I'm showing a data table from MySQL, let say in x.php like
<?php $sql = "SELECT * FROM tblname";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$id=$row["id"];
echo "<tr><td>". $row["0"]. "</td><td>". $row["1"]."</td><td>".$row["2"]."</td><td>".$row["3"]."</td><td>"."<a href='y.php?=$id'>More details</a>"."</td></tr>";
$_SESSION["id"]=$id;
}
} else {
}?>
But where I go to y.php class to see more details it shows data only from last row of MySQL records.
How can I fix it?
What's wrong is you are using the $_SESSION to fetch the detail.
What you want isn't to store the id in the $_SESSION, you want to put it in each link, so that when a user clicks a link he has access to the row detail.
Basically, parameters passed in a link (like index.php?param1=value1) are found in $_GET global variable.
So I guess in y.php you use $_SESSION['id'] to fetch the row detail, use $_GET['id'] instead.
Also, in x.php edit this :
"<a href='y.php?=$id'>More details</a>"
to :
"<a href='y.php?id=$id'>More details</a>"
And remove : $_SESSION['id'] = $id;
Related
I wanna build a presence check for our choir in the style of tinder but not as complex.
The database contains names and file paths of pictures of the members. When you click on the "present" or "not present" button, the next picture and name should be shown. In the background, the database table should be updated with true/false for presence. (this will be done later)
My problem is that it almost works, but instead of showing one member, it shows all members with their pictures in one single page.
I understand that I could fire with Javascript to continue and paused php-function but I don't get the clue how.
I tried "break" in the php and call the function again but that didn't work.
<?php
$conn = new mysqli(myServer, myUser, myPass, myDbName);
$sql = "SELECT * FROM mitglieder";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<img class='pic' src='" .$row["folder"]. "/" .$row["img"]. "'><br>" ;
echo "<div id='name'>" .$row["vorname"]. " " .$row["name"]. "</div> <br>";
echo "<img src=''img.png' id='present' onclick='isPresent()'>";
}
} else {
echo "0 results";
}
$conn->close();
?>
<html>
<script>
$( document ).ready(function() {
console.log("Ready");
);
</html>`
You can use php function
mysqli_fetch_all()
assign it on the variable outside the while loop and loop or access the indexes in your code.
For Example:
$data = mysqli_fetch_all();
echo $data[0]['name'];
foreach($data as $item)
{
echo $item['name'];
}
You need a way to establish a "state" between your web page and the PHP backend so that you can step through the data. I suggest something like this:
Use an auto-increment integer primary key for the database. That way you can access the data in index order. Let's name the column id
Have your JS code send a form variable - named something like LAST_ID to the PHP in your get. i.e http://someurl.com/script.php?LAST_ID=0
On your first call to the server, send LAST_ID = 0
In your PHP code, fetch the value like this: $id = $_GET('LAST_ID');
Change your SQL query to use the value to fetch the next member like this:
$sql = sprintf("SELECT * FROM mitglieder where id > %d limit 1", $id); That will get the next member from the DB and return only 1 row (or nothing at the end of data).
Make sure to return the id as part of the form data back to the page and then set LAST_ID to that value on the next call.
You can use a HTTP POST with a form variable to the server call that sets that member id to present (maybe a different script or added to your same PHP script with a test for POST vs GET). I suggest a child table for that indexed on id and date.
I hope that puts you in a good direction. Good luck
I have a table with registered users. My code is suppose to delete a row when clicking delete in the table.
This is in the database.php
.....
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
echo '<tr><td align="left">' . $row['Id'] . '</td><td align="left">Delete</td></tr>';
}
......
So, I'm getting the id when clicking delete. So far, this part works but when I tried to run the delete query it doesn't work.
delete.php
<?php
session_start();
include 'connection.php';
if (isset($_POST['Id']) && is_numeric($_POST['Id'])){
$id = mysqli_real_escape_string($conn, $_POST['Id']);
$result = mysqli_query("DELETE FROM table_name WHERE Id= '$id' ")
or die(mysqli_error());
echo "<h3><br><br><a href=database.php> <b> Go Back</a></h3>";
echo "Data Deleted";
}else {
echo "Error";
echo "<h3><br><br><a href=database.php> <b> Go Back</a></h3>";
}
?>
I just get "Error" and it doesn't remove the row. How can I fix it?
Edit:
<?php
session_start();
include 'connection.php';
if (isset($_GET['Id']) && is_numeric($_GET['Id']))
{
$id = mysqli_real_escape_string($conn, $_GET['Id']);
$result = mysqli_query("DELETE FROM User_reg WHERE Id= '$id' ")
or die(mysqli_error());
echo "<h3><br><br><a href=AdminLog.php> <b> Go Back</a></h3>";
echo "Data Deleted";
}else {
echo "Error";
echo "<h3><br><br><a href=AdminLog.php> <b> Go Back</a></h3>";
}
?>
Still getting the same result with the delete query not working.
Also "Id" name is set in the same way as in the database.
$_POST['Id'] is not set, because you got to that script via a link.
Delete
links are GET requests, not POST requests. So, $_GET['id'] (note that it is $_GET['id'] rather than $_GET['Id'] because you used id in your link) should be set, but it's not really safe to use a link to delete things to begin with.
