** Here are the Codes that i use**
** I use this for recording of the scores, there 5 zeros cause there are more 3 students
$Students_C1= array(0,0,0,0,0);
$Students_C2= array(0,0,0,0,0);
** These are for inputting numbers
<td>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" id="inputs" name="SampleUP" value="<?php $Students_C1[0]; ?>" placeholder="<?php echo $Students_C1[0] ?>">
<input type="hidden" name="UpdateC1" >
</form>
</td>
<td>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" id="inputs" name="SampleUPs" value="<?php $Students_C2[0]; ?>" placeholder="<?php echo $Students_C1[1] ?>">
<input type="hidden" name="UpdateC2" >
</form>
** And these are for updating the php array after i input those numbers
<?php
if( isset($_GET['UpdateC1']) )
{
$val1 = htmlentities($_GET['SampleUP']);
$Students_C1[0]=$val1;
}
else{}
if( isset($_GET['UpdateC2']) )
{
$val2 = htmlentities($_GET['SampleUPs']);
$Students_C2[0]=$val2;
}
else{}
?>
The flow should be if i input a number in C1 and C2 next to the student's name. that number should stay as i input. The problems is that after i input C2 number (After C1), both of the numbers reset to zero, but the C1 works fine
Any Potential solution is heartly welcomed. A complete beginner in PHP
Student Grading System
Okay, I can't see your output but firstly both your input IDs are the same name. Then, you do not have an input type as submit instead it is hidden. So how do you submit your form then? If you do not submit it, then there is no value passing through the GET request. Your form does not have a method as well. Next is you're having two forms on one page so I'm guessing that if you have two forms then u need to enter one form and submit it first before submitting the other? I guess that's the reason why only one of it is updated. You should include your webpage output as well to make it easier to debug.
Related
At the bottom of this code you'll see an 'Accept Offer' button, when I click on that another piece of code gets executed as you can see on the bottom of this post.
For example this project has 3 bidders, so 3 times bidder_id and writer_bid so I use 'foreach' and load it in divs, works fine, but now I need to store those variables in a database, which technically works but it doesn't store the bids from the row I pull them from, it just takes the data from the last row, that is if I place the code at the bottom of this thread in my header.
However when I put it inside the loop it executes three times, I saw that when I got an error message that I had to close 3 times cause there are 3 rows in the database table that I pull the data from.
How can I prevent this, and either have it load once when the code is inside the foreach loop, or have it pull the correct writer_bid and bidder_id to store.
<div class="WB-Bottom-block lefts">
<?php $getBidders=" AND project_id=$project_id"; $bidders=getBidder($getBidders); foreach($bidders as $bidder) {
$bidder_id=$bidder['writer_id'];
$writer_bid=$bidder['writer_bid'];
?>
<div class="findwriters_boxes">
<div class="findwriters_right">
<div style="float:right;margin-top:6px;width:170px;">
<input type="hidden" name="writer_bid" id="writer_bid" value="<?php echo $writer_bid; ?>" />
<input type="hidden" name="bidder_id" id="bidder_id" value="<?php echo $bidder_id; ?>" />
<input type="submit" class="homebtn11" name="submit" id="submit" value="Accept Offer"/>
</div>
</div>
</div><?php } ?>
Below the code that needs to be executed and that results in issues, whether I place it inside the foreach loop, or inside the header instead.
As you can see I tried to store it in input fields so that it stays there so the header can pull it on refresh of the page / click of the button.
<?php if(isset($_POST['todo']) && $_POST['todo']=='submit_project') {
$balance=get_client_balance_info($current_user->ID);
$writer_bid=$_POST['writer_bid'];
$bidder_id=$_POST['bidder_id'];
if($balance >= $_POST['writer_bid']) {
global $wpdb;
$sql3="UPDATE `wp_project` SET `writer_id` = '".$bidder_id."' WHERE `id` =". $project_id;
$wpdb->query($sql3);
$sql4="UPDATE `wp_project` SET `price` = '".$writer_bid."' WHERE `id` =". $project_id;
$wpdb->query($sql4);
$sql5="UPDATE `wp_project` SET `status` = '2' WHERE `id` =". $project_id;
$wpdb->query($sql5);
$success_msg="You accepted a bid, the money will be deducted from your account.";
}
else $fail_msg="Your balance is not sufficient.";
I think you should make a form for each div that you are adding right now you are putting the bidder_id in the different inputs but the same name.
