I'm trying to use AJAX to make my website a little slicker than it currently is.
I have a block of code that shows the following
if($guess == 0){
echo "Enter a guess:;
*AJAX form* ---> inserts a MySQL database record with the users guess
<div to show results of script for form> - not necessarily needed
}else{
echo "Update your guess:";
*SAME AJAX form* ---> updates the MySQL database record with the users new guess
<div to show results of script for form> - - not necessarily needed
}
The problem I'm having is that I want part of the webpage to show:
Your latest guess is: £x.xx
However because a user will start off with 0 guesses, and AJAX sends the form "behind" the scenes, I'm struggling in knowing how to show the above line once the user has made a guess.
So also that way when they revisit the page at a later date, it shows the last guess they had.
Those are the only 2 elements of the page I want to be able to refresh, the rest of the information doesn't have to refresh.
Some more code here:
<?php
if($guessess_open == 1){
echo "<h2>guesses are CLOSED</h2>";
}
else{
//it's this part that I want to have always shown on the page, but when I visit the page fresh it doesn't show me any results.
<div id="results"></div>
<form name="myform" id="myform" action="" method="POST">
<!-- The all important guess field -->
<label for="guess" id="guess_label">Guess<br></label>
<input type="text" name="guess" id="guess" size="10" value=""/>
<?php
echo "<input type='hidden' name='user_id' id='user_id' size='10' value='$user'/>";
echo "<input type='hidden' name='item_id' id='item_id' size='10' value='$itemID'/>";
echo "<input type='hidden' name='title' id='title' size='100' value='$title2'/>";
echo "<input type='hidden' name='owner_id' id='owner_id' size='10' value='$ownerid'/>";
echo "<input type='hidden' name='guesses_open' id='guesses_open' size='10' value='$guesses_open'/>";
echo "<input type='hidden' name='exist' id='exist' size='10' value='$exist'/>";
?>
<!-- The Submit button -->
<br>
<input type="image" name="submit" src="URL" width="150px" height="100px">
</form>
The only way I can get it to show something in that part of the page is to have some code in which checks a database for a guess and if it exists display the result, but then when I submit a new form, because it only updates the results it doesn't update the part of code for example could include
else{
include("URL.php?item_id=" . $itemID. "&user=" .$user. "");?>
//it's this part that I want to have always shown on the page, but when I visit the page fresh it doesn't show me any results.
<div id="results"></div>
which shows the latest guess when a user visits the page, but then when a new guess is made, this doesn't update... I don't have to have my forms php up display the result if the above part of code would update each time I make a guess... I hope this is clearer?
Related
I have seen many questions related to auto fill, but none of them worked for me. I'm not even sure if a solution to my problem lies in HTML or PHP. I am new in both of them and I'm still not used to them. I'm working on a simple chat app. What I have now is chat window, text field, and nick name field. Both of them pass the values to the text file (which is how I want this to work, no change here). Problem is that both these fields work as a form, and each time I submit, nick name field refreshes. What I would want is auto completing nickname field so it stays the same (it'd be the best if it stayed even after browser refreshes, but it will be okay if it only goes through the form submit)
code if needed:
PHP:
<?php
$action = $_GET["action"];
$myText = $_POST["mytext"];
$nick = $_POST["nick"];
if($action = "save") {
$targetFolder = "/var/www/html/xami/";
file_put_contents($targetFolder."htmlinput.txt", $nick.">".$myText);
}?>
HTML:
<form action="?action=save" name="myform" method="post">
<label for="nick">Nick:</label>
<input type=text id="nick" name="nick" placeholder="Nick" value="Name" required><br>
<input type=text name="mytext" placeholder="Text" required>
<input type="submit" value="Send.">
</form>
I was fooling around with autocomplete but no positive results.
I'm leaving post for tomorrow, I'll reply then.
Assuming your HTML is on a .php page you could have...
if(0<strlen($nick)){
echo "<input type=hidden id='nick' name='nick' value='$nick'>";
}else{
echo "<input type=text id='nick' name='nick' placeholder='Nick' value='Name' required><br>";
}
Or, use AJAX as #Rishi says.
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 $_SESSION['id']
I have a PHP code which generates a dynamic list inside a form like the following, note that the list is built dynamically from database:
echo '<form name="List" action="checkList.php" method="post">';
while($rows=mysqli_fetch_array($sql))
{
echo "<input type='password' name='code' id='code'>";
echo "<input type='hidden' name='SessionID' id='SessionID' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
}
What I need is to POST the data corresponding to the user choice when he clicks on the button for that row to another page.
If we use hyperlinks with query strings there will be no problem as I'll receive the data from the other page using a GET request and the hyperlinks would be static when showed to the user.
Also I need to obtain the user input from a textbox which is only possible with POST request.
Simply from the other page (checkList.php) I need these data for further processing:
$SessionID=$_POST['SessionID'];
$Code=$_POST['code'];
As I have a while loop that generates the fields, I always receive the last entry form the database and not the one corresponding to the line (row) that the user chosed from the LIST.
