For some reason I'm trying to display all of the members from the database in a list in order to access each of their profiles when I click on them, but I'm only getting the link of the last person in the database, any help?
include_once "../../mysql_server/connect_to_mysql.php";
//This code is used to display friends in the box of friends
$sql = mysql_query("SELECT * FROM myMembers");
$numberofRows = mysql_num_rows($sql);
$memberDisplayList = 'There are ' . $numberofRows .' members<br /><br />';
while($row = mysql_fetch_array($sql)) {
$id = $row['id'];
$firstname = $row["firstname"];
$lastname = $row["lastname"];
/////// Mechanism to Display Pic. See if they have uploaded a pic or not
$check_pic = "../../members/$id/image01.jpg";
$default_pic = "../../members/0/image01.jpg";
if (file_exists($check_pic)) {
$user_pic = "<img src=\"$check_pic?$cacheBuster\" width=\"80px\" />";
} else {
$user_pic = "<img src=\"$default_pic\" width=\"80px\" />";
}
$memberDisplayList = '' . $firstname .' '. $lastname .'<br />';
}
// ------- END WHILE LOOP FOR GETTING THE MEMBER DATA ---------
I think instead of
$memberDisplayList = '<a href= (...etc)
you meant to type
$memberDisplayList .= '<a href= (...etc)
which would append the new links to your string.
Also you don't seem to be echoing your $user_pic and $memberDisplayList strings anywhere.
Its because your overwriting the variables on each iteration, you need to hold the data within an array then do another foreach loop where ever you output:
<?php
while($row = mysql_fetch_array($sql)){
/////// Mechanism to Display Pic. See if they have uploaded a pic or not //////////////////////////
$check_pic = "../../members/{$row['id']}/image01.jpg";
$default_pic = "../../members/0/image01.jpg";
if (file_exists($check_pic)) {
$user_pic = "<img src=\"$check_pic?$cacheBuster\" width=\"80px\" />";
} else {
$user_pic = "<img src=\"$default_pic\" width=\"80px\" />";
}
$user[] = array('id'=>$row['id'],
'firstname'=>$row["firstname"],
'lasname'=>$row["lastname"],
'user_pic'=>$user_pic,
'display_list'=>'' . $row["firstname"] .' '. $row["lastname"] .'<br />');
}
?>
Where are you actually constructing the HTML? You're setting a bunch of variables in the code you presented, and that looks okay for what it is. So it's probably in the presentation logic. If that's in the loop, you're in good shape. But if it's outside of the loop, then I can't imagine where you'd ever display anything but the last row.
Related
How it looks:
https://jsfiddle.net/jef2L8m6/
How it should look:
https://jsfiddle.net/jef2L8m6/1/
I know it looks really bad, this is just for testing purposes only.
Some of the Backend Code:
<?php //Selects all of the logged in users messages.
$name = $_SESSION["name"];
$con = mysqli_connect('localhost','root','','chat');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM `chat` ORDER BY date";
$result = mysqli_query($con,$sql);
$numrows = mysqli_num_rows($result);
if( $numrows == "0" or !isset($_SESSION["name"])){
echo "<div class='msg'>You are not registered (Or there are no messages to display)</div>";
exit();
}else{
echo "";
}
echo "<div class='msg_container'>";
while($row = mysqli_fetch_array($result)) {
echo "<div class='msg_user'>";
echo "<div class='username_user'><span>" . $row['username'] . "</span></div>";
echo "<div class='message_user'><span>" . $row['message'] . "</span></div>";
echo "</div>";
}
echo "";
mysqli_close($con);
?>
Thank you so much for taking your time to read this.
I am trying to figure out how I would change the div tags of each separate user depending on their name?
Is there any way to do this using PHP, I have tried doing 2 separate query's of one that selects just the users messages and another that selects everyones (excluding the users)
But none of them worked due to it not ordering them correctly.
Could I somehow change the div's using PHP if the username that comes out is not equal to the username in the session?
Thank's so much, if you don't think I explained this very well please give me some feedback and I will change/add what you need, THANK YOU!
Thank you so much "u_mulder", you have been very helpful in making me think of a simple way to solve this problem.
I was thinking way too complex for something so simple!