So it will get the last inputs, maybe it's better to specify the inputs with the row id or to separate the forms or make the input names as array.
I hope this helps you.
I fixed it with the help of Diar Selimi like this:
<div style="float:right;margin-top:6px;width:170px;">
<form action="" name="frmeditor" method="post" id="frmeditor" >
<input type="hidden" name="todo" id="todo" value="submit_project" />
<input type="hidden" name="writer_bid" id="writer_bid" value="<?php echo $writer_bid; ?>" />
<input type="hidden" name="writer_id" id="writer_id" value="<?php echo $writer_id; ?>" />
<input type="submit" class="homebtn11" name="submit" id="submit" value="Accept Offer"/>
</form>
Before that my form and value="submit_project" tags were scattered all over the place!
I've created a registration form that successfully passes its variables from the registration page (go-gold.php) to a summary/verfication page (go-gold-summary.php). The data appears correctly on the second page.
However, I want to able to use an image button to return back to the registration page, in case the user made an entry error. Going back, the original form should now be populated with the data that was first entered.
The problem is that I cannot re-send/return the data from the second page, back to the first. My text fields appear blank. I do NOT want to use Session variables.
The code is truncated from the entire page.
Registration Page (go-gold.php):
<?php
$customer_name = $_POST['customer_name'];
?>
<form action="go-gold-summary.php" method="post">
Name: <input type="text" name="customer_name" id="customer_name" value= "<?php echo $customer_name ?>" />
<input name="<?php echo $customer_name ?>" type="hidden" id="<?php echo $customer_name ?>">
</form>
Summary Page (go-gold-summary.php)
<?php
$customer_name = $_POST['customer_name'];
?>
<form action="go-gold.php" method="post">
Name: <?php echo $customer_name ?> <input type="hidden" id="<?php echo $customer_name ?>" name="<?php echo $customer_name ?>">
<INPUT TYPE="image" src="images/arrow_back.png" id="arrow" alt="Back to Registration"> (Button to go back to Registration Page)
</form>
Thanks!
go-gold-summary.php should be changed like this.
<?php
$customer_name = $_POST['customer_name'];
?>
<form action="go-gold.php" method="post">
Name: <?php echo $customer_name ?> <input type="hidden" value="<?php echo $customer_name ?>" name="customer_name">
<INPUT TYPE="submit" src="images/arrow_back.png" id="arrow" alt="Back to Registration"> (Button to go back to Registration Page)
</form>
notice how I've changed this line
<input type="hidden" id="<?php echo $customer_name ?>" name="<?php echo $customer_name ?>">
into this
<input type="hidden" value="<?php echo $customer_name ?>" name="customer_name">
$_POST is an associative array and as you submit the form it will be populated like this:
$_POST["index"] = value;
where "index" is the text field "name" and value is the text field value.
You've missed that one in your code. Just update it with my code and it will work
Why you would not want to use the php session? Please give any reason for not to use it. I am asking this way since my reputation does not allow me to comment questions or answers any other than my own. Plese do not -1 for this.
Another way could be using cookies to store the data temporarily, but that and posting the data back and forth in the post request is really insecure compared to session.
there are very few ways to maintain variables across pages. The alternative is to have separate form on the second page with hidden text fields containing the $_POST data, and the submit button calls the previous page. No way of getting around the "back button" on a browser though unfortunately.
