I'm using autocomplete search in my project where user inputs an integer, submits it and it shows in same page what integer was passed in. I want users to be able - after submitting first integer - to submit another one, just can't figure out the solution. My code:
<form action='' method='post'>
<p>Add integrent:</p><input type='text' name='sastavdalja'value=''class='auto'>
<input type="submit" name="submit" value="Submit" />
<br/>
<p>Added integrents</p>
<?php
$int = $_POST["sastavdalja"];
echo $int;
Put this at the top of the page before any output or html
<?php session_start();?>
then use this:
<?php
$_SESSION['int'][]=$_POST["sastavdalja"];
foreach($_SESSION['int'] as $int){
echo $int;
}
Now you're saving each time the user posts something into a session array, and then you're looping through this array and printing each entry in it.
Related
I am trying to create a quiz, in order to do this I am using a form which users will input information into, once they submit this form it should add the input to an array/list.
The user should then be able to enter information and the process would repeat.
The finished product would be an array with each element corresponding to the order the answers were given.
-
I have tried so far using both array_push() and declaring elements, eg: $my_array[0] = $input;.
The current problem I am experiencing is that each time I submit the form, the $count variable doesn't seem to increment.
Instead it simply stores the data in the first element and overwrites which was previously there.
I am inclined to believe this is a problem with the posting of the submit button.
-
Here is my code:
<html>
<body>
<form action="" method="POST">
<input type="text" name="INPUT" placeholder="Input something"; required /><br><br>
<input type="submit" name="Submit" /><br><br>
<?PHP
$my_array = array();
$count = 0;
if(isset($_POST['Submit'])){
global $count;
$input = $_POST['INPUT'];
$my_array[$count] = $input;
print_r($my_array);
echo "Count:" . $count;
$count++;
}
?>
</form>
</body>
</html>
The crux of the issue here is that variable values do not persist across PHP requests. Every time you submit the form, you are throwing away your old $count and $my_array variables and initializing new variables with the same names.
Here is a working version of your code snippet, which takes advantage of the PHP $_SESSION variable to have persistent information between requests:
<?php
session_start();
if (!isset($_SESSION["my_array"])) {
$_SESSION["my_array"] = array();
}
?>
<html>
<body>
<form action="" method="POST">
<input type="text" name="INPUT" placeholder="Input something"; required /><br><br>
<input type="submit" name="Submit" /><br><br>
<?php
if(isset($_POST['Submit'])){
array_push($_SESSION["my_array"], $_POST['INPUT']);
print_r($_SESSION["my_array"]);
echo "Count:" . count($_SESSION["my_array"]);
}
?>
</form>
</body>
</html>
I want to create a checkbox and check whether checkbox is checked or not in another PHP page so I created form like :
<form action="heartbeat.php" method="post">
<input type="checkbox" name="keepme">
<input type="submit" value="log in">
</form>
and then I tried to extract this input on heartbeat.php like:
<?php
/*
* A PHP file for laying down a heartbeat JavaScript call.
*/
if(!isset($_POST["keepme"])){
$auto_logout = 10;
}
else{
$auto_logout = 1000;
}
?>
but it never gets $_POST["keepme"] value. Any idea??
If you don't pass value to checkbox tag, the POST array will be empty. You need to do something like that
<input type="checkbox" name="keepme" value='1'>
Make Sure you have submit button
If(isset($_POST['submit-btn'])){
echo $_POST['keepme'];
}
Suppose I have a form. After I submit my form, the data is submitted to dataprocess.php file.
The dataprocess.php file processes the variable sent via form and echoes desirable output.
It seems impossible to echo to a specified div in specified page only using PHP (without using AJAX/JavaScript as well). I do not want to use these because some browsers might have these disabled.
My concern is that I want to maintain the same formatting of the page that contained the form element. I want the form element to be there as well. I want the query result to be displayed below the form.