I'm going to recommend that you clean up the names of variables so that your code can
at least tell us what it's supposed to do. It should be rare that someone looks at your code
and has a lot of trouble trying to see what you're trying to accomplish :P, ESPECIALLY when you need help with something ;]. I'm going to try some things and hope that it makes doing what you want easier to comprehend and perhaps get you your answer.
It's good to try your best to not echo large amounts of HTML unnecessarily within a script , so firstly I'm going to remove the
echos from where they are not necessary.
Secondly, I'm going to use a mysql function that returns an easier to process result.
$user = mysqli_fetch_assoc($sql)
Third, I don't know if form having a name actually does anything for the backend or frontend of php, so I'm
just going to remove some of the extra crust that you have floating around that is either invalid HTML
or just doesn't add any value to what you're trying to do as you've presented it to us.
And yes, we "note" that you're building something from the database because the code looks like it does =P.
I'm also sooo sad seeing no recommendations from the other answers in regard to coding style or anything in regard to echoing html like this :(.
<?php while($user = mysqli_fetch_assoc($sql)): ?>
<form action="checkList.php" method="post">
<input type='password' name='code' value='<?php echo $user['code'] ?>' />
<input type='hidden' name='SessionID' value='<?php echo $user['id'] //Whatever you named the field that goes here ?>' />
<input type='submit' value='Take Survey' />
</form>
<?php endwhile; ?>
i not sure this is correct
echo '<form name="List" method="post">';
while($rows=mysqli_fetch_array($result))
{
echo "<input type='password' name='code' id='code'>";
echo "<input type='button' value='Take Survey' onclick=show($rows[0])>";
echo "<br>";
}
and javascript
<script>
function show(id)
{
alert(id);
window.location="checkList.php?id="+id;
}
</script>
On checkList.php
$id=$_GET['id'];
echo $id;
You can just check in checkList.php whether $_POST['code'] exists and if exists retrieve $_POST['SessionID'] which will be generated from database. But one thing, if You have all hidden fields in one form, they all will be sent, so You need to think how to correct that - maybe seperate forms for each hidden field, submit button and maybe other POST fields.
And afterwards, You will be able to get data in the way You need - $SessionID=$_POST['SessionID'];
I suppose it is the easiest way to solve that.
You can try something like this:
while($rows=mysqli_fetch_array($sql))
{
$index = 1;
echo "<input type='password' name='code' id='code'>";
//Attach $index with SessionID
echo "<input type='hidden' name='SessionID_$index' id='SessionID' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
}
On checkList.php
<?php
$num = explode('_', $_POST['SessionID']);
$num = $num[1];
//in $num you will get the number of row where you can perform action
?>
$form = 1;
while($rows=mysqli_fetch_array($sql))
{
echo '<form name="List_$form" action="checkList.php" method="post">';
echo "<input type='password' name='code' id='code_$form'>";
echo "<input type='hidden' name='SessionID' id='SessionID_$form' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
echo '</form>';
$form++;
}
I have this form which allows the input of any product quantity from 1-10:
<form method='post' action='cart.php'>
<input type='number' name='quantitychange' size='2' min='1' max='10' value=".$_SESSION["itemsSelected"][$i][1].">
<input type='hidden' name='ProductID' value=".$_SESSION["itemsSelected"][$i][0].">
<input type='submit' value='Update'>
</form>
And another form (button) to display a selection of payment modes:
<form action='cart.php' method='post'>
<input type='hidden' name='next'>
<input type='submit' value='Select Payment Mode'>
</form>
What I want to happen is that when a user did not input anything (1st form), ex. null or 0, I want to display an alert box that says 'Product quantity can't be null or 0'.
Here's my code for that:
if (isset($_POST['next'])) {
if ($_POST['quantitychange']==null || $_POST['quantitychange']==0) {
?>
<script type='text/javascript'>
alert('Product quantity can't be null or 0.');
</script>
<?php
}
else {
echo "
//Payment modes here
";
}
}
The error is that even when a user inputs a quantity bet. 1 to 10, it still displays the alert message. Any help? Thank you.
By the way, the input type "number" only works in Google Chrome browser.
Use a small javascript (or jQuery) function to validate the form before posting it. Have this function throw up the alert if your condition isn't met and then return false. If the condition is met, return true, and it gets submitted.
Edited to add since this might get googled, I'll help a bit with code snippet I have used. The below example is jQuery and was used in production for a web application I made for my employees. document.form.doit.submit(); should be the pure javascript way of submitting the form.