There are various ways to get around this issue. One way is to have the delete link in your table direct you to a intermediate confirmation page that posts to the actual delete script.
it would not work because you are sending a get parameters and checking for post and note the comment above for prepared statement and also try not to use get to delete data because a programmer can easily change the id and delete another user info use post instead because it cant be tweaked that is why social use let me callm it ajax to delete, because a deleted cannot be retrieved unless you create an alternative so use POST METHOD instead change this
if (isset($_POST['id']) && is_numeric($_POST['id'])){
$id = mysqli_real_escape_string($conn, $_POST['id']);
to this
if (isset($_GET['id']) && is_numeric($_GET['id'])){
$id = mysqli_real_escape_string($conn, $_GET['id']);
This should work
I have made a connection to mysql database and echoing values from a table.
while($data = mysql_fetch_array($res))
{
?>
<a href="nextpage.php"<?php echo $data['rowname'] ?></a>
<?php
}
?>
Problem is when I click on the particular link, on the nextpage.php, it should display only the result of the value of a href clicked. So on the nextpage.php, I have defined something like SELECT * from tablename where rowname = 'a href value'.
What's happening now is that it displays only the last rowname value regardless of whichever link I click on, very obvious!
I have tried forms, arrays, SESSIONS but to no avail. How do I fix this?
the href should be like this
<?php echo $data['rowname']; ?>
and then on next page you can use $_GET['val'] and pass it to SELECT query
Try example as below
page1.php
while($data = mysql_fetch_array($res))
{
echo "<a href='nextpage.php?id=".$data['rowname']." >". $data['rowname'] ."</a>";
}
?>
your href will look like nextpage.php?id=[your value]
nextpage.php
$qry = mysql_query("select * from [table_name] where id = ".$_GET['id']."",$con);
while($data = mysql_fetch_array($res))
{
echo $data[0];
}
on nextpage pass value using $_GET['id'] to sql query.
Here is my code:
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;
$conn=mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name'],$conn);
$exec_query =mysql_query($query) or die(mysql_error());
$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
The Problem is that I am not getting anything in $row I cant get into the while loop, nothing shows up when I try to echo the value of $row, No error Nothing. Can you help me to find a problem in my code ?
Ps : The database is their. I have checked for the query for the corresponding value of $campagin_id. and also when i tried to echo $exec_query it echoed this : Resource id #8
PPS : The database have more than 7 record for each id so it doesn't matter if I call mysql_fetch_array($exec_query) more than once before going in to the while loop. and for the $campagin_id in the session their are many records present in the database.
You have written $row=mysql_fetch_array($exec_query) and then you are echoing something. and you are using the same in while.
Instead of:
$row=mysql_fetch_array($exec_query);
echo "<br> row = ".$row;
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
Use this (as per my knowledge you should not use $row=mysql_fetch_array() once you have used before while):
while ($row=mysql_fetch_array($exec_query)){
echo "I am In";
}
If the query returns Resource id #8 then that means it was successful - ie there were no errors. There were probably no rows returned by that query, so no rows in your table that match the given campagin_id.
You are also calling mysql_fetch_array() twice separately, you shouldn't do that because your while loop will skip the first row because calling this moves the pointer in the result set forward by one.
Also you can't echo an array as you are trying to, if you want to see the contents of an array use print_r() or var_dump().
I suggest adding some code to handle no rows found:
if($exec_query && mysql_num_rows($exec_query) > 0)
{
while ($row=mysql_fetch_array($exec_query)){
echo "Row: " . print_r($row, true);
}
}
else
{
echo 'None found';
}
Try this code.
<?
$campagin_id = $_SESSION['campagin_id_for_camp'];
$query = "SELECT * FROM survey_result where campagin_id = ".$campagin_id;
mysql_connect($dbconfig['db_hostname'],$dbconfig['db_username'],$dbconfig['db_password']) or die(mysql_error());
mysql_select_db($dbconfig['db_name']);
$exec_query =mysql_query($query) or die(mysql_error());
while ($row=mysql_fetch_assoc($exec_query)) {
echo "<br/> row = <pre>".print_r($row)."</pre><br/>";
}
?>
I'm trying out my hand at php at the moment - I'm very new to it!
I was wondering how you would go about selecting all items from a mySQL table (Using a SELECT * FROM .... query) to put all data into an array but then not displaying the data in a table form. Instead, using the extracted data in different areas of a web page.
For example:
I would like the name, DOB and favorite fruit to appear in one area where there is already say 'SAINSBURYS' section hardcoded into the page. Then further down the next row that is applicable to 'ASDA' to appear below that.
I searched both here and google and cant seem to find an answer to my strange questions! Would this involve running the query multiple times filtering out the sainsburies data and the asda data where ever I wanted to place the relevant
echo $row['name']." ";
echo $row['DOB']." "; etc etc
next to where it should go?
I have got php to include data into an array (I think?!)
$query = "SELECT * FROM people";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
echo $row['name']." ";
echo $row['DOB']." ";
echo $row['Fruit']." ";
}
?>
Just place this (or whatever your trying to display):
echo $row['name']." ";
Anywhere you want the info to appear. You can place it within HTML if you want, just open new php tags.
<h1>This is a the name <?php echo $row['name']." ";?></h1>
If you want to access your data later outside the while-loop, you have to store it elsewhere.
You could for example create a class + array and store the data in there.
class User {
public $name, $DOB, $Fruit;
}
$users = new array();
$query = "SELECT * FROM people";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)) {
$user = new User;
$user->name = $row["name"];
$user->DOB = $row["DOB"];
$user->Fruit = $row["Fruit"];
$users[$row["name"]] = $user;
}
Now you can access the user-data this way:
$users["USERNAME"]->name
$users["USERNAME"]->DOB
$users["USERNAME"]->Fruit