I currently have made a page that shows all the users, but I just want to only display the currently logged in user, Please help me im begging you I spent 4 hours of nothing T_T im new at php and im very lost, im so hopeless right now
<?PHP
$customerquery=mysql_query("select * from customerinfo");
while($customerrows=mysql_fetch_array($customerquery)){
?>
<tr>
<td>Id</td><td>First Name</td><td>Last Name</td><td>Address</td><td>Contact No</td> <td>Username</td><td>Password</td><td>Edit</td>
</tr>
<tr>
<td><?PHP echo $customerrows['id'];?></td>
<td><?PHP echo $customerrows['fname'];?></td>
<td><?PHP echo $customerrows['lname'];?></td>
<td><?PHP echo $customerrows['address'];?></td>
<td><?PHP echo $customerrows['contactno'];?></td>
<td><?PHP echo $customerrows['username'];?></td>
<td><?PHP echo $customerrows['password'];?></td>
<td>edit</td>
</tr>
<?PHP } ?>
You need to specify a WHERE condition in your query. That will fetch only one row, so you will not need a while loop:
$id = (int) $_GET['id']; // assuming you pass user id in URL
$customerquery = mysql_query("select * from customerinfo WHERE id = $id");
$customerrows = mysql_fetch_array($customerquery);
// rest of your html
Related
Okay, my problem is that how can I set different hyperlinks to each row of a table in php?
so far i've done this -->
<?php
$conn = mysqli_connect("localhost", "root", "", "makaut_colleges");
$result = mysqli_query($conn, "SELECT * FROM collegelist");
while ($row = mysqli_fetch_assoc($result)):
?>
<tr>
<td><?php echo $row['College Name']; ?></td>
<td><?php echo $row['College Address']; ?></td>
<td><?php echo $row['College Code']; ?></td>
<td><?php echo $row['Review']; ?></td>
</tr>
<?php endwhile; ?>
How can i add different hyperlinks to each row which will be extracted from database? i'm using 'College Code' as primary key.
You can try something like this
Try this for making tr clickable
$dynamic_link = "dynamic.php";
<tr onclick="document.location = '<?php echo $dynamic_link ?>'">
td code---------
</tr>;
where $dynamic_link is dynamic based on your requirement
Thier should be some logic or linking with url either save in table or create some link
I'm having some trouble generating output from an sql query via php. When I execute the query "SELECT * from projectdb WHERE name = 'Crys' or ID = 14142" on phpmyadmin it returns valid results, but attempting to do so by passing a post value yields an empty table. See code below:
<html>
<title>Search result</title>
<body>
<table border="1px">
<tr>
<td>name</td>
<td>ID</td>
<td>position</td>
<td>job scope</td>
<td>contact_no</td>
<td>days_off</td>
<td>wages</td>
</tr>
<?php
if (isset($_POST['value']))
{
$ID=$_POST['staff ID'];
$name=$_POST['staff name'];
$admincon =mysqli_connect("localhost","root","","projectdb");
/* Query I need to execute and print in table*/
$sqlsrch2 =mysqli_query($admincon, "select * from staff where name='".$name."' or ID='".$ID."'");
while($result=mysqli_fetch_assoc($sqlsrch2)){
?>
/* table where results needs to be printed */
<tr>
<td><?php echo $result['name'];?></td>
<td><?php echo $result['ID'];?></td>
<td><?php echo $result['position'];?></td>
<td><?php echo $result['job_scope'];?></td>
<td><?php echo $result['contact_no'];?></td>
<td><?php echo $result['days_off'];?></td>
<td><?php echo $result['wages'];?></td>
</tr>
<?php
}
}
?>
</table>
</body>
</html>
A few points to note:
I'm just a beginner in PHP and SQL; I'm just looking for a simple answer.
Security is not at all a concern here; this is just a rough demo.
If this is marked as duplicate, do help redirect me to a link where I can get the solution. Thanks!
