enter image description herei am working on project CMS with php and one of my while loops doesn't work and i cant find why.
i am echo on (td) inside of loop and nothing showed on the page but when i echo outside of while loop it work very well.
i have commentet the loop to see you.
Can you give me some help where i have the problem?
<table class="table table-bordered table-hover">
<thead>
<tr>
<th>Id</th>
<th>Author</th>
<th>Comment</th>
<th>Email</th>
<th>Status</th>
<th>In Response to</th>
<th>Date</th>
<th>Approve</th>
<th>Unapprove</th>
<th>Delete</th>
</tr>
</thead>
<tbody>
<?php
$query = "SELECT * FROM comments ";
$select_comments = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($select_comments)) {
$comment_id = $row['comment_id'];
$comment_post_id = $row['comment_post_id'];
$comment_author = $row['comment_author'];
$comment_content = $row['comment_content'];
$comment_email = $row['comment_email'];
$comment_status = $row['comment_status'];
$comment_date = $row['comment_date'];
echo "<tr>";
echo "<td>$comment_id</td>";
echo "<td>$comment_author</td>";
echo "<td>$comment_content</td>";
/
echo "<td>$comment_email</td>";
echo "<td>$comment_status</td>";
//THIS IS THE LOOP DOESN'T WORK
$query = "SELECT * FROM posts WHERE post_id = $comment_post_id";
$select_post_id_query = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($select_post_id_query)) {
$post_id = $row['post_id'];
$post_title = $row['post_title'];
echo "<td><a href='../post.php?p_id=$post_id'>$post_title</a></td>";
}
echo "<td>$comment_date</td>";
echo "<td><a href='comment.php?approve=$comment_id'>Approve</a></td>";
echo "<td><a href='comment.php?unapprove=$comment_id'>Unapprove</a></td>";
echo "<td><a href='comment.php?delete=$comment_id'>Delete</a></td>";
echo "</tr>";
}
?>
</tbody>
</table>
<?php
//approve posts
if(isset($_GET['approve'])){
$the_comment_id = $_GET['approve'];
$query = "UPDATE comments SET comment_status = 'approved' WHERE comment_id = $the_comment_id ";
$approve_comment_query = mysqli_query($connection, $query);
header("Location: comment.php");
}
//unapprove posts
if(isset($_GET['unapprove'])){
$the_comment_id = $_GET['unapprove'];
$query = "UPDATE comments SET comment_status = 'unapproved' WHERE comment_id = $the_comment_id ";
$unapprove_comment_query = mysqli_query($connection, $query);
header("Location: comment.php");
}
//delete posts
if(isset($_GET['delete'])){
$the_comment_id = $_GET['delete'];
$query = "DELETE FROM comments WHERE comment_id = {$the_comment_id} ";
$delete_query = mysqli_query($connection, $query);
header("Location: comment.php");
}
?>
You are overwriting variable $row in your second loop. You should change it eg. to $subRow to avoid such situation.
i have find the missing part on of my db with the connection.
THNX all you friends :D
Related
I want to add this line
a href="r.php?id=<?php echo $row['id']; ?>" >Details>
which will display details of this id on another page.
but i don't know how to write this line correctly. can you help me please thank you.
