I have this code that fetches information in a file and shows the result. That all works as expected. What I'm not able to do is to export the button value in the function when it is pressed so the value of the pressed button will transfer to a variable in my php function. How can I do that?
Here the code:
<h2>Demmarage script torrent</h2>
<form action="search.php" method="POST">
<input type="text" name="input_value">
<input type="submit" name="submit">
<?php
echo "<br>";
if (isset($_POST['submit'])){
$findme = $_POST['input_value'];
$findme1 = str_replace (" ", ".", $findme);
$savedarr = unserialize(file_get_contents('torrent.bin'));
foreach ($savedarr as $val1){
$mystring = $val1['title'];
if((stripos($mystring, $findme) !== false) or (stripos($mystring, $findme1) !== false)) {
echo "Show trouve: ";
echo $mystring;
?>
<button type="submit" value="<?php echo $val1['link']; ?>" name="editId">Telecharger</button>
<?php
echo "<br>";
}
}
}
if (isset($_POST['editId'])){
//Here i want to import the value of the pressed button to do something
echo "download start";
}
?>
you should use a hidden field, like
<input type="hidden" name="link" id="link" value="<?php echo $val1['link'] ?>" />
After the submit you will be able to get the value with
$_POST['link']
Also, this should be in a form,otherwise the submit will post EVERY field ...
like
foreach(...)
{?>
<form action="" method="POST">
<input type="hidden" name="link" id="link" value="<?php echo $val1['link'] ?>" />
<input type="submit" />
</form>
<?php}
The problem is that you are setting the value twice by <button>value1</button> and by <button value="value2">... remember that the value of a submit button is always the text that shows up inside the button.
You should do replace this:
<button type="submit" value="<?php echo $val1['link']; ?>" name="editId">Telecharger</button>
with this:
<form method="POST">
<input type="hidden" value="<?=$val1['link']?>" name="editId" />
<input type="submit" value="Telecharger" />
</form>
Related
I wanted to know why I am not able to fetch the value of radio button into the PHP Variable. Thanks in advance!
PHP CODE:
$Type = (isset($_GET['type'])?$_GET['type'] : null);
echo "<b>Type</b>";
echo $Type;
Here no value in the type variable is getting displayed.
HTML CODE:
<input type = 'radio' name='type' value='Residential' checked<?PHP print $Type;>> Residential<br>
<input type = "radio" name="type" value="Commercial" <?PHP print $Type;?>> Commercial<br>
First, you need a form with a method (whether you want to use post or get) and an action, which will be the URL where to go to, in this case it's the same page (<?php echo $_SERVER['PHP_SELF']; ?>)
<?php
$Type = (isset($_GET['type'])?$_GET['type'] : null);
echo "<b>Type</b>";
echo $Type;
?>
<form name="form1" method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
<input type='radio' name='type' value='Residential' checked<?php echo $Type;?>> Residential<br>
<input type="radio" name="type" value="Commercial" <?php echo $Type;?>> Commercial<br>
<input type="submit" value="submit" />
</form>
And you forgot the submit button to submit the form? Or you didn't, but just posted incomplete code. Anyway, above code should work.
<?php
print_r($_GET);//print values from form
?>
<form method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
<input type='radio' name='type' value='Residential'> Residential<br>
<input type="radio" name="type" value="Commercial"> Commercial<br>
<input type="submit" value="submit" />
</form>
can we use $_POST 2 times with different page ?
this example ..
action.php
<form action="next.php" method="post">
<input name="test" value="test" />
<button type="submit" name="test">submit</button>
</form>
next.php
<?php
$test=$_POST['test'];
?>
<form action="next1.php" method="post">
<input name="test1" value="<?php echo $test; ?>" />
<button type="submit" name="test1">submit</button>
</form>
next1.php
<?php
echo $test2=$_POST['test1']; >> in here i didnt get test1 value . why ?
?>
Within next.php and action.php you need to change your input type like as
<input type="text" name="test" value="test" />
<input type="text" name="test1" value="<?php echo $test; ?>" />
Within next.php and action.php you were missing the type attr of input tag
and you have same name attributes for submit and input within next.php need to remove or change the value from respective
<input type="submit" value="submit"/>
Below code i have tried my local server and it executed successfully.
action.php:-
<form action="next.php" method="POST">
<input type="text" name="test"/>
<input type="submit" value="submit"/>
</form>
next.php:-
<?php
$test = $_POST['test'];
echo $test;
?>
<form action="next1.php" method="POST">
<input name="test1" value="<?php echo $test; ?>" />
<input type="submit" value="submit"/>
</form>
next1.php:-
<?php
$test2=$_POST['test1'];
echo "------".$test2;
?>
Hope this will help.I think this is achieved your requirement.
