I am a newbie working on a screen which has a conditional button which should appear only if $step==2. In this case, the button is shown and on click, redirects me to another page within my project. However, the url to which it should redirect me is not fixed, but has a parameter within it. I am having trouble getting the onclick to work!
I have tried defining a $path variable in which I store the relocation string. I do this in the following way:
<?php $path = '"window.location=\'{{ url("/location/'.$location->id.'/description") }}\'"'?>
When I echo this $path, I get exactly the string I need, which is:
"window.location='{{ url("/location/2/description") }}'"
(Note: The $location->id in my test case is 2)
However, I then try using this $path for the onclick of the button and it's not working! Here is the whole thing:
<?php if ($step == 2) : ?>
<?php $path = '"window.location=\'{{ url("/location/'.$location->id.'/description") }}\'"'?>
<button class="button_s_red" onclick=<?php echo $path; ?>>{{ trans('setup_process.STEP2_Button') }}</button>
<?php endif; ?>
When I inspect this button in chrome, it seems that on the onclick, the / character is not getting through. But this is odd since when I echo the $path I do get this character, and / doesn't seem to be an escape character from what I've googled.
I've tried pretty much everything and googled every possible help but can't seem to make it work! Thanks in advance for your help!
id.'/description") }}\'"'?> >{{
trans('setup_process.STEP2_Button') }}
I think u must use " in Your code like this.
<?php if ($step == 2) : ?>
<?php $path = '"window.location=\'{{ url("/location/'.$location->id.'/description") }}\'"'?>
<button class="button_s_red" onclick=("<?php echo $path; ?>")>{{ trans('setup_process.STEP2_Button') }}</button>
<?php endif; ?>
Related
enter the html code once a condition is satisfied but then i aim to use php values and html tags together.
tried tag etc and got most of the part in running just unable to deal with href beacuse it is referencing to some values.
not sure where to use "" or ``.
<?php.....
echo `Yes`;
?>
"yes" should work as a hyperlink but at the moment the php values are not processed that`s why no result.
You are using wrong sequence of quote
<?php
...
?>
<?php
echo '<a href="limitdatabase.php?Dropdown=' .
$_GET['Dropdown'].
';&search='.
$search_name .
';&wise='.
$_GET['wise'].
';">Yes</a>';
?>
and you should use single quote and not backtics for string
You are using incorrect concatenation. Whenever you want to add any PHP variable then you need to close the braces and need to start after PHP variable as below:
<?php
echo 'Yes';
?>
Also you should wrap Dropdown and wise in braces as it will not work directly in get as you have used. Hope it helps you!
You can use combination of html with php inside:
<a href="limitdatabase.php?Dropdown=<?PHP echo $_GET[Dropdown] . "&search="
. $search_name . "&wise=" . $_GET[wise]; ?>">Yes</a>
Also, you can input whole link in string:
<?php
$mylink = "limitdatabase.php?Dropdown=" . $_GET[Dropdown] . "&search=" . $search_name . "&wise=" . $_GET[wise];
?>
YES
I have some php that connects to a database and creates a drop down list. I have a specific form in the HTML that I'd like to put the list in.
<body>
<form>
// some text inputs
// where i'd like the drop down to go
<?php makeList(parameter1, parameter2); ?>
// submit button
</form>
<?php
// connect to database
function makeList(arg1, arg2) {
echo '<select>';
while ($row = mysqli_fetch_array($result)){
echo "<option">;
echo $row[$column];
echo "</option>";
echo '</select>';
}
</body>
The only languages I'm allowed to use (apart from the sql) are php, html and javascript. As it is right now, makeList() returns an empty list. When I include opening and closing form tags in the function it returns a fully functional list, but then it acts as it's own form and I need to to be a part of the original form. Thanks.
EDIT: Sorry, forgot to mention the makeList function works fine when called within the php tags. It's when I call it in the HTML that it returns an empty list.
Firstly, you have some syntax issues with your script. It's not a valid HTML file, not a valid PHP file, and not a valid JS file.
If it were up to me, I'd define the PHP function at the stop of my script. Be careful to balance your opening and closing PHP tags. Something like this:
<?php
// connect to database
function makeList($arg1, $arg2) {
echo '<select>';
while ($row = mysqli_fetch_array($result)){
echo "<option">;
echo $row[$column];
echo "</option>";
echo '</select>';
}
?>
And only after that would I start to output my HTML.
Now there are a couple of important things to note about that script I just posted:
the database code is not in here...I don't see any connection or query getting run or anything
In your script, this function doesn't look valid. arg1 and arg2 need a $ in front of each to be a valid PHP function. If it's a JS function you want then well, you are very confused and probably need to go back and figure out why this is not a valid JS function.
