("localhost","username","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("dbname", $con);
$random= rand(10, 30);
mysql_query("UPDATE winners SET pool= '$random'");
mysql_close($con);
And here is the script to call up the values.
php?
$db = mysql_connect('localhost','username','pass') or die("Database error");
mysql_select_db('dbname', $db);
$query = "SELECT pool FROM winners";
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($result))
if ( $row % 2 )
{
echo "<h4>Result 1</h4>";
echo "$row";
echo "<br />";
}
else
{
echo "<h4>Result 2</h4>";
echo "<br />";
}
I've tried every possible variation I can think of to this code but all I can get is it to echo either Array or Resource #9. Any insight into this would be greatly appreciated.
You do not want to echo the content of $row, which contains all data returned for the current line you've fetched from the database when calling mysql_fetch_array().
Instead, you want to access the content of $row's pool item :
echo $row['pool'];
You should probably take a closer look at the manual page for mysql_fetch_array(), and the examples it contains.
Note that you'll probably also want to modify the condition in the following line :
if ( $row % 2 )
You probably don't want the test to be done on $row, but on an item it contains.
$query = "SELECT pool FROM winners";
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
if ( $row['pool'] % 2 )
{
echo "<h4>Result 1</h4>";
echo "$row['pool']";
echo "<br />";
}
else
{
echo "<h4>Result 2</h4>";
echo "<br />";
}
}
Hope this helps.
Seems pretty straightforward, but not getting an error or result.
<html>
<body>
<?php
$con = mysql_connect("localhost","***","***");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("ubook247", $con);
$result = mysql_query("SELECT * FROM buzz_data
WHERE index=4");
while($row = mysql_fetch_array($result))
{
echo $row['buzz_img'] . " " . $row['buzz_title'];
}
?>
</body>
</html>
screenshot of db:
Index is a keyword in SQL, you'll need to escape it for the query to work. Try this:
SELECT * FROM buzz_data WHERE `index` = 4
Try editing the following row:
while($row = mysql_fetch_array($result))
into becoming like this:
while($row = mysql_fetch_assoc($result))
This makes php fetch an array with "labels" for the different fields, instead of naming them 0, 1, 2 and so on.
I've been trying for hours to figure out how to put a link into the following text output via PHP echo(). I basically want to make the title field I'm pulling from my events table (as seen in #4 in the code below) to come back into the browser as a link instead of just text...
the original code that brings back the event title:
<?php
// 3. Perform database Query to bring list of events
$result = mysql_query("SELECT * FROM events", $connection);
if (!$result) {
die("Database query failed: " . mysql_error());
}
// 4. Use returned Data
while ($row = mysql_fetch_array($result)) {
echo $row ["eventtitle"]."<br/>".$row["eventdesc"]."<br/>";
}
?>
How would I go about getting that $row["eventtitle"] to appear in the browser as a link? Let's say if the link was just "eventprofile.php". This is probably an easy fix, but I've been getting a million errors with trying different things with <a href>s.
<?php
$result = mysql_query("SELECT * FROM events", $connection);
if (!$result) {
die("Database query failed: " . mysql_error());
}
while ($row = mysql_fetch_array($result)) {
echo "".$row["eventtitle"]."<br/>".$row["eventdesc"]."<br/>";
}
?>
add an anchor tag for it!!
echo "" . $row['eventtitle'] . '' . "<br />" .$row["eventdesc"]."<br/>";
And thats it.