I work with this code:
<form method="get" name="MobileDetails">
<input name="brand" id="brand" value="<?php echo $brand;?>" type="hidden">
<input name="brid" id="brid" value="<?php echo $brandid;?>" type="hidden">
<button type="button" name="submitButton" value="get Details" onclick="getDetails()">
</form>
java script
<script type="text/javascript">
function getDetails(){
var brand = document.getElementById('brand').value;
var brandid = document.getElementById('brid').value;
document.MobileDetails.action = 'details.php?brand='+brand+'&id='+brandid;
document.MobileDetails.submit();
}
</script>
But it does not work in while loop. Whats the problem? My code is given below.
When i click on the button it do not do anything. But the code work great with out while loop given on the top.
<?php
require_once('connection.php');
$SQL= "SELECT*FROM mobile ORDER BY price ASC LIMIT 10";
$result= mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)){
$brand=$db_field['brand'];
$id=$db_field['id'];
$model=$db_field['model'];
echo "<form method='get' name='MobileDetails'>";
echo " <input name='brand' id='brand' value='". $brand ."' type='hidden'>";
echo" <input name='brid' id='brid' value='". $id ."' type='hidden'>";
echo" <input name='mod' id='mod' value='". $model ."' type='hidden'>";
echo" <button type='button' name='submitButton' value='get Details' onclick='getDetails()'/>
</form> ";
echo "CLICK HERE";
}
?>
You're using several times the same id. Ids have to be unique.
Uou are dealing with multiple id's. The job of an ID is to be unique identifier for the element. I suggest just using
<form action="details.php" type="get">
this will do exactly what you are trying to achieve without using the function.
The thing with element ID's is that they need to be unique for the page; however, as you may see, not required for HTML to be displayed. When calling your JS function getDetails(), it grabs the element by ID but when you have multiple ID's in the page, this will fail.
So what can you do? Well, in your loop, you create a new form for each 'brand'. You can pass a reference of the form to the grabdetails and then, by NAME, grab the values from that form.
Rather than using Javascript to generate a link based on given details put in a hidden field, you should just generate the action at the PHP level.
echo "<form method='get' name='MobileDetails' action='details.php?brand=$brand&id=$brandid'>";
But since you do have hidden fields, using just action='details.php' the form will take the user to
details.php?brand={brand}&brid={id}&mod={model}
You should look into POST or making your button into a plain link rather than having a form.
Related
So I have a php page that looks like this
<?php
echo "<table border='1' width= 300px >
<tr>
<th>Friend Names</th>
<th>Remove Friends</th>
</tr>";
while($row = mysqli_fetch_assoc($result2))
{
$friend_id_got = $row['friend_id2'];
$query3 = "SELECT profile_name
from friends
where friend_id = '$friend_id_got' ";
$result3 = $conn->query($query3);
$final3 = mysqli_fetch_assoc($result3);
echo "<tr>";
echo "<td>" . $final3['profile_name'] . "</td>";
echo "<td>"
?>
<form action="friendlist.php" method= "POST">
<button id="add-friend-btn" type= 'submit' name = 'submit'>Unfriend</button>
</form>
<?php
"</td>";
echo "</tr>";
}
echo "</table>";
When I press the button, I need the corresponding name to delete it. The only problem I'm facing is that How do I get the name corresponding to the button.
I think we need to relate the buttons and the name somehow, so when a specific button is pressed i get the corresponding name
From the question and comments, and a glance at your code it sounds like you probably actually need two pieces of data to be submitted to the server when your button is clicked:
The action which should be undertaken in response to the request (i.e. unfriending)
The ID of the person being unfriended
To achieve that you can add some hidden fields to your form. These are invisible to the user but will be available to PHP in the $_POST data when the form is submitted.
Something like this:
<form action="friendlist.php" method= "POST">
<input type="hidden" name="unfriend_id" value="<?=$friend_id_got ?>" />
<input type="hidden" name="action" value="unfriend" />
<button id="add-friend-btn" type="submit" name= "submit">Unfriend</button>
</form>
Following on from the comment from #ADyson:
<form action="friendlist.php" method= "POST">
<input type="hidden" name="cancel_id" value="<?=$friend_id_got ?>" />
<button id="add-friend-btn" type="submit" name="submit">Unfriend</button>
</form>
By including a hidden field in the form, you're able to store more information.
You can see that I'm storing the ID of the friend you're unfriending in the value of the hidden field, so when the form is submitted (the button is clicked) you'll have access to "cancel_id" in the POST data, which will obviously contain the ID of the friend to unfriend.