$conn = new mysqli('localhost', 'root', '', 'dbmtc');
$perPage = 10;
$page = 0;
if (isset($_POST['page'])) {
$page = $_POST['page'];
} else {
$page=1;
};
$startFrom = ($page-1) * $perPage;
$sqlQuery = "SELECT fname, lname, dept, email
FROM staff ORDER BY fname ASC LIMIT $startFrom, $perPage";
//echo $sqlQuery;
$result = mysqli_query($conn, $sqlQuery);
$paginationHtml = '';
while ($row = mysqli_fetch_assoc($result)) {
$paginationHtml.='<tr>';
$paginationHtml.='<td>'.$row["fname"].' '.$row["lname"].'</td>';
$paginationHtml.='<td>'.$row["dept"].'</td>';
$paginationHtml.='<td>'.$row["email"].'</td>';
**this line should be here**
$paginationHtml.='</tr>';
}
$jsonData = array(
"html" => $paginationHtml,
);
echo json_encode($jsonData);
?>
This part is in index.php
<table class="table table-hover table-bordered">
<thead>
<tr>
<th>Name</th>
<th>Department</th>
<th>Email</th>
<th>Operations</th>
</tr>
</thead>
<tbody id="content">
</tbody>
</table>
This is the table
And when i insert
$paginationHtml .= '<td>Details</td>';
I got this I don't why, that why I said it's not good
enter image description here
First of all: You didn't select the ID in your sql query. Please add it in the selected fieds.
These are the possibilities to write this and you can choose one of them:
1- $paginationHtml .= '<td><a href="r.php?id='.$row['id'].'" >Details</a></td>';
2- $paginationHtml .= "<td><a href=\"r.php?id={$row['id']}\" >Details</a></td>";
I created a table which is updated through a form and each row gets assigned a specific number.
When viewing this table, I want to click on that assigned number and get a page where all the details of that row are displayed.
If I do $sql = "SELECT * FROM clients WHERE nif_id='114522';"; - where the nif_id is the assigned number - I get the values for that number, but I need it to change with every number in the table.
Any ideas?
UPDATE
This is the table code:
<div class="card card-body">
<table class="table">
<thead>
<tr>
<th>NIF</th>
<th>Nome</th>
<th>Apelido</th>
<th>Telemóvel</th>
<th>E-mail</th>
</tr>
</thead>
<tbody>
<?php
include_once '../includes/db.inc.php';
$sql = "SELECT * FROM clients ORDER BY nif_id ASC;";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if ($resultCheck > 0) {
while ($row = mysqli_fetch_assoc($result)) {
$first = $row["prm_nome"];
$last = $row["apelido"];
$phone = $row['nmr_tlm'];
$email = $row['mail'];
$nif = $row['nif_id'];
echo '<tr>';
echo '<td>'.$nif.'</td>';
echo '<td>'.$first.'</td>';
echo '<td>'.$last.'</td>';
echo '<td>'.$phone.'</td>';
echo '<td>'.$email.'</td>';
echo '</tr>';
}
}
?>
</tbody>
</table>
</div>
You can use the get request parameters.
ex: www.myapp.com/table?id=3920393
add functionality in your PHP file as follows
if(isset($_GET["id"])){
$id = $_GET["id"];
$sql = "SELECT * FROM clients WHERE nif_id='".$id."';";
//make db call & display HTML
}
This is a very simple implementation and does not implement any security or SQL injection security. This was more of a conceptual answer as to how you can tackle your problem.
This is quite a common scenario for web-based systems.
<div class="card card-body">
<table class="table">
<thead>
<tr>
<th>NIF</th>
<th>Nome</th>
<th>Apelido</th>
<th>Telemóvel</th>
<th>E-mail</th>
</tr>
</thead>
<tbody>
<?php
include_once '../includes/db.inc.php';
$sql = "SELECT * FROM clients ORDER BY nif_id ASC;";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if ($resultCheck > 0) {
while ($row = mysqli_fetch_assoc($result)) {
$first = $row["prm_nome"];
$last = $row["apelido"];
$phone = $row['nmr_tlm'];
$email = $row['mail'];
$nif = $row['nif_id'];
echo '<tr>';
echo '<td>'.$nif.'</td>';
echo '<td>'.$first.'</td>';
echo '<td>'.$last.'</td>';
echo '<td>'.$phone.'</td>';
echo '<td>'.$email.'</td>';
echo '</tr>';
}
}
?>
</tbody>
</table>
</div>
where the detail.php is another page to query specific details regarding the query nifid.
As a reminder, if the data type of the column is INT, there is no need to use single quotes to surround the value in the SQL statement.