If you want to do it within one file e.g. index.php then here is the code
<?php
if(isset($_POST['test'])){
if(isset($_POST['test_text'])){
echo 'Output from NEXT: '.$_POST['test_text'];
}
}
elseif(isset($_POST['test1'])){
if(isset($_POST['test_text1'])){
echo 'Output from NEXT1: '.$_POST['test_text1'];
}
}
?>
<table style="width:100%;">
<tr >
<td style="width:50%;">
<form action="" method="post" name="next">
<input type="text" name="test_text" value="<?php echo isset($_POST['test_text']) ? $_POST['test_text']:''; ?>" />
<button type="submit" name="test">submit test1</button>
</form>
</td>
<td style="width:50%;">
<form action="" method="post" name="next1">
<input type="text1" name="test_text1" value="<?php echo isset($_POST['test_text1']) ? $_POST['test_text1']:''; ?>" />
<button type="submit" name="test1">submit test1</button>
</form>
</td>
</tr>
</table>
I hope this will help.
You need to send the POST value with the second form again, e.g. in a hidden field, otherwise the value won't be sent.
Furthermore your submit button shouldn't have the same name as the input field you want the content from. So change the name of your submit button, e.g.:
action.php
<form action="next.php" method="post"> <input name="test" value="test" /> <button type="submit" name="submit">submit</button> </form>
next.php
<form action="next1.php" method="post">
<input name="test1" value="<?php echo $test; ?>" />
<button type="submit" name="submit">submit</button>
</form>
I have 4 different pages both with one form each.
I want to gather all the entries on each of the pages and submit once.
Here is code.
Page 1
<form action="page2" method="POST">
<input type="text" name="sex">
<input type="submit" value="Submit">
</form>
Page 2
<form action="page3" method="POST">
<input type="text" name="size">
<input type="hidden" name="sex" value="<?php echo $_POST['sex'] ?>" >
<input type="submit" value="Submit">
</form>
Page 3
<form action="page4" method="POST">
<input type="text" name="colors">
<input type="hidden" name="size" value="<?php echo $_POST['size'] ?>" >
<input type="submit" value="Submit">
</form>
Page 4
<form action="verNote.php" method="POST">
<input type="text" name="likes">
<input type="hidden" name="colors" value="<?php echo $_POST['colors'] ?>" > <input type="submit" value="Submit">
</form>
Then i will like to get all the infos on verNote.php
<?php
echo $_POST['sex'];
echo '<br>';
echo $_POST['size'];
echo '<br>';
echo $_POST['color'];
echo '<br>';
echo $_POST['likes'];
?>
This code above dont seem to post entries from both pages 1 and 2, just for 3 and 4 alone gets submitted.
Will appreciate immediate assistance form anyone who understands my question.
Regards!
You need to load the hidden fields again each time
Page 3
<form action="B.php" method="POST">
<input type="text" name="colors">
<input type="hidden" name="size" value="<?php echo $_POST['size'] ?>" >
<input type="hidden" name="sex" value="<?php echo $_POST['sex'] ?>" >
<input type="submit" value="Submit">
</form>
Page 4
<form action="B.php" method="POST">
<input type="text" name="likes">
<input type="hidden" name="colors" value="<?php echo $_POST['colors'] ?>" >
<input type="hidden" name="sex" value="<?php echo $_POST['sex'] ?>" >
<input type="hidden" name="size" value="<?php echo $_POST['size'] ?>" >
<input type="submit" value="Submit">
</form>
I didn't understand 100% what you're trying to achieve, but have you tried using sessions?
Do this in B.php:
<?php
session_start();
if( isset($_POST['sex']))
$_SESSION['sex'] = $_POST['sex'];
if( isset($_POST['size']))
$_SESSION['size'] = $_POST['size'];
if( isset($_POST['color']))
$_SESSION['color'] = $_POST['color'];
if( isset($_POST['likes']))
$_SESSION['likes'] = $_POST['likes'];
?>
Then you can retrieve the values from any other file, just call session_start(); and use the $_SESSION superglobal.