I could echo exact html code with some modification but that's memory expensive and I want it systematic.
Is it possible to process the form within the same page? Instead of asking another .php file to process it? How does one implement it?
The above is just for knowledge. It will be long and messy to include the PHP script within the same HTML file. Also, that method might not be efficient if I have same process.php file being used by several forms.
I am actually looking for efficient methods. How do web developers display query result in same page? Do the echo all the html formatting? also, does disabling JavaScript disable jQuery/AJAX?
Yes it is possible to process the form on the same page.
<?php
if (isset($POST))
{
//write your insert query
}
?>
<html>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<!-- Your form elements and submit button -->
</form>
<table>
<?php
//your select query in a while loop
?>
</table>
</body>
</html>
But if you choose this technique instead of ajax, you have to refresh all the page for each insert action.
An example
<div id="dialog-form">
<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">
<table>
<tr>
<td>Job</td>
<td>
<input type="text" name="job" />
</td>
</tr
</table>
<input type="submit" value="Insert" />
</fieldset>
<input type="hidden" name="doProcess" value="Yes" />
</form>
</div>
<?php
$myQuery= $db->prepare("INSERT INTO Jobs (job) VALUES (:p1)");
if (isset($_POST['doProcess']) && $_POST['doProcess'] == 'Yes')
{
$myQuery->bindValue(":p1", $_POST['job'], PDO::PARAM_STR);
$myQuery->execute();
}
?>
if you really dont want to use ajax (which i think you should). You can do something like this.
<form action="" method="POST">
<input type="text" value="something" name="something_name"/>
<?php
if(isset($_POST['something_name'])){
echo '<div id="display_something_name_if_exists">';
echo $_POST['something_name'];
echo '</div>';
}
?>
</form>
Basically what it does is submits to itself and then if there is a submission (tested with isset), it will echo a div with the correct information.
i am retrieving data form database using a search query.
PHP code (which I'm using in search query to display search results)
echo "<span style='background-color= #FFFF00'>$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 $_REQUEST variable
<form id="enter" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" onsubmit="return validateForm(this);" >
<p>
<input id="submitBtn" name="submitDetails" type="submit" value="Submit Details" onClick="myClickHandler(); return false;" />
</p>
</form>
<script type="text/javascript">
function myClickHandler(){
if(validation()){
showConfirm();
}
}
</script>
<?php
session_start();
$outputDetails = "";
$outputDetails .= "<table id='sessionDetails' border='1'>
<tr>
<th>Number of Sessions:</th>
<th>{$_POST['sessionNum']}</th>
</tr>";
$outputDetails .= "</table>";
echo $outputDetails;
?>
Above is the code for my form. What I am trying to do is that if the user submits the form, then it will go back to its own page. But if the "SessionNum" equals '1', then instead of posting the form to itself, it should post the form or in other words navigate to the "session_marks.php' page but it is not idng this, if sessionNum equals 1 then it still submits form or navigate back to its own page, what am I doing wrong?
Also lets say it displays a number for the sessionNum and then I submit the form and it submits the form back to itself, the number disappears, how do I keep the number displayed when submitting the form to itself?
Thanks
Where is the conditional logic to change the target of the form post? All I see in the form tag is this:
action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>"
This will always set the form's action to be the current PHP file, not any other PHP file. If you want to conditionally post to a different file, you'll need to add conditional logic in there. Something like this (though there may be better ways to do it, keep in mind that I'm very out of practice with PHP):
action="<?php $_POST['sessionNum'] == 1 ? echo 'session_marks.php' : echo htmlentities($_SERVER['PHP_SELF']); ?>"
As for the number disappearing, I don't see any form element with the name sessionNum. If there isn't such a form element, then there will be nothing in $_POST['sessionNum'], so the number will "disappear" because there's no value to be displayed.
If the above is your actual code, session_start(); has to be placed before ANY other output (html, php's echo, print etc...)