Sample detail.php:
<?php
if(!isset($_GET['nifid']) || (int)$_GET['nifid'] <= 0) {
// Invalid or missing NIFID
header('Location: table.php');
}
include_once '../includes/db.inc.php';
$id = (int)$_GET['nifid'];
$sql = "SELECT * FROM clients WHERE nif_id=$id";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
// TODO: display the result in whatever way you like
?>
I currently have this code set up:
$sql = "SELECT * FROM homework WHERE class = '$class'";
$result = mysqli_query($conn, $sql);
$data_exist = false;
if (mysqli_num_rows($result) > 0) {
// output data of each row
$data_exist = true;
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
$teacher_set = $row["teacher_set"];
$class = $row["class"];
$name = $row["name"];
$description = $row["description"];
}
}
And then:
<?php if ($data_exist){?>
<p><?php echo $id ?></p>
<p><?php echo $teacher_set?></p>
<p><?php echo $name?></p>
<p><?php echo $description?></p>
<?php
}?>
However, the issue is if there is multiple results in the database it only outputs one of them, how can I prevent this from happening and output two?
I want to make it so every row has their own section, like this: http://prntscr.com/hcgtqn so if there is only one result, one one will show etc.
You have to echo data in a loop. Right now you are reassigning values in while($row = mysqli_fetch_assoc($result)) iterations and printing just the last one.
You need to print each time you read a row from the database.
about the styles, you can represent it in many ways. In the code below I present it in a table.
<table>
<thead>
<tr>
<th>id</th>
<th>teacher set</th>
<th>name</th>
<th>description</th>
</tr>
</thead>
<tbody>
<?php
$sql = "SELECT * FROM homework WHERE class = '$class'";
$result = mysqli_query($conn, $sql);
$data_exist = false;
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_array($result)) {
$id = $row["id"];
$teacher_set = $row["teacher_set"];
$class = $row["class"];
$name = $row["name"];
$description = $row["description"];
// you need to print the output now otherwise you will miss the row!
// now printing
echo "
<tr>
<td>".$id."</td>
<td>".$teacher_set."</td>
<td>".$name."</td>
<td>".$description."</td>
</tr>";
}
}
else // no records in the database
{
echo "not found!";
}
?>
</tbody>
</table>
</body>
</html>
Firstly, the headings are stored in h1 tags. This is taken from a separate table named "menu_type". Which is linked through a "menu" table.
I am trying to display data on the base like this:
HEADING
Table Data
2nd Heading
Second Data
--- In a loop until it is all completed ---
Here is a like page of what it is doing
I believe I have the methods correct and can see what it is doing, it is printing the first heading, then a blank table, then the second heading and then the data from the first table.
See my code:
<?php
$query = "SELECT * FROM menu_type";
$result = mysqli_query($connect, $query);
$result_array = array();
$numRows = mysqli_num_rows($result); // returns the num of rows from the query above, allowing for dynamic returns
while($row = mysqli_fetch_assoc($result)){
$menuType = $row['type'];
$result_array[] = $menuType; // This array holds each of the menu_types from DB
}
$countArray = count($result_array);
for($i = 0; $i < $numRows; $i++){
$query = "SELECT * FROM menu WHERE type_id='$result_array[$i]'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
echo "
<div id='hide'>
<h1 id='wines' class='head-font text-center head'>$result_array[$i]</h1>
<table class='table table-hover table-responsive'>
<thead>
<tr>
<th>
Item
</th>
<th class='text-right'>
Price
</th>
</tr>
</thead>
<tbody>
<tr>
";
$menuQuery = "SELECT * FROM menu, menu_type WHERE type_id='$i' AND menu.type_id = menu_type.id";
$menuResult = mysqli_query($connect, $menuQuery) or die(mysqli_error($connect));
while ($row = mysqli_fetch_assoc($menuResult)) {
$name = $row['name'];
$description = $row['description'];
$price = $row['price'];
echo "
<td>$name - <small>$description</small></td>
<td class='text-right'>£$price </td>
";
}
echo
"
</tr>
</tbody>
</table>
";
}
// print_r($result_array[2]);
?>
You don't need these anymore, it's like you're repeating the query. It looks incorrect also.