I missed the bold text about the session variables - disregard if this does not apply:
one way to maintain variables across pages on the server side is to use $_SESSION
first include the following at the top of your PHP pages to keep a session active:
session_start();
once you submit the for and move to page 2, add the following:
$_SESSION['customer_name'] = $_POST['customer_name'];
As well, on the first page, you could change the form element as such:
<input type="text" name="customer_name" value="<?PHP if isset($_SESSION['customer_name'] || !empty($_SESSION['customer_name'])) { echo $_SESSION['customer_name']; } ?>">
this will keep the filled in data and display it when the user returns tot he page, and if they put in something different it will be updated when they hit page 2 again.
i am retrieving data form database using a search query.
PHP code (which I'm using in search query to display search results)
echo "<span style='background-color= #FFFF00'>$query</span><br>";
$count=$dbo->prepare($query);
$count->execute();
$no=$count->rowCount();
if($no > 0 ){echo " <span>No of records = ".$no."</span>";
echo "<table><tr><th>PHONE NUMBER</th><th>OWNER NAME</th></tr>";
foreach ($dbo->query($query) as $row){
echo "<tr><td>$row[ROLLNO]</td><td>$row[CNAME]</td></tr>";
}
echo "</table>";
i want to do like this,
when a user clicks on a phone number, it should redirect to a new page and in that new page, my input box should be filled with this phone number and should be submitted.
Input Box Code (which I'm using in page 2)
<form name="phone_number_form" id="phone_number_form" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" onsubmit="return vali()" >
<input type="text" name="number" id="number" />
<input type="submit" name="submit" value="Submit" />
</form>
Looking at w3schools PHP has a $_POST variable which is used to collect values from a form sent with method="post". There's also $_GET and $_REQUEST which seems to merge both post and get data. http://www.w3schools.com/php/php_post.asp
There are a couple of options like making a request (get) from your first page or post the data from your first page.
REQUEST Method
Heres how I think the request way to do it would work
PAGE 1
Amend the foreach that renders the table row to include an hyperlink to your second page
foreach ($dbo->query($query) as $row){
echo "<tr><td>$row[ROLLNO]</td><td>$row[CNAME]</td></tr>";
}
PAGE 2
Amend the textbox to be populated with the phone number from the request variable
<input type="text" name="number" id="number" value="<?php echo $_REQUEST["phoneno"]; ?>" />
POST Method
In your first page using javascript when the user clicks the phone number set a hidden field and submit the form to the second page. Again you should be able to read the hidden fields value from the $_REQUEST variable
I am trying to write a dynamic form using PHP. I'd like to have a single webpage that contains two forms:
The upper form allows to search for an element in the mysql database, e.g., for a name
The lower form shows the data that is associated with this name in the database
If I press on the "Search" button of the upper form, then the the lower form is shown and the text fields are filled with data from the database that belong to this name. If I change the user name to some other value and press again "Search", then the data that is associated with the new record is shown and so on.
The lower form also has a button "Update" which allows to transfer changes made to the text boxes (in the lower part) to the database.
Now, I have the following problem: In my script I set initially the value of name (from the upper form) to "". When I then press the "Search" button, then the lower part of the form is shown and the corresponding data is shown in the lower part. When I then press the "Update" button, then the text field associated with name is set to the empty string. This is because in my script I set initially name to the "". I'd like that in this case the data entered in the upper form is not changed, i.e., it stays the same.
I guess, I am missing something here. There is probably an easy solution for this and I am doing something fundamentally wrong. It'd be great if you could help me.
That's what I tried... I deleted lots of details, but I guess that can give you an idea what I am trying to do. Notice that the whole code is in the file update.php.
<?php
function search_bus($mysql, $name)
{
// do some stuff here...
}
function update_bus($mysql, $b_id)
{
// do some stuff here...
