I have products stored in a database. These items have ID, NAME, QUANTITY, and STOCK_NUMBER fields. I want to make an href for all of the data in the database:
$Result=mysql_query("SELECT * FROM Products");
while($row=mysql_fetch_array($Result)){
href .....
}
I want the result like this
item_id=1&name_1_tv&quantity_1=2&stock_number_1=1411
&item_id=2&name_2_mobile&quantity_2=5&stock_number_2=5894
&item_id=3&name_3_radio&quantity_3=2&stock_number_3=18541
&item_id=4&name_4_tv&quantity_4=2&stock_number_4=1025
&item_id=5&name_5_computer&quantity_5=1&stock_number_5=1455
&item_id=6&name_6_cd&quantity_6=2&stock_number_6=5888
all these under the link
Use http_build_query().
while($row=mysql_fetch_assoc($Result)) // note the change to assoc
{
$query = http_build_query($row);
echo "&$query<br>";
}
Note that GET requests have very low maximum length limits. IE chokes on URLs longer than 2 kilobytes, for example.
Related
I'm quite new to php and mysql so forgive me if I'm doing this completely wrong. I am making a printing balance application and the code below is a part of it.
$command="SELECT itemname FROM items";
$results = mysql_query($command);
while($row = mysql_fetch_assoc($results))
{
foreach ($row as $key => $value) {
print "<input type='radio' name='itemtype' value='$value'>".$value."</input><br />";
}
}
This here is supposedly the price printing form where the user chooses between SHORT BOND PAPER and LONG BOND PAPER (the column itemname from items). The options appear as radio buttons. It works but now I'm stuck with the issue of being able to fetch the price as inserted in their respective rows. Since the itemname and their price are all user-inputted into the database, I'm not supposed to declare their specific price into the code itself, and should be able to retrieve it from the database. I want to be able to get the prices based on the item chosen by the user once they click submit, because I'd need to do this to compute for the price of printing when multiplied with the number of pages later.
I think it's definitely possible but I'm not quite sure how. Theoretically, it would be along the lines of SELECT itemprice FROM items WHERE itemname = $value but ha, I don't think it works that way.
solution edit: here's the complete solution for reference. $compute is just a sample to test if it works while 5 is a sample number of pages that would be entered.
if ($command = mysql_query("SELECT `itemprice` FROM `items` WHERE `itemname` LIKE '" . mysql_escape_string($_POST['itemtype']) . "'"))
{
while($row = mysql_fetch_assoc($command))
{
$compute = $row['itemprice'] * 5;
echo $compute;
}
}
else
{
echo (mysql_error ());
}
It would be something like that indeed.
SELECT itemprice FROM items WHERE itemname = $_POST['itemtype']
assuming that itemprice is the actuial colum of the price. HOwever, doing it like this, makes your code vulnerable to mysql injections. So before you contine, consider reading up a little.
Ive been battling away with the following problem
Ive got a page where I pull names from players specific to their positions in a sport squad.
Example: I will display all the Wings in the squad using a dropdown where a coach can then pick his wing for the game.
There are dropdowns for each different position
The aim of the page is to let the coach quickly select his team for a fixture
After the coach selected his team he will, select the opponents for which the selected team will play against.
When he clicks submit the selected oppents and players will get stored in two arrays which will get called to display the team selected and their opponents on a new page. (After which it will get uploaded to the DB.)
I am having trouble getting the values from the select list to display on the new page.
I guess I have to do something like this on the new page:
foreach ($_REQUEST['opponents'] as $opponents){
print $opponents;
echo'<br>';
}
but it is not giving the desired results.
Strangely what gets printed is the variable name from the previous page select menu.
Upon further inspection I did a vardump on the new page and it says that $opponenets gets passed a value of string which is the variable name and not the value thereof?
My page looks like this
My question is how would I go abouts getting the values from the select dropdowns
if(isset($_POST["submit"]))
{
foreach ($_REQUEST['opponents'] as $against){
var_dump($against);
print $against;
echo'<br>';
}
}
else
{
echo'<h1>Select your Team</h1>';
$x = array("THP", "HKR", "LHP", "LH", "FLH"); //players positions gets assigned to x which will be used to query the database
echo '<form name="playerselect" method="post" action="">';
//query database with different query after each loop
for ($i = 0; sizeof($x) > $i; $i++)
{
//query where position field equeals variable x
$result = mysql_query("SELECT `name`, `position` FROM `player_info`
WHERE `position` = '$x[$i]'") or die(mysql_error()) ;
//Gets data from DB and assigns values to arrays below
while($row = mysql_fetch_array($result))
{
$playername[] = $row['name'];
$position[] = $row['position'];
}
//print player position
print $position[0];
echo'<br>';
//unset the array so that it is empty for the next query in the new loop
unset($position);
echo '<select name="players[]" >' ;
foreach ($playername as $name )
{
//Put playernames relevant to the position in the option element
echo'<option value="$name" selected="selected">'.$name;'</option>';
echo'<br>';
}
echo'</select>';
//unset array so that its contents is empty for the next query in the new loop
unset($playername);
echo'<br>';
}
You cannot. Your submit will only transmit select values. This is not a bug, it is a feature. You do not want to send data back and forth from/to the server/client which is known to both of them.
On the server you are free to query your database at any time. You can also cache your select list into the $_SESSION variable in your initial list read. However this is advanced fittling as your cache list may become outdated and also your server memory utilization must leave space for file caching (the SESSION cache goes to files).