I've been turning in circles for a while now trying to understand what I have done wrong with lines 68 and 110. It comes up with:
"expects parameter 1 to be resource, boolean given"
but I don't understand where my mistake is. Would someone be able to point me in the right direction or explain my error so I can better understand?
I'm really new to PHP (only started learning it last week) so I've been mostly doing my best with tutorials and what I can find online.
Line 68 onwards:
while($row = mysql_fetch_array($query)){
$card_number = $row['card_number'];
$card_id = $row['card_id'];
$card_name = $row['card_name'];
$card_mana_img = $row['card_mana_img'];
$card_type = $row['card_type'];
$card_rarity = $row['card_rarity'];
$card_set = $row['card_set'];
}
?>
Lines 99 onwards:
<table>
<tr>
<td>Number</td>
<td>Name</td>
<td>Type</td>
<td>Mana</td>
<td>Rarity</td>
<td>Set</td>
</tr>
<?php while ($row = mysql_fetch_array($query)) { ?>
<tr>
<td><?php echo $card_number; ?></td>
<td><?php echo $card_name; ?></td>
<td><?php echo $card_type; ?></td>
<td><?php echo $card_mana_img; ?></td>
<td><?php echo $card_rarity; ?></td>
<td><?php echo $card_set; ?></td>
</tr>
<?php } ?>
</table>
<br>
<?php echo $paginationCtrls; ?><br>
<?php echo $textline2;?><br>
<?php echo $textline1;?>
Don't use this: $sql = "SELECT number FROM magicorigins_cardset";
Use this:
$sql = "SELECT count(*) FROM magicorigins_cardset";
i have a table with id, name, address, sector, financiar, link
on the link i when i press it i want to show me 2 tables from the id of row selected, ex: id 1.
http://postimg.org/image/khelg1m0z/
and the result: http://s28.postimg.org/srvcwj065/Capture2.jpg
now it's a static page with search clause where by id 1, but i need an automatically link show by id on each row.
<?php
include "connect.php";
$sql = "select * from studenti where id='1'";
$query = mysql_query($sql) or die (mysql_error());
?>
<table width="70%" cellpadding="5" cellspace="5">
<tr><td>Id</td>
<td>Nume</td>
<td>Localitate</td>
<td>Judet</td>
<td>Sector Financiar</td>
<td>Link</td></tr>
<?php while ($row = mysql_fetch_array($query)) { ?>
<tr>
<td><?php echo $row['id']; ?></td>
<td><?php echo $row['nume']; ?></td>
<td><?php echo $row['localitate']; ?></td>
<td><?php echo $row['judet']; ?></td>
<td><?php echo $row['sector_financiar']; ?></td>
<td><?php echo $row['link']; ?></td>
<?php } ?>
</table>
<?php
include "connect.php";
$sql1 = "select * from certificari where id='1' ";
$query = mysql_query($sql1) or die (mysql_error());
?>
<table width="70%" cellpadding="5" cellspace="5">
<tr><td>Id</td>
<td>Denumire certificare</td>
<td>Serie si numar certificare</td>
<td>Data certificarii</td>
<td>Valabilitate certificare</td>
<td>Sector Financiar</td></tr>
<?php while ($row = mysql_fetch_array($query)) { ?>
<tr>
<td><?php echo $row['id']; ?></td>
<td><?php echo $row['nume']; ?></td>
<td><?php echo $row['serie_numar']; ?></td>
<td><?php echo $row['data']; ?></td>
<td><?php echo $row['valabilitate']; ?></td>
<td><?php echo $row['sector_financiar']; ?></td>
<?php } ?>
</table>
You should not use the mysql_ functions since they are depricated and are very vulnerable to SQL injection attacks. Instead, I will use MySQLi in this answer, but you could also use PDO if you want to.
To display records for different ID's based on what the user selects you can pass an ID in the page URL. When you link to the page you add the desired ID to the address like this:
Link
The value of the variable id will now be available in page.php as $_GET['id'].
