I have that code to create a button
$button = "<input type='submit' id='liga' value='liga'>";
echo $button;
I have the php function
function liga(){
....}
as I do so by clicking the button it calls the function?
Using html this code works, but I really need use the php button, how can I make it?
<input type="submit" name="liga" value="liga" />
if (isset($_REQUEST['liga'])) {
liga();
} elseif (isset($_REQUEST['desliga'])) {
desliga();
}
This little snippet will take an array with the index being the name of the button and the value as the label the button should have. It will then make a button for each element of the array.
<?php
$foo = array('name'=>'label', 'name2'=>'label2');
foreach ($foo as $k=>$v)
echo "<input type=\"submit\" name=\"$k\" value=\"$v\" />\n";
?>
You can also put the PHP variable right inside the HTML code using <?=...?>:
<input type="submit" name="<?= $myPhpVar ?>" value="<?= $myOtherPhpVar ?>" />
Or you can put complicated expressions (or whole programs) using the typical <?php...?> tags inside HTML brackets - whatever it echod becomes the content of that html tag:
<input type="submit" name="<?php echo $myPhpVar; ?>" value="<?php echo "LABEL: ".$myOtherPhpVar; ?>" />
I found a solution
echo "<input type='submit' name='liga' value='Liga'>";
echo "<input type='submit' name='desliga' value='Desliga'>";
Related
<?php
if(isset($_POST['btnLogin'])){
$myVariable = $_POST['fieldParameter'];
if(condition){
//do something
}else{
echo "
<form method='POST' action='submit.php'><br/>
<input type='hidden' name='myVariable' value='<?php echo $myVariable; ?>'/>
<br/>
<input type='submit' name='btnSubmit' id='submit' value='Submit'>
</form>
";
}
}
?>
Notice that the variable $myVariable is contained in the main IF block. I'm trying to send the value of $myVariable to submit.php as hidden field.
Also, i enclosed all the html tags using one echo statement with double quotes.
I found related questions here in SO but can't find similar to embedding php within a long echo of html tags
I tried to put value='<?php echo $studentNo; ?>' with no success.
I want to access it in a submit.php file like this,
submit.php
<?php
$aVariable = $_POST['myVariable'];
echo $aVariable;
?>
How can I pass the value contained in $myVariable as hidden field? Is there something wrong with the way I use double and single quotes?
If you are already echoing a string you shouldn't put <?php echo "" ?> inside it again. You should concatenate your string instead. But in your case you don't even need to do that, because you're using double quotes for echoing which means you can simply just write your variable in it.
echo "<form method='POST' action='submit.php'><br/>
<input type='hidden' name='myVariable' value='$myVariable;'/>
<br/>
<input type='submit' name='btnSubmit' id='submit' value='Submit'>
</form>";
If you were using single quotes for your echo, it would look like this:
echo '<form method="POST" action="submit.php"><br/>
<input type="hidden" name="myVariable" value="' . $myVariable . '"/><br/>
<input type="submit" name="btnSubmit" id="submit" value="Submit">
</form>';
You just need to type $myVariable instead of in your string. Double quotes "" only creates a string literal. It doesn't directly output data like inline HTML. As you can see from the syntax coloring in StackOverflow, the
You can try these variants (simplified):
// code before
echo "<input type='hidden' name='myVariable' value='$myVariable'/>";
// code after
// OR //
// code before
?>
<input type='hidden' name='myVariable' value='<?= $myVariable ?>'/>
<?php
// code after
Note that the quotes you use in HTML don't affect PHP, as long as you escape them properly (use \" and \' where appropriate).
I have got a while loop that runs through all the records of the database printing them on a table. Now i also have some checkboxes within that very loop that I want to use to submit a form when clicked. Now when I click the checkbox it will indeed submit the form thanks to a Jquery script I found, BUT whenever i submit it it submits with the ID of the first record of the table.This Image
shows the table, as you see the first record has ID 34. Now every checkbox I click will send the $id 34.
This does not happen with normal submit buttons.
Is there a way I can submit with the individual userID's
while ($openResInfo = mysql_fetch_array($openResQuery))
{
$id = $openResInfo[0];
$complete = $openResInfo[7];
?>
<form id='resComplete' action='dashboard_openReserveringen_PHP.php' method='GET'>
<?php
echo "<input type='hidden' name='userID' value='$id'>";
?>
<input type="hidden" name="complete" value="0" >
<input id='complete' type='checkbox' name='complete' value='1' onchange='$("#resComplete").submit();' <?php if($complete == 1){echo "checked";}?>>
</form>
I'm sorry if i'm not very clear with the explanation it is quite hard to explain this situation. Thank you guys!
Probably your problem is that you are submiting the same form always and its because you create a form for each row but it has the same id
For you the easy way is to put each form with the id cointaining the unique value of the row and doing submit with that.
Something like this
while ($openResInfo = mysql_fetch_array($openResQuery))
{
$id = $openResInfo[0];
$complete = $openResInfo[7];
?>
<form id='resComplete_<?php echo $id; ?>' action='dashboard_openReserveringen_PHP.php' method='GET'>
<?php
echo "<input type='hidden' name='userID' value='$id'>";
?>
<input type="hidden" name="complete" value="0" >
<input id='complete' type='checkbox' name='complete' value='1' onchange='$("#resComplete_<?php echo $id; ?>").submit();' <?php if($complete == 1){echo "checked";}?>>
</form>
It looks like the <form> your creating has a static id, so ALL forms will have id='resComplete'. The jQuery submit function will grab the first element with id='resComplete' and submit it. You need to make it unique for every form and make the onchange='$("#resComplete").submit();' code match it.