<script type="text/javascript">
function subForm() {
// document.form.doit.submit();
if( test condition passes ) {
$('#save_order').submit();
}
}
</script>
<form id="save_order" action="oms_db.php" method="POST">
<input id="doit" type="button"
value="i am a button" onClick="subForm();">
</form>
I think you have some error in your forms. Instead of the below:
<input type='number' name='quantitychange' size='2' min='1' max='10' value=".$_SESSION["itemsSelected"][$i][1].">
<input type='hidden' name='ProductID' value=".$_SESSION["itemsSelected"][$i][0].">
you should be using something like this:
<input type='number' name='quantitychange' size='2' min='1' max='10' value="<?php echo $_SESSION["itemsSelected"][$i][1]; ?>">
<input type='hidden' name='ProductID' value="<?php echo $_SESSION["itemsSelected"][$i][0]; ?>">
The value parameters in the hidden input fields needs to be echoed from PHP. What you have now is like the value is simple strings ".$_SESSION["itemsSelected"][$i][0].".
I suggest you use
if(empty($_POST['quantitychange'])) { echo 'yourerror'; }
As it is far cleaner then your script. (http://php.net/manual/en/function.empty.php)
Update:
Also, you can't use two seperate forms like you do, your browser only posts whats between
<form>
</form>
Using only one will fix your problem.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
table updates empty spaces when user do not enter anything to the textbox
greetings :)
i am having problems updating my database whenever the user clicks on the submit button.
i am going to show you the flow of my program,i already tried figuring out the problem,but i just can't find solutions. i hope someone could help me.
i have 2 problems encountered here:
my database won't update after clicking the submit button
the user may choose which to update,if the textbox is empty,it will update the data with empty spaces.and i want the data to remain as it is if the textbox is empty.
in my program,if you want to update the employee information,you must click the name that contains a link in the page. (in my program its the employee name that needs to be clicked) when clicked,a pop up will open.
the link in my index.php contains the following code:
<td class="sub" width="100" align="center">
<a href="" onclick = javascript:newPopup('empinfo.php?emp=<?php echo $eid ?>');><?php echo$ename?></a>
</td>
NOTE the empinfo.php is my pop up window,it calls the pop up when clicked. emp isthe name i assign to pass in the empinfo.php it contains the employee ID. NO PROBLEM HERE,I JUST WANT TO SHOW YOU THE FLOW
when the empinfo.php appears,it will show this format:
Employee name: //textbox here
Position: /textbox here
Department: /textbox here
Employee Tag: /textbox here
**SUBMIT BUTTON**
when the user clicks the submit button, it should have updated the database with the inputted values,but mine won't update :(
here is the codes i used:
<?php
$con=mysql_connect('localhost','root','mariel') or die(mysql_error());
mysql_select_db('intranet',$con);
if(isset($_POST['submitted']))
{
$qry = "UPDATE gpl_employees_list SET emp_nme = '".$_POST['name']."', emp_pos = '".$_POST['pos']."', emp_dep = '".$_POST['dep']."', emp_tag = '".$_POST['tag']."' WHERE emp_id = '".$_GET['emp']."' ";
mysql_query($qry) or die (mysql_error());
}
?>
this is the content code in my form,together with the submit that i used:
<form action="index.php" method="POST">
<input type='hidden' name='submitted' id='submitted' value='1'/>
<input type='hidden' name='eid' id='eid' value= '<?php echo $_GET['emp']?>' />
<fieldset>
<div class='container'>
<label for='ename' >Employee name:</label><br/>
<input type='text' name='ename' id='ename' value='' maxlength="50" /><br/><br/>
</div>
<div class='container'>
<label for='pos' >Position:</label><br/>
<input type='text' name='pos' id='pos' value='' maxlength="50" /><br/><br/>
</div>
<div class='container'>
<label for='dep' >Department/Division:</label><br/>
<input type='text' name='dep' id='dep' value='' maxlength="100" /><br/><br/>
</div>
<div class='container'>
<label for='tag' >Employee Tag:</label><br/>
<select name="tag" id="tag">
<option value="Y">Yes</option>
<option value="N">No</option>
</select> <br/><br/>
</div>
<div class='container'>
<input type='submit' name='Submit' value='Submit' onclick = "location.reload();window.close()"/>
</div>
</fieldset>
</form>
i hope someone could clear it up for me
MisaChan
It's not updating because you probably need to refer to $_POST['eid'] instead of $_GET['emp'] because you don't have it in index.php like index.php?emp=1. You already have that field so use that:
<input type='hidden' name='eid' id='eid' value= '<?php echo $_GET['emp']?>' />
Also you don't need to do this:
onclick = "location.reload();window.close()"
Type submit reloads the page by default.
Lastly, consider #Sam152's pointers :)
There could be a number of things wrong, but these points should help you debug your script.
Firstly you need to escape your post
variables to ensure things like
apostrophes don't mess up your query,
it's also a security vulnerability.
Secondly, make sure your form action is pointing to the PHP script. Maybe put a print statement at the top of the script to make sure PHP is actually receiving the data.
Then assign the value of the SQL query to a variable and print it out before you run it. You can then easily see what's being sent to the SQL server. Maybe run it in an SQL management tool my phpMyAdmin and observe any errors with it.
Hope this helps. Feel free to update your question with new information as it comes.