Here is the final code for anyone who this may help:
while($row = mysqli_fetch_array($result)) {
$class_msg = "msg";
$class_username = "username";
$class_message = "message";
if ($row['username'] == $_SESSION['name']) {
$class_msg = "msg_user";
$class_username = "username_user";
$class_message = "message_user";
}
echo "<div class='$class_msg'>";
echo "<div class='$class_username'><span>" . $row['username'] . "</span></div>";
echo "<div class='$class_message'><span>" . $row['message'] . "</span></div>";
echo "</div>";
}
while($row = mysqli_fetch_array($result)) {
$class = 'msg';
if ($row['username'] == $_SESSION['name']) {
$class = 'msg_user';
}
echo "<div class='" . $class . "'>";
// other codes here
}
Thank you for reading this, I am building a shopping cart with PHP session, when I click the submit button on the form, the $_POST was able to pass the value to the $_SESSION[''] array, but when I click the submit button again, the $_SESSION array only contain my last $_POST variable and the previous variables are all gone. As a shopping cart, the session array suppose to contain every data obtain from the POST value.
I checked with the SESSION_ID, which was able to shows the same ID when I submit the form, the var_dump shows the current SESSION array works (except only showing the last item). Please help me what is wrong in my code in order to keep the value into the SESSION array, thank you.
here is the full code, the sqli_query was replaced to specify item_id for troubleshooting, also this php will be included in another php that have an id in url, which seems irrelevant to this matter, but just for your information.
<?php if(!isset($_SESSION)){session_start();}?>
<?php
//if(isset($_GET['id']) && !empty($_GET['id'])){
require 'connecttosql.php';
$result = mysqli_query($con,"select COLOUR_EN, COLOUR_ZH, SIZE FROM mydb.item join mydb.colour using(item_id) join mydb.size using(item_id) WHERE ITEM_ID='T76'")
or die("Error: " . mysqli_error($con));
while($row = mysqli_fetch_array($result)){
$size[] = $row['SIZE'];
$colour_zh[] = $row['COLOUR_ZH'];
$colour_en[] = $row['COLOUR_EN'];
}
mysqli_free_result($result);
mysqli_close($con);
for($x=0;$x<count($colour_zh);$x++){
$colour[$x] = $colour_zh[$x] . "/" . $colour_en[$x];
}
echo "<form action='' method='POST'>";
echo "<ul>";
echo "<li>size: </li>";
echo "<li><select name = 'size'>";
foreach(array_unique($size) as $sizeli){
echo "<option value = '" . $sizeli . "'>" . $sizeli . "</option>";
}
echo "</select></li>";
echo "<li>colour: </li>";
echo "<li><select name = 'colour'>";
foreach(array_unique($colour) as $COLOURli){
echo "<option value = '" . $COLOURli . "'>" . $COLOURli . "</option>";
}
echo "</select></li>";
echo "<li><input type='SUBMIT' name='submit' value='ADDTOCART'></li>";
echo "</ul>";
$_SESSION['size'] = array();
$_SESSION['colour'] = array();
if(isset($_POST['submit'])) {
$_SESSION['size'][] = $_POST['size'];
$_SESSION['colour'][] = $_POST['colour'];
// $_SESSION['id'] = $_GET['id'];
}
echo SESSION_ID();
var_dump($_SESSION['size']);
var_dump($_SESSION['colour']);
// var_dump($_SESSION['id']);
/*
}else{
include 'index.php';
die();
}
*/
?>
You reinitialize (and therefore reset) the arrays at every request:
$_SESSION['size'] = array();
$_SESSION['colour'] = array();
Add a check like this:
if(!isset($_SESSION['size'])) {
$_SESSION['size'] = array();
}
if(!isset($_SESSION['colour'])) {
$_SESSION['colour'] = array();
}
it looks like you are resetting your session variables before it gets to this line...
if(isset($_POST['submit'])) {
try checking for existance of these before resetting...
$_SESSION['size'] = array();
$_SESSION['colour'] = array();
While doing this:
$_SESSION['size'] = array();
$_SESSION['colour'] = array();
you are doing a reset to $_SESSION['size'] and $_SESSION['colour'].
You can replace it with:
if(empty($_SESSION['size'])) $_SESSION['size'] = array();
if(empty($_SESSION['colour'])) $_SESSION['colour'] = array();
or just delete these two entries.
i'm using the following code to show clickable pictures, but underneath the picture it shows a description for the picture, in this part:
<br>".$row['optie'];"
but some lines in the table are too long to display the picture and description correctly next to each other.
So how can i show these descriptions and add a line break?