If you go for the database query you may need some ID as sort of anchor. Just put the into the $_SESSION variable - eg.:
$_SESSION['positions']=$x;
In your example the $x seems to be static, which obviously reduces the need to cache it into the $_SESSION - however on other occasions you may need this method.
I'm creating a small project with PHP/MYSQL but i can't get my query working the way i need it. I have 2 tables
Table 1 (char):
Id, name.
Table 2 (spells):
Id, char, spell_name.
I'm getting the output:
Name Spell1
Name Spell2
Name Spell3
But I need it to be:
Name Spell1
Spell2
Spell3
Here's my query:
$query = "SELECT char.name AS name, spells.spell_name AS spell
FROM char, spells
WHERE (char.id = spells.spell_name)";
Any ideas?
I think you're gonna have to first get the ID of the character to query, and then pull the spells s/he has access to. Example:
$char_id = 0; // value would be assigned arbitrarily.
$query = "SELECT *
FROM 'spells' s
WHERE s.char = $char_id;";
$result = $pdo->query($query);
while($row = $result->fetchObj()){
// do something with the spells obj here
}
With SQL, you need to grab full rows at a time, so I believe the situation you want isn't possible.
As Goldentoa11 wrote. Make two selects, or create query with two result sets (more selects in one command), or accept current state (is normal and you can verify data consistency). I prefer current state, but sometimes use any of described solution (based on query frequency, size of result etc.).
If you need to list such data, you can than use something like this:
$currentName = null;
while ($row = mysql_fetch_object($result))
{
if ($currentName != $row->name)
{
echo "<b>" . $row->name . "</b><br />";
$currentName = $row->name;
}
echo $row->spell_name . "<br />";
}
I have a table that stores images of three categories: cars, bikes, buses. These images appear on two pages. On the first page, you should see an image of each category, defined so random. On the second page, you should see the remaining pictures. How can I do this using php?
I would prefer to do this in MySQL, you can set the order to:
ORDER BY MD5(ImageNo+UserRandomNumber)
SQL FIddle
You can generate a single random number per user, store this in the session then apply this to the ordering and as long as the number supplied stays the same, the order will remain the same too.
You can indeed use a session to store which images were shown on the first page, and then leave them out on the second page. Without knowing your exact table structure, something like this should work:
<?php
// page 1
session_start();
function get_random_image_id($category)
{
// SELECT id FROM images WHERE category = $category ORDER BY RAND() LIMIT 1
}
function show_image($id)
{
// SELECT * FROM images WHERE id = $id
// echo <img ...>
}
$_SESSION['image_ids'] = array(
get_random_image_id("cars"),
get_random_image_id("bikes"),
get_random_image_id("buses"),
);
foreach($_SESSION['image_ids'] as $image_id)
{
show_image($id);
}
?>
<?php
// page 2
session_start();
function show_images($exclude_ids)
{
$ids = implode(",", $exclude_ids);
// SELECT * FROM images WHERE id NOT IN ($ids)
}
$exclude = array();
if(isset($_SESSION['image_ids'])) // has the user visited page 1?
{
$exclude = $_SESSION['image_ids'];
}
show_images($exclude);
?>
Been trying to get my head around while loops for the last few days but the code seems very inefficient for what I'm trying to achieve. I'm assuming I'm over-complicating this though nothing I've tried seems to work.
Each topic in my forum can have related topic IDs stored in a separate table. A post ID is also stored in this table, as that specific post references why they are considered related.
DB Table contains only: topic_id, related_id, post_id
// Get related IDs and post IDs for current topic being viewed
$result = $db->query('SELECT related_id, post_id FROM related_topics WHERE topic_id='.$id.'');
// If related topics found, put both of the IDs into arrays
if ($db->num_rows($result)) {
while($cur_related = mysql_fetch_array($result)){
$reltopicarray[] = $cur_related['related_id'];
$relpost[] = $cur_related['post_id'];
}
// If the first array isnt empty, get some additional info about each related ID from another table
if(!empty($reltopicarray)) {
$pieces = $reltopicarray;
$glued = "\"".implode('", "', $pieces)."\"";
$fetchtopics = $db->query('SELECT id, subject, author, image, etc FROM topics WHERE id IN('.$glued.')');
}
// Print each related topic
while($related = mysql_fetch_array($fetchtopics)){ ?>
<?php echo $related['subject']; ?> by <?php echo $related['author']; ?>
// Id like to show the Post ID below (from the array in the first while loop)
// The below link doesnt work as Im outside the while loop by this point.
<br />View Relationship
<?php } ?>
The above currently works, however I'm trying to also display the post_id link below each related topic link, as shown above.
if you change the second while loop to something like this:
<?php
$i = 0;
while($related = mysql_fetch_array($fetchtopics)){
//show view link
// [...]
//show the view related link
?>
View Relationship
<?php
//increment the i so that you can get the next post in the next iteration of the loop
$i++;
}
?>
[sidenote]
You probably should not be doing database queries in the same location you are generating the html for future-you's sanity.
[/sidenote]
[edit]
You could also do it all as one query:
SELECT related_topics.related_id,
related_topics.post_id,
related_topics.topic_id,
topics.subject,
topics.author,
topics.image,
topics.etc
FROM related_topics
LEFT JOIN topics ON topics.id = related_topics.topic_id
WHERE topic_id= $id
Then you only have to loop through it once for all of the links.