Eg.
<?php
while ($openResInfo = mysql_fetch_array($openResQuery))
{
$id = $openResInfo[0];
$complete = $openResInfo[7];
?>
<form id='resComplete-<?php echo $id; ?>' action='dashboard_openReserveringen_PHP.php' method='GET'>
<?php
echo "<input type='hidden' name='userID' value='$id'>";
?>
<input type="hidden" name="complete" value="0" >
<input id='complete' type='checkbox' name='complete' value='1' onchange='$("#resComplete-<?php echo $id; ?>").submit();' <?php if($complete == 1){echo "checked";}?>>
</form>
Better yet, use jQuery to find out what form it's in by chaning the onchange to something like:
<input id='complete' type='checkbox' name='complete' value='1' onchange='$(this).closest('form').submit();' <?php if($complete == 1){echo "checked";}?>>
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>';
$form++;
}
I want to pass the same boolean value "isProvena" several times in two PHP files. The first time I pass the value from utilitizer.php to hraprint.php by using the codes as follows:
if ($_POST['type'] == 'printphys' || $_POST['type'] == 'printprovenahpa')
{
echo "<form action='/content/822' method=post>";
echo "<input type=hidden name=filename value='$filename'>";
if($_POST['type'] == 'printprovenahpa') {echo "<input type=hidden name=isProvena value='1'>";}
echo "<input type=hidden name=content value='";
if ($_POST['type'] == 'printphys') echo 751;
else if ($_POST['type'] == 'printprovenahpa') echo 520;
echo "'>";
echo "<input type=submit value='Start Job'></form>";
}
And then I get the value "isProvena" from hraprint.php and post(get) again:
$isProvena = false;
extract($_REQUEST, EXTR_IF_EXISTS);
$isProvena = (boolean)$isProvena;
<form action="/content/822" method="GET">
<input type="hidden" name="isProvena" value="<?php echo ($isProvena) ? '1' : '0' ?>" />
<tr>
<td><label for="showOnlyScreening">Print Only Screenings:</label></td>
<td><input id="showOnlyScreening" type="checkbox" name="showOnlyScreening" value="1" <?php echo ($isProvena) ? 'checked="checked"' : ''?>/></td>
</tr>
<tr>
</table>
</form>
And post again:
<form action="/content/822" method="POST">
<input type="hidden" name="isProvena" value="<?php echo ($isProvena) ? '1' : '0' ?>" />
</table>
And I do the judgement here:
if($isProvena){
.........
}
The reason I need to post(get) several times is that there are several page redirect action happens in the same PHP file(hraprint.php). When I was trying to get the value which is supposed to be 'true' from if($isProvena){} and execute the function, I failed.
Anyone can help me to have a look and tell me what is wrong?
It would be easier if you simply use sessions for that. Sessions are made specifically for that purpose - passing variables easily from one page to another.
And it is not yet established in the last code block of your answer that $isProvena already exists because I do not see any extract() there.
P.S. Use the $_POST and $_GET variables instead of extracting the $_REQUEST. The code is vulnerable to the problems caused by register_globals
I want to send data from one page to the other via a form in PHP. In the initial page I have included the following PHP script that generates an input form with hidden fields. The names of these hidden fields are generated by the PHP:
<?php
echo "<form class='available-form' name='available_os' method='get' action='process-results.php'>";
echo "<input type='hidden' name='$software'></input>";
echo "<input type='hidden' name='$version'></input>";
echo "<input type='submit' name='available-button' value='Find Available Libraries for this Software'></input>";
echo "</form>";
?>
In the second page, named process-results.php, I would like to get the names of these hidden fields via the $_GET method but of course using $_GET[$software] and $_GET[$version] wouldn't work...Can someone tell me if there is a solution to this issue or if there is a better alternative? Thanks in advance
Instead of
"<input type='hidden' name='$software'></input>";
you should use
"<input type='hidden' name='software' value='".$software."'></input>";
for each. This way, you can use $_GET['software'] to retrieve the value. Do this for each of your hidden inputs.
I think you may want something like:
<form ... >
<input type="hidden" name="software" value="<?php echo $software ?>" />
<input type="hidden" name="version" value="<?php echo $version ?>" />
</form>
and then
$_GET['software'];
$_GET['version'];
I'm not sure what you're trying to accomplish, but this looks odd to me. Isn't the below code more of what you're looking for?
<?php
echo "<form class='available-form' name='available_os' method='get' action='process-results.php'>";
echo "<input type='hidden' name='software' value='$software'></input>";
echo "<input type='hidden' name='version' value='$version'></input>";
echo "<input type='submit' name='available-button' value='Find Available Libraries for this Software'></input>";
echo "</form>";
?>
That way you will get a query string in form of ?software=yoursoftwarename&version=yourversion and it will be available via $_GET["software"] and $_GET["version"] on the next page.
You could iterate over each of the items in the $_GET array on process-results.php. The problem is that the keys for the value will be whatever $software and $version are set to on the first page. Try something like this:
foreach($_GET as $key=>$string) {
// Do stuff with them
}
Add
enctype="multipart/form-data"
To the tag... so it looks like
<form enctype="multipart/form-data" method......
If you really need to have the dollar-sign inside the name, escape it:
echo "<input type='hidden' name='\$software'>";
or put the string in single-quotes:
echo '<input type="hidden" name="$software">';
Otherwise PHP is looking for a variable named "$software", if you look inside the browser-source you will see that the name-attributes are empty(except you're having those variables defined somewhere).