I am trying to pass input tag value to the another page to simply multiply the value of qty with price and return the updated price.The problem is when i pass the value of qty with get method it passes the value of qty only but does not pass another values.
cart.php
echo" <form method='get' name='form1' action='update_cart.php?id={$id}&name={$name}&price={$price}&qty=$_GET['qty']'>";//the problem comes here.
echo"<input type='number' name='qty' max='10'>
<input type='submit' value='update'></form>";
update_cart.php
$id = isset($_GET['id']) ? $_GET['id'] : "";
$name = isset($_GET['name']) ? $_GET['name'] : "";
$qty=isset($_GET['qty'])? $_GET['qty']: "";
$price=isset($_GET['price'])? $_GET['price']: "";
$price=$price*$qty;
header('Location: cart.php?action=quantity_updated&id=' . $id . '&name=' . $name . '&price='.$price . '&qty='.$qty);
when i click on update button after giving qty a value it shows something like this.
http://localhost/abc/cart.php?action=quantity_updated&id=&name=&price=0&qty=2
form method="get" passes variables automatically.
You do (and can) not need to append query string to the action attribute.
You do however need to represent those data in your form, if with <input type="hidden" name="qty" value="<?=$qty?>" /> style.
if you want pass some values , you can choose 2 way for it :
you can use the input with hidden type in you form.
if you use hidden type input you can use data just on one page.
<form method='get' action='?????'>
<input type='hidden' name='?????' value='?????'>
<input type='hidden' name='?????' value='?????'>
....
...
..
<input type='number' name='qty' max='10'/>
<input type='submit' value='submit'/>
</form>
you can use session for save data and use it in another page.
if you use session, you can use data in all php pages.
on page 1:
<?php
session_start();
$_SESSION['name']='value';
?>
on page 2:
<?php
session_start();
echo $_SESSION['name']; // value
?>
Your syntax for this line contains errors.
echo" <form method='get' name='form1' action='update_cart.php?id={$id}&name={$name}&price={$price}&qty=$_GET['qty']'>";
Change it to...
echo" <form method='get' name='form1' action='update_cart.php?id=".$id."&name=".$name."&price=".$price."&qty=".$_GET['qty']."'>";
When i run into a glitch, I always find find the answer on StackOverflow, but this time, although I'm sure the fix is easy, I just can't seem to get it right !
Basically, i'm trying to add a "delete" button next to each row fetched from my mysql database. The users should be able to delete a specific post, if needed.
When i hit the delete button, it's always the latest row that gets deleted. So i guess there's something wrong with the value passed in each row : seems like they're overridden by the latest one.
Below's my code:
<?php
$table = query("SELECT post, postid FROM post_list WHERE id = ? ORDER BY
time DESC LIMIT 15", $_SESSION["id"]);
foreach ($table as $row)
{
$post = $row["post"];
$postid = $row["postid"];
echo ("<table>");
echo ("<tr>");
echo("<td>" . $post . "</td>");
echo("</td>")?>
<div id="posteraser">
<form action='' method='post'>
<input type='hidden' name='postid' value='<?php echo $postid?>'>
<input type='submit' name='posteraser'>Delete</input>
</form>
</div>
<?php
echo ("</td>");
echo ("</tr>");
echo ("</table>");
echo '<hr>';
}
?>
And below on the same page, there's the delete button code:
<?php
if(isset($_POST['posteraser']))
{
$sql = query("DELETE FROM post_list WHERE postid = '$postid' ");
redirect ('home.php');
}
?>
Any help/tips will be much appreciated !
Thanks a lot !
You have to pass here the $_POST['postid']
if(isset($_POST['posteraser'])){
$postid = $_POST['postid'];
$sql = query("DELETE FROM post_list WHERE postid = '$postid' ");
redirect ('home.php');
}
OR as procedure way
$sql = query("DELETE FROM post_list WHERE postid = ? ",$postid);
A developer should always be aware of the HTML code they create with their PHP code.
It's essential thing.
As a matter of fact, HTML code is the very result of our efforts. NOT nice picture on can see in the browser windows - it's browser's job - but the very HTML code.
So, if you bother to see into generated code, you would discover something that can be boiled down to
<input type='hidden' name='postid' value='1'>
<input type='hidden' name='postid' value='3'>
<input type='hidden' name='postid' value='4'>
<input type='hidden' name='postid' value='5'>
<input type='hidden' name='postid' value='9'>
Do you have any questions why you have only last value?
Speaking of solutions, you have two choices
create a separate form for the every row
mark the very Delete button with id.
<input type='submit' name='posteraser[<?php echo $postid?>]'>Delete</input>
for example
let's check the logic from select statement.
you are selecting postid and assigning it to a hidden element and when you press delete button
that hidden id is sent to server.
so form creating under for loop is
<div id="posteraser">
<form action='' method='post'>
<input type='hidden' name='postid' value='<?php echo $postid?>'>
<input type='submit' name='posteraser'>Delete</input>
</form>
</div>
but hidden element is creating with same name for each row.
so when you press delete button . first hidden id is sent to server.
and this hidden id is already newest as from your select statement.
so what's the solution for it..
either you should sent postid through get attaching it in your url so that you can identify
which delete button is pressed.
or create a logic to send only that id on which delete is pressed.