$menuQuery = "SELECT * FROM menu, menu_type WHERE type_id='$i' AND menu.type_id = menu_type.id";
$menuResult = mysqli_query($connect, $menuQuery) or die(mysqli_error($connect));
The menu items are already in this query, you just have to loop through it;
$query = "SELECT * FROM menu WHERE type_id='$result_array[$i]'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
UPDATE 1: this code is incorrect, it doesn't insert the value to the array. I updated the code (after "try this").
$result_array[] = $menuType;
UPDATE 2: the $result is repeatedly used in mysqli functions, the index is being moved. What I did is copied the initial $result to $resultCopy. Try code again, haha
Use array_push($array,$value_you_insert) function, for inserting elements to an array.
Try this;
<?php
$query = "SELECT * FROM menu_type";
$result = mysqli_query($connect, $query);
$resultCopy = $result;
$result_array = array();
$numRows = mysqli_num_rows($result); // returns the num of rows from the query above, allowing for dynamic returns
while($row = mysqli_fetch_assoc($resultCopy)){
$menuType = $row['type'];
array_push($result_array,$menuType);
}
for($i = 0; $i < $numRows; $i++){
echo "
<div id='hide'>
<h1 id='wines' class='head-font text-center head'>".$result_array[$i]."</h1>
<table class='table table-hover table-responsive'>
<thead>
<tr>
<th>Item</th>
<th class='text-right'>Price</th>
</tr>
</thead>
<tbody>
";
$query = "SELECT * FROM menu WHERE type_id='$result_array[$i]'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
while ($row = mysqli_fetch_assoc($result)) {
$name = $row['name'];
$description = $row['description'];
$price = $row['price'];
echo"
<tr>
<td>".$name." - <small>".$description."</small></td>
<td class='text-right'>£".$price."</td>
</tr>
";
}
echo
"
</tbody>
</table>
";
}
?>
I've made this code using a tutorial that allows me to upload users to the database. The whole thing works great, but the only problem is that it starts to show 2 of the same user over and over, the list starts expanding 5 every user i add... What could be the problem causing this?
Item in index that lays out the whole list:
<h2>Names:</h2>
<table border='1'>
<tr>
<th>ID</th>
<th>Username</th>
</tr>
<?php
$sql_list = "SELECT * FROM names ORDER BY username ASC";
$results = mysqli_query($db, $sql_list) or die(mysql_error());
$names = "";
if(mysqli_num_rows($results) > 0) {
while($row = mysqli_fetch_assoc($results)) {
$id = $row['id'];
$user = $row['username'];
$names .= "<tr><td>$user</td></tr>";
echo $names;
}
} else {
echo "No Users Found";
}
?>
</table>
Either output the one record per iteration; or build the whole HTML block, then output the block. I think the simplest would be:
while($row = mysqli_fetch_assoc($results)) {
$id = $row['id'];
$user = $row['username'];
echo "<tr><td>$user</td></tr>";
}
... alternative approach
<h2>Names:</h2>
<table border='1'>
<tr>
<th>ID</th>
<th>Username</th>
</tr>
<?php
$sql_list = "SELECT * FROM names ORDER BY username ASC";
$results = mysqli_query($db, $sql_list) or die(mysqli_error($db));
$names = "";
if(mysqli_num_rows($results) > 0) {
while($row = mysqli_fetch_assoc($results)) {
$id = $row['id'];
$user = $row['username'];
$names .= "<tr><td>$user</td></tr>";
}
} else {
$names = "No Users Found";
}
echo $names;
?>
</table>
Also you can't use mysql_* functions with mysqli_*. See http://php.net/manual/en/mysqli.error.php.
Simplest example of the issue: https://eval.in/627250