Your function refers to a variable, $result, that you have not bothered to define. It is not mentioned anywhere else in your script. It is most certainly not mentioned anywhere inside your function. For $result to be defined inside your function, you either need to pass it in as an array or declare it as a global:
global $result
Your function doesn't return anything at all. It just echoes stuff. This doesn't mean you can't use it, but it does mean that the function has no return value. Echoing the result of makeList won't output anything at all
So after that script above, you might have something like this:
<body>
<form>
// some text inputs
<?php makeList($parameter1, $parameter2); ?>
// submit button
</form>
Depending on what your parameters ($parameter1 and $parameter2) are this should work.
<body>
<form>
// some text inputs
<?php echo makeList($parameter1, $parameter2); ?>
// submit button
</form>
<?php
// connect to database
function makeList($arg1, $arg2) {
echo '<select>';
while ($row = mysqli_fetch_array($result)){
echo "<option>";
echo $row[$column];
echo "</option>";
echo '</select>';
}
</body>
I have a giant image/button in my header on every page that leads my users to a form they fill out, I want to remove that image/button on the form page itself, hoping to use css. I'm using wordpress, but this form is not a wordpress page, its located in a different directory and its called index.php. How do I go about hiding the image?
Within index.php, the first few lines there is a call to get_header() which loads the header so i thought to put this bit of code right after it:
<?php if ( is_page('mysite.org/the-index-page/') ) { ?>
<style type = "text/css">.order-btn { display: none; }</style>
<?php } ?>
however nothing happens. Am I doing this right or is there a better way to do it?
A better solution is probably to remove the button from that specific page. It sounds as if the button markup is inside your header.php file. You could try this:
<?php if ( !is_page( 'the-index-page' ) ) : ?>
// Button markup here
<?php endif; ?>
And if it's not working, this should:
$url = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];
<?php if ( !( $url == 'mysite.org/the-index-page/' ) ) : ?>
// Button markup here
<?php endif; ?>
Captcha works with no problem, but I have no idea why it's not generating a new code to dislay? I've looked into the documents but could find something that could solve my problem.
Is there something here that I'm missing. thanks.
public function actions()
{
return array(
'captcha'=>array(
'class'=>'CCaptchaAction',
'backColor'=>0xFFFFFF,
),
);
}
my view file:
<?php echo $form->labelEx($model,'verifyCode'); ?>
<?php $this->widget('CCaptcha'); ?>
<?php echo $form->textField($model,'verifyCode'); ?>
<?php echo $form->error($model,'verifyCode'); ?>
This is a known bug, that would most likely be fixed in Yii2.
On the Yii forums, user Black suggests:
My solution was to remove the session key on my controller action on get. Be careful not to remove it in any other place because it will probably fail on server validation.
$session = Yii::app()->session;
$prefixLen = strlen(CCaptchaAction::SESSION_VAR_PREFIX);
foreach($session->keys as $key)
{
if(strncmp(CCaptchaAction::SESSION_VAR_PREFIX, $key, $prefixLen) == 0)
$session->remove($key);
}
Another way to workaround would be to use JavaScript to click on the refresh link on every page load as mentioned by Soph:
$(function() {
$('#yw0_button').click();
});
I am trying to update the $_SESSION['state'] whenever the user clicks on the anchor tag by firing the onclick event. The variable $state_names_array as all the names of the states of a specific country. But problem is that no matter whichever of the anchor tag I click on, the $_SESSION['state'] is always getting updated by the last element of the $state_names_array.
<?php
foreach($state_names_array as $value){
$temp = '<?php $_SESSION["state"]=$value;?>';
?>
<?php echo $value ?><br>
<?php
}
?>
I will not code whole code, I will just show you an example, how to do it:
<?php
foreach ($state_names_array as $value) {
echo ''.$value.'<br>';
}
?>
Then you will have javascript function change_value:
here you will send ajax call to some PHP file, where you will process $_SESSION["state"]=$value;
Also you didnt say, what you want after anchor is clicked, because you will be able to access new $_SESSION["state"] only after refreshing your site...
Also maybe you just want to redirect to index.php, after clicking on some anchor, and that will be enough for you, then just use value as $_GET parameter, f.e.:
<?php
foreach ($state_names_array as $value) {
echo ''.$value.'<br>';
}
?>
and add to you index.php file this lines:
if (isset($_GET['new_value'])) {
$_SESSION["state"] = $_GET['new_value'];
}