}
// some global variables
$b_id = 0;
$username = ""; // username of business
// get b_id that corresponds to username
if (isset($_REQUEST['search']))
{
$b_id =0; // business id
if (isset($_POST['user']))
{
$username = $_POST['user'];
$b_id = search_bus($mysql, $username);
}
}
elseif(isset($_REQUEST['update']))
{
update_bus($mysql, $b_id);
}
?>
<h2>Search:</h2>
<form name="search_bus" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
Username: <input type="text" name="user" value="<?= htmlentities($username) ?>"/>
<input type="submit" value="Suchen" name="search"/>
</form>
<?php
if($b_id != 0)
{
?>
<h2>Data:</h2>
<form name="business_design" method="post" action="<?php echo $_SERVER['PHP_SELF'];?>">
<-- some form follows here -->
<?php
}
?>
I think what you're missing is to create a HTML Hidden field to keep the value of Name variable.
<input type="hidden" name="name" value="<?php print $nameVar ?>" />
Add this input to both forms so you can keep the value no matter what button the user clicks.
Hope this helps.
Adding code to verify the
<h2>Search:</h2>
<form name="search_bus" method="post"
action="<?php echo $_SERVER['PHP_SELF'];?>">
Username: <input type="text" name="user" value="<?= htmlentities($username) ?>"/>
<input type="hidden" name="b_id" value="<?php print $b_id?>" />
<input type="submit" value="Suchen" name="search"/>
</form>
<?php if($b_id != 0) { ?>
<h2>Data:</h2>
<form name="business_design" method="post" action="<?php echo $_SERVER['PHP_SELF'];>">
<input type="hidden" name="b_id" value="<?php print $b_id?>" />
<-- some form follows here -->
<?php } ?>
Dont initialize $b_id if it already comes into the http request.
if (!isset($_POST['b_id']))
{
$b_id = 0;
}
else
{
$b_id = $_POST['b_id'];
}
This way you can alway remember the last selected value of b_id.
Hope this can help you.
I have spent quite some time making a function and the last 15-20 minutes trying to figure this out. I need help!
I am selecting multiple rows from the database and then running them in a while loop.
They are available on a dropdown menu.
<form method="POST" action="adminprocess.php">
<fieldset>
<p>
<label class="left2">League:</label>
<select name="league" class="combo">
<?php
$q = $database->selectAllLeagues();
while($row=mysql_fetch_assoc($q))
{
$theid = $row['id'];
extract($row);
?>
<option value="<? echo $theid; ?>">
<? echo $format.'_'.$game.'_'.$name.'_Season '.$season;?>
</option>
<?
}
?>
</select>
</p>
<p>
<input type="hidden" name="replaceleague" />
<input type="hidden" name="format" value="<? echo $format; ?>" />
<input type="hidden" name="game" value="<? echo $game; ?>" />
<input type="hidden" name="name" value="<? echo $name; ?>" />
<input type="hidden" name="season" value="<? echo $season; ?>" />
<input type='submit' class="button" value='Select league' />
</p>
</fieldset>
</form>
$theid seems to be working fine dependning on which row i select on the dropdown menu.
However, I cant get the values below in the hidden inputs to pass through the correct values from the row selected in the dropdown box.
It seems to always pass through the 4 variables from the first row of the database.
So basically, I need it to select the right row and use that data.
What am i doing wrong!!!
Thanks for reading!
Your hidden fields are initialized outside the loop, so they'll use the values that were left over from the last iteration of the while loop. (i.e. the last fetched row)
Why do you actually need the hidden fields in the first place? When you submit the form, the league field will contain the ID of the row selected in the drop-down box. Using the ID, you can fetch the other fields from the database when processing the form.
To directly answer your question about the while loop, it's because the hidden inputs are echoed outside the loop, after which data the last-iterated row from your database is used by PHP to output to those hidden inputs.
But I suggest that instead of using hidden form elements like that, you submit your form with the <option> with the value a user chooses, read the value (as in $_POST['league']), and fetch the row from your database with that ID and use it accordingly. (You may wish to keep the replaceleague hidden input if your application needs it of course.)
It's much easier, plus it ensures the information about the row a user chooses is coming from your database and not tampered with. In fact, for most applications this is the right way to go.