This looks wrong:
echo ("<tr>");
echo("<td>" . $post . "</td>");
echo("</td>")?>
The trailing </td> shouldn't be there. Something else perhaps?
Also, you don't show how postid gets into $_SESSION['id']
I have a pop up box which checks if the user is signed in or not. If he is, I'm echoing out a small form which the user will press a button and it will submit to the DB. The variables are displayed on the popup but when pressed submit, they do not pass to the submit php file.
$add_wish = "<form action='memWishList.php' method='post' id='memWishList'>
<h3>Add this item to your Wish List?</h3><br>
<input type='hidden' name='title' value='".$title."'>".$title."</input><br>
<input type='hidden' name='link' value='".$link."'></input><br>
<input type='submit' name='submit' value='Add'/><button id='cancel'>
Cancel</button>
</form>";
echo $add_wish;
I want to pass the values title and link to be submitted to the DB. Here's my memWishList.php file:
if (isset($_POST['submit'])){
//get member id
$title = mysqli_real_escape_string($_POST['title']);
$link = mysqli_real_escape_string($_POST['link']);
$mysql = "INSERT INTO wish_list (memNum, title, link, date) VALUES ('$memnum', \
'$title', '$link', now())";
$myquery = mysqli_query($mysqli_connect, $mysql);}
Doing it this way, I only get the member id and the date inserted, not the title and the link. What's the problem? The reason why I'm echoing out this form is there's an if/else statement for logged in users and non logged in. Would be much easier to do it in html but can't...
DB: memnum(varchar), title(longtext), link(longtext), date(date). I have other tables where long links and titles are inserted just fine as longtext. They're coming from rss feeds.
please check documentation: mysqli_real_escape_string function expect the string as 2nd parameter if you use a procedural approach. It could be i.e.:
$link = mysqli_real_escape_string($mysqli_connect, $_POST['link']);
You have some markup errors. Your hidden input tags should look like:
<input type='hidden' name='link' value="<?php echo $link ?>">
Update your HTML file to look like this and all of the values will be sent to the $_POST variable:
<form action='memWishList.php' method='post' id='memWishList'>
<h3>Add this item to your Wish List?</h3><br>
<input type='hidden' name='title' value="<?php echo $title ?>"><?php echo $title ?><br>
<input type='hidden' name='link' value="<?php echo $link ?>"><br>
<input type='submit' name='submit' value='Add'/><button id='cancel'>Cancel</button>
</form>
Trying to perform a very simple task here.
I have an <ol> that contains 4 rows of data in some handy <li>s. I want to add a delete button to remove the row from the table. The script in delete.php appears to have finished, but the row is never removed when I go back and check dashboard.php and PHPMyAdmin for the listing.
Here's the code for the delete button (inside PHP):
Print "<form action=delete.php method=POST><input name=".$info['ID']." type=hidden><input type=submit name=submit value=Remove></form>";
Moving on to delete.php:
<?
//initilize PHP
if($_POST['submit']) //If submit is hit
{
//then connect as user
//change user and password to your mySQL name and password
mysql_connect("mysql.***.com","***","***") or die(mysql_error());
//select which database you want to edit
mysql_select_db("shpdb") or die(mysql_error());
//convert all the posts to variables:
$id = $_POST['ID'];
$result=mysql_query("DELETE FROM savannah WHERE ID='$id'") or die(mysql_error());
//confirm
echo "Patient removed. <a href=dashboard.php>Return to Dashboard</a>";
}
?>
Database is: shpdb
Table is: savannah
Ideas?
It's refusing to stick because you're calling it one thing and getting it with another. Change:
"<input name=".$info['ID']." type=hidden>"
to
"<input name=ID value=".$info['ID']." type=hidden>"
because in delete.php you're trying to access it with:
$id = $_POST['ID'];
You should really quote attribute values as well ie:
print <<<END
form action="delete.php" method="post">
<input type="hidden" name="ID" value="$info[ID]">
<input type="submit" name="submit" value="Remove">
</form>
END;
or even:
?>
form action="delete.php" method="post">
<input type="hidden" name="ID" value="<?php echo $info['ID'] ?>">
<input type="submit" name="submit" value="Remove">
</form>
<?
Please, for the love of the web, don't built an SQL query yourself. Use PDO.
Just another point I'd like to make. I'm 95% sure that you can't give an input a numeric name/id attribute. It has to be like "id_1" not "1".
Also with php you can do arrays.
So you could do this
<input name="delete[2]">
then in your php
if(isset($_POST['delete']))
foreach($_POST['delete'] as $key=>$val)
if($_POST['delete'][$key]) delete from table where id = $val