<?php
$query = "SELECT * FROM $tabel";
$result = mysql_query($query);
$aantal = 0;
echo "<table>";
echo "<tr>";
while( ($row = mysql_fetch_array($result)))
{
$src = $row['afbeelding'];
echo '<div class="Image">';
echo "<td align='center'><h2><a href='pagina3.php?lang=" . $_SESSION['lang'] . "&naam=" . $naam . "&postcodehuisnummer=" .$postcodehuisnummer ."&fietskeuze=" . $fietskeuze . "&opties=" . $row['optie'] . "&optieid=" . $row['opties_id'] . "' '><img src=".$src." width='400px'><br>".$row['optie'];"</a><br /><br /> ";
echo '</div>';
$aantal++;
if ($aantal==3) {echo "<tr>"; $aantal=0;}
}
echo "</tr></tr>";
echo "</table>";
?>
You should simply concatenate your value onto the end of your URL.
$row['optie'] = str_replace('\r', '<br>', $row['optie']);
I'm new at php and JavaScript so I hope you can help me!
below is a php code! it take data from my db and display my photos and photo's title. what I want to do is if you click on one of this photos and than a new php expe: photo.php opens and you can see the same photo alone not with the other ones.
<?php
$link = mysql_connect("localhost", "root","");
mysql_select_db("test", $link);
$query = "select * from post order by id DESC;";
$results = mysql_query($query, $link) or die ("you comand can't be executed!".mysql_error());
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
echo $row['title']. "<img src=/1/" . $row['location']. " width=580px > " . "<br /><br /><br />";
}
}
?> '
<?php
$link = mysql_connect("localhost", "root","");
mysql_select_db("test", $link);
$query = "select * from post order by id DESC;";
$results = mysql_query($query, $link) or die ("you comand can't be executed!".mysql_error());
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
//send photo_id (id of photo from your table) from url NOTE: I've put link here
echo $row['title']. "<a href='index.php?photo_id=".$row['id']."'><img src=/1/" . $row['location']. " width=580px ></a> " . "<br /><br /><br />";
}
if(isset($_GET['photo_id'])) {
//here you get only one id of photo now retrieve the data from database and display the image
}
}
?>
You will get the photo by the help of the url value here photo_id.
do this way you are getting results from database and you are
displaying them with tile and image your aim is to click on
image or image title go to a new page and display image alone
the image title as link in first page
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
<a href='new page url?id=<?php echo $row["id"] ?>'>
echo $row['title'];
echo "</a>"
}
}
where $row['id'] is database unique id if have no it is always preferred
to create a database with unique id and in new page get the uid
using $id=$_GET['id'] and
do as below
if($results){
$i = 1;
while ($row = mysql_fetch_assoc($results)){
if($id==$roe['id']{
echo $row['title']. "<img src=/1/" . $row['location']. " width=580px > ";
}
}
}
I am setting up a webpage for a student organization with bios for the officers along with pictures and whatnot.
the first page simply is html and css. it has a picture, name under it and a link to the full bio where it links to "bio.php?id=" and then the id in my SQL database for that person.
now i am trying to make the php page to allow a simple template php page using the user's id. unfortunately when i do everything that I think is right, I get an odd error.
here is my code
<html>
<body>
<?php
//connection to database
//specify database
$id= $GET['id'];
$sql = " SELECT * FROM Members_table WHERE Id='$id' ";
$result = mysql_query($sql) or print ("Can't select entry from table bloghomepage.<br />" . $sql . "<br />" . mysql_error());
WHILE($row = mysql_fetch_array($result)) {
$name = $row['Name'];
$position = $row['Position'];
$major = $row['Major'];
$hometown = $row['Hometown'];
$awards = $row['Awards'];
$bio = $row['Description'];
$act = $row['Activities'];
$pic = $row['Picture'];
$misc = $row['other'];
?>
<h1><?php print $name; ?></h1>
<p><?php print '<img src="' . $pic . '"'; ?>
<?php } ?>
</body>
</html>
This is what i see on my webpage:
" . $sql . "
" . mysql_error()); WHILE($row = mysql_fetch_array($result)) { $name = $row['Name']; $page_id= $id; $position = $row['Position']; $major = $row['Major']; $hometown = $row['Hometown']; $awards = $row['Awards']; $bio = $row['Description']; $act = $row['Activities']; $pic = $row['Picture']; $misc = $row['other']; ?>
and thats all. any ideas what i am doing wrong?
you just don't have PHP enabled on your host.
Hint: always see page source, not picture rendered by browser. It's HTML code being result of your PHP script, so, you have to check HTML code, not a picture rendered from it.
The PHP isn't being parsed, presumably because the necessary module/content handler isn't set up within your web server.
It's not directly related to the topic but you might want to cast the value of the GET parameter as an integer before reusing it in a query to prevent basic SQL injection
$id = intval( $_GET['id'] );