The actual PHP in your page would then look something like the code below. I have left out some of your HTML for brevity, but you can just add it int.
//Connect to the DB. Might want to put this in your connect.php and include it.
$db = new mysqli('localhost', 'user', 'pass', 'db');
//Prepare the statement. The ? will be your ID.
$statment = $db->prepare("SELECT * FROM studenti WHERE id = ?");
//Bind the parameters, so that the first (and only) question mark is turned into your ID.
//The 'i' means integer, if you store the id as a string your should use 's' instead.
$statement->bind_param('i', $_GET['id']);
//Execute the query.
$statement->execute();
//Get the results.
$result = $statement->get_result();
//Iterate over it to create the output.
while ($row = $result->fetch_assoc()) {
?>
<tr>
<td><?php echo $row['id']; ?></td>
<td><?php echo $row['nume']; ?></td>
<td><?php echo $row['localitate']; ?></td>
<td><?php echo $row['judet']; ?></td>
<td><?php echo $row['sector_financiar']; ?></td>
<td><?php echo $row['link']; ?></td>
</tr>
<?
}
//Then do the same thing for the table certificari.
//Note that you only need to connect to the DB once.
This beginners guide to MySQLi is very helpful if you need some more guidance.
I'll try to explain the problem straight away. I have one HTML form which takes input just like a comment form and it saves the xyz data into a MySQL database using PHP. Now, what I want is to create and display links for those comments on a page.
I mean the comments which have been saved including the user's email and name, should be opened by clicking a link.
I don't want to display all the details on a single page from the database for all the users. There should be a page on which links are shown, when a user click a link, the full post should be displayed in next page.
There is not something which I know about this process. Please help me out.
// $rows = set of result from your database query
foreach($rows as $row){
echo '<a'
. ' href="my_link_to_display_comment?id='.$row['id'].'">'
. 'Comment from '.$row['user_name']
. '</a>';
}
First a page to display all the links like the below example -
$result = mysql_query("SELECT * FROM calendar WHERE sort_month='11'");
while($row = mysql_fetch_array($result))
{echo
"".$row['event_name'].""
;}
and then in event.php(the next page after clicking link)
$id = $_GET['id'];
$sql = "select * from calendar where id = $id";
$result = mysql_query($sql, $con);
if ($result){
$row = mysql_fetch_row($result);
$title = $row[12];
$content = $row[7];} ?>
<?php echo $title ?>
<?php echo $content ?>
If you want to show details of a single user just do this.
You can make a search box by using a form.
eg. like if I want to display a details of a student, I will search him by using his roll number and run these queries.
<?php //to search student
require_once './secure.inc.php';
$status = 0;
if(isset($_POST['submit'])){
$roll_number = $_POST['roll_number'];
$query = "select * from students where roll_number=$role_number";
require_once '../includes/db.inc.php';
$result = mysql_query($query);
if(mysql_num_rows($result)==1){
$status = 1;
$row = mysql_fetch_assoc($result); //mysql_fetch_array - both numeric and key index
}else{
$status=2;
}
}
?>
//to display
<?php } else if($status==1) { ?>
<table>
<tbody>
<tr>
<td>Roll Number : </td>
<td><?php echo $row['roll_number']; ?></td>
</tr>
<tr>
<td>Name : </td>
<td><?php echo $row['name']; ?></td>
</tr>
<tr>
<td>Gender : </td>
<td><?php echo $row['gender']; ?></td>
</tr>
<tr>
<td>Email : </td>
<td><?php echo $row['email']; ?></td>
</tr>
<tr>
<td>Mobile Number : </td>
<td><?php echo $row['mobile_number']; ?></td>
</tr>
<tr>
<td>Course : </td>
<td><?php echo $row['course']; ?></td>
</tr>
</tbody>
</table>
<?php } ?>