I need each row to have a unique button beside it which will delete the message. I need it to only be able to be deleted by the receiver. My first thought was to echo a HTML form inside the while loop. I don't think that would work though. If it does work what would the SQL statement inside if(isset()) look like? Each message has an ID so could I use that? If you need to see rest of the code let me know.
$Select = "
SELECT * FROM Messages
WHERE Receiver='$Identifier'
ORDER BY ID
DESC
LIMIT 10";
$Result = $Connect->query($Select);
if (mysqli_num_rows($Result) > 0) {
while ($Row = mysqli_fetch_assoc($Result)) {
echo nl2br("Sender = " . $Row["Sender"] . " Message = " . $Row["Message"] . "\n");
}
}
why don't try this:
echo "Sender = " . $Row["Sender"] . " Message = " . $Row["Message"] . "<button id=".$Row["ID"].">Delete</button>";
Related
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.
I have a MYSQL table containing tournament data for players. Each player has stats like games_won, games_lost, etc. I am trying to display this information in a table.
My PHP code looks like this:
//Connecting to $db
//...
//Store table data into array
$result = mysqli_query($db, "SELECT * FROM ladder");
while($row = mysqli_fetch_array($result)) {
echo $row['id'] . " " . <a href=$row['link']$row['username']</a> . " " . $row['tourney_wins'] . " " . $row['game_wins'] . " " . $row['game_losses'] . " " . $row['last_played'];
echo "<br>";
I get an error (Unexpected "<") on
<a href=$row['link']$row['username']</a>
I am trying to display each username in the table as a hyperlink to their respective profile on another site. Can anyone give me a correction? Thanks!
This is how it should be quoted (assuming I interpreted how you wanted the link properly):
//Connecting to $db
//...
//Store table data into array
$result = mysqli_query($db, "SELECT * FROM ladder");
while($row = mysqli_fetch_array($result))
{
echo( $row['id']." <a href='".$row['link']."'>".$row['username']."</a> ".$row['tourney_wins']." ".$row['game_wins']." ".$row['game_losses']." ".$row['last_played'] );
echo("<br>");
}
Change
<a href=$row['link']$row['username']</a>
to
"<a href=$row['link']$row['username']</a>"
You forgot to quote your html:
echo $row['id'] . " " . <a href=$row['link']$row['username']</a> . " " ...snip...
^--here ^--here
Since it's not quoted, PHP is interpreting it <a as concatenate less than undefined constant 'a', which makes no sense.
While dynamically generating a page with different types of contents , the "post" content is appearing above the static content which is being generated.I want it the other way around. Does there appear to be anything in my code that would make this happen, or do you think the problem has something to do with my database? Thanks.
$query = "SELECT * FROM content WHERE pages LIKE '%$pageID%'";
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) {
// Display pages's static content
if ($row['type'] == "static") {
echo "
<h2>" . $row['tile'] . "</h2>
<content>" . $row['body'] . "</content>
";
}
// Display pages's posts
else {
echo "
<h2>" . $row['tile'] . "</h2>
<content>" . $row['body'] . "</content>
";
}
SELECT * FROM content WHERE pages LIKE '%$pageID%' ORDER BY type desc
Add this to the end of your query:
ORDER BY CASE WHEN type = 'static' THEN 0 ELSE 1 END
This is the part of the PHP code I am having the issue:
$query = "SELECT * FROM clients where idcard = '$idcard'";
$result = mysqli_query($dbc, $query)
or die("Error quering database.");
if(mysqli_fetch_array($result) == False) echo "Sorry, no clients found";
while($row = mysqli_fetch_array($result)) {
$list = $row['first_name'] . " " . $row['last_name'] . " " . $row['address'] . " " . $row['town'] . " " . $row['telephone'] . " " . $row['mobile'];
echo "<br />";
echo $list;
}
Even if I insert an existing idcard value I get no output when there is the if statement, an incorrect idcard displays "Sorry, no clients found" fine. However if I remove the if statement if I enter an existing idcard the data displays ok.
Can you let me know what is wrong with the code please ?
Thanks
Use mysqli_num_rows to count the results:
if(mysqli_num_rows($result) == 0) echo "Sorry, no clients found";
mysqli_fetch_array() fetches an item from the database.
This means your if() code fetches a first item from the database.
Then, when you call mysqli_fetch_array() again from the while() condition, the first item has already been fetched, and you are trying to fetch the second one ; which does not exist.
You must ensure that you use the result from mysqli_fetch_array() and not call it one time just for nothing ; or, as an alternative, you could use the mysqli_num_rows() function (quoting) :
Returns the number of rows in the result set.
$query = "SELECT * FROM clients where idcard = '$idcard'";
$result = mysqli_query($dbc, $query)
or die("Error quering database.");
if(mysqli_num_rows($result) == 0) {
echo "Sorry, no clients found";
}else{
while($row = mysqli_fetch_array($result)) {
$list = $row['first_name'] . " " . $row['last_name'] . " " . $row['address'] . " " . $row['town'] . " " . $row['telephone'] . " " . $row['mobile'];
echo $list . "<br />";
}
}
Try this.
EDITED: Added closing bracket.
Use mysqli_num_rows() to test if there is anything returned.
Imagine you put some money in your pocket.
Eventually an idea came to your mind to see if you are still have the money.
You are taking it out and count them. All right.
Still holding them in hand you decided to take them from pocket. Oops! The pocket is empty!
That's your problem.
To see if you got any rows from the database you can use mysqli_num_rows(). It will return the number of bills, without fetching them from the pocket.
The problem is, that you try to use mysqli_fetch_array to queck for the number of results. mysqli_fetch_array will fetch the first result, compare it to false and then discard it. The next mysqli_fetch_array will then fetch the second result, which is not existing.
If you want to check if any clients where found, you can use mysqli_num_rows like this:
$idcard = mysqli_escape_string($dbc, $idcard); // See below: Prevents SQL injection
$query = "SELECT * FROM clients where idcard = '$idcard'";
$result = mysqli_query($dbc, $query) or die("Error quering database.");
if(mysqli_num_rows($result) == 0) {
echo "Sorry, no clients found";
} else {
while($row = mysqli_fetch_array($result)) {
// Do whatever you want
}
}
If $idcard is a user supplied value, please look out for SQL injection attacks.
I'm working at this thing that takes information from a MySQL database that comes from a contact form and displays it all to the user. Well, I got that far without issues using:
$result = mysql_query("SELECT * FROM contact");
while($row = mysql_fetch_array($result))
{
echo $row['name'] . "<br/>" . " " . $row['email'] . "";
echo "<br/>";
echo $row['message'];
echo "<br />";
}
Now I would like to add a button next to each item that would single them out, basically taking me to a page where only the selected row shows up, not all of them, as happens on this page.
The thing is that i don't know how to do that. So, does anyone have any ideas?
Thanks!
Add a button that refreshes the page with a "GET" parameter, such as yourpage.php?uid=123; Then do
if ($_GET['uid'] !== null) {
$id = (int) $_GET['uid'];
$result = mysql_query('SELECT * FROM contact WHERE id='.$id);
...
}