EDIT
Using sessions, you verNote.php file could be something like this:
<?php
session_start();
echo $_SESSION['sex'];
echo '<br />';
echo $_SESSION['size'];
echo '<br />';
echo $_SESSION['color'];
echo '<br />';
echo $_SESSION['likes'];
echo '<br />';
?>
I'm writing a form and this is its structure:
HTML:
<form method="post" action="script.php">
<input type="text" name="nombre" >
<input type="submit">
<input type="submit">
</form>
PHP:
<?php
if(isset($_POST))
$options = array();
$options[] = "";
// here I use the value input in "name" text input,
// so I need to get it from the form
for ($i=0;$i<=count($buscar)-1;$i++) {
$options[] = "<option value='{$i}'>{$dato}</option>";
}
echo '<select class="" id="articles" size="1" name="articles">';
echo implode("\n", $options);
echo '</select>';
?>
Is there any way to tell the first submit to let the php be executed (because it creates a select item)?
Then, when the select is selected, click on the second submit and send the complete form?
You can do in two ways:
Different Naming:
<input type="submit" name="sub1" />
<input type="submit" name="sub2" />
And you can check it using:
isset($_REQUEST["sub1"])
isset($_REQUEST["sub2"])
Passing Value:
<input type="submit" name="sub" value="Submit Here" />
<input type="submit" name="sub" value="Submit There" />
And you can check it using:
($_REQUEST["sub"] == "Submit Here")
($_REQUEST["sub"] == "Submit There")
I am encountering a strange phenomena: When I submit a form in my code, then beside doing what is there in that form, it looks like it also trigger the comment form. So when the page is reloading, I get some blank comments or duplicate comments. In my code I have several forms with submit, and one of that is the form for input comment:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="a" value="<?php echo $_GET['id'] ?>">
<input type="hidden" name="b" value="<?php echo $_SERVER['REMOTE_ADDR'] ?>">
<input type="text" name="c" value="Name"><br>
<textarea name="d">
</textarea>
<input type="submit" />
</form>
Could any of you point out for me where it gets wrong?
As requested, I post the whole (updated) code here:
<h3>Comments</h3>
<!-- <p>Put your comments here: </p> -->
<?php
//a: commenters
$i = addslashes($_POST['a']);
$ip = addslashes($_POST['b']);
$a = addslashes($_POST['c']);
$b = addslashes($_POST['d']);
if(isset($_POST['form1'])){
if(isset($i) & isset($ip) & isset($a) & isset($b))
{
$connector= new DbConnector();
$r = mysql_query("SELECT COUNT(*) FROM `databasename`.`ban` WHERE ip=$ip"); //Check if banned
$r = mysql_fetch_array($r);
if(!$r[0]) //Phew, not banned
{ // echo "a: ".$a." b: ".$b." ip: ".$ip." i: ".$i."";
$Date4=date('Y-m-d H:i:s');
if(mysql_query("INSERT INTO `databasename`.`Comments` VALUES ('$a', '$b', '$ip', '$Date4')"))
{
?>
<script type="text/javascript">
window.location="/index.php?id=".<?php echo $i; ?>;
</script>
<?php
}
else echo "Error, in mysql query";
}
else echo "Error, You are banned.";
}
}
$x = mysql_query("SELECT * FROM `databasename`.`Comments` ORDER BY i DESC ");
while($r = mysql_fetch_object($x)) echo "<div class='c'>".$r->a."<p>".$r->b."</p> </div>";
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="a" value="<?php echo $_GET['id'] ?>">
<input type="hidden" name="b" value="<?php echo $_SERVER['REMOTE_ADDR'] ?>">
<input type="text" name="c" value="Name"><br>
<textarea name="d">
</textarea>
<input type="submit" name="form1" />
</form>
You need use isset()
in HTML submit button you need to use name attribute for each and every form,
<input type="submit" name="form1" />
IN PHP,
<?php
if(isset($_POST['form1']){
echo $_POST['a'];
}
?>
Your code,
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="a" value="<?php echo $_GET['id']; ?>">
<input type="hidden" name="b" value="<?php echo $_SERVER['REMOTE_ADDR']; ?>">
<input type="text" name="c" value="Name"><br>
<textarea name="d"></textarea>
<input type="submit" name="form1" />
</form>
I would use:
if(!empty($_POST['form1']){
echo $_POST['a'];
}
empty() will return true if the variable is an empty string, false, array(), NULL, “0?, 0, and an unset variable.
Read more: http://www.virendrachandak.com/techtalk/php-isset-vs-empty-vs-is_null/