This is the code I am executing inside PHP condition
<?php
if(is_null($model)) {
echo "<script>window.parent.closeIframe(true)</script>";
}
?>
It doesn't work even If I put alert or calling a function inside my PHP condition.
Is it possible to run a Javascript code inside PHP ?
Is there any PHP extensions needed to be enabled ?
Was working on a recent project to execute JavaScript after PHP was finished executing. Try placing the script tag after the PHP tag:
test.php:
<?php
echo "hello";
include("example.html") // To insert HTML page
?>
<script>
alert("hello");
</script>
This worked for me. You can try it. I forced $model to be null.
<?php
$model = null;
if(is_null($model)) {
echo "<script>alert('Hello World')</script>";
}
?>
You can't call close from client Javascript from server PHP.
but you can call from HTML client. put close in bottom of HTML body tag
and make sure Javascript is turned on in your browser.
Hi if you are executing a query and then checking if result is null then this script to run then please correct. PHP is_null is true when it get NULL php datatype and a query never returns this. So if i am right then you can use this one this is my example which i used to check your method.
<?php
$con = mysqli_connect("localhost","root","","user") or die("error");
$res = mysqli_query($con, "select * from user where id =-99");
$model = mysqli_num_rows($res);
if(empty($model)) {
echo "<script>window.parent.closeIframe(true)</script>";
}
?>
just to check you can refer
$model = 0;
<?php
if(empty($model)) {
echo "<script>alert("hello")</script>";
}
?>
Related
I want to use javascript variable as php variable. I am echo php variable then its print. but when i am use for fetching data from database its show an error
Notice: Undefined index: document.write(i)
here my code
javascript
var i=0;
function inc()
{
i+=1;
}
<?php $foo="<script>document.write(i)</script>"; ?>
php
code work for
echo $foo
code not work for
$i=$foo;
$query="select * from TABLE where id = $i";
$result=mysqli_query($conn,$query);
while($row=mysqli_fetch_row($result))
{
echo $row[0];
}
Then It show This Error Notice: Undefined index: document.write(i)
PHP is server-side code that is run to generate a page. Javascript is client-side code that is run after the page is sent to the visitor's browser. Javascript can't affect the server-side code because the server code is done running by the time the Javascript runs. If you want to have a user's selection change the behavior of the PHP code the next time the form is loaded, pass a variable through a $_POST variable when the form is submitted.
If you want your PHP and Javascript code to be using the same value, have the PHP code write a Javascript variable initialization into the page's <head> section before any Javascript would run that would need to use it.
<script>
var i=0;
function inc()
{
i+=1;
return i;
}
</script>
<?php
$foo = '<script type="text/javascript">document.write(inc());</script>'; //Script function call which return the var i value to php variable
echo $foo;
?>
I have find the way to print current window size with using js and php but when i execute with function it won't works. Any one please help.
<?php $currentwidth = " <script>document.write(window.innerWidth); </script>";
echo $currentwidth; //works fine but belowis not working
if ( $currentwidth <= "480" ) { echo "TRUE"; } else { echo "FALSE"; }; ?>
You are mixing up javascript and php. javascript runs in browser and php runs in server. If you want php to print something you have to send the value to the server.
I have a javascript method in separate js file that I want to call from a HTML page with a php argument.
<?php
include "connection.php";
$selectedpatient = $_POST['patient_dropdown'];
$myquery = "SELECT * FROM `patient_info` patient_info.Name = '$selectedpatient' ";
$query = mysql_query($myquery);
if ( ! $query ) {
echo mysql_error();
die;
}
$data = array();
for ($x = 0; $x < mysql_num_rows($query); $x++) {
$data[] = mysql_fetch_assoc($query);
}
$tempdata = json_encode($data);
?>
<script> data_arrival($_tempdata); </script>
And I defined the source for the javascript file in the HTML header. It displays two errors:
1) Parse error in the javascript file - that's understandable as I included the javascript file in header and the file gets executed before the php actually retrieves any data.
2) data_arrival method is undefined
How can I fix this ?? I want to pass the $tempdata (after its populated by php) to data_arrival method as an argument.
Thanks in advance !
First of all: it's PHP that's executed first, not JavaScript. It can't be the other way round in your example.
data_arrival is undefined... because either you haven't defined it at all, or because it is defined after it's called.
To pass the value from PHP to JavaScript in your case, you can use:
data_arrival(<?php echo $_tempdata; ?>);
It will generate something like:
data_arrival([a, b, c, d, ...]);
Of course, data_arrival function need to be defined prior to its execution.
Edit
And maybe it's good to use the same variable name: $_tempdata vs $tempdata.
This is a php file below named getques.php at server. In this php file, the variables subnm1, chpno1 and qnumber1 are posted from a html-javascript file from client side. With these, a database at server is accessed and the output of SQL query is stored in a variable $px.
<!DOCTYPE html>
<html>
<body>
<?php
$x = $_POST["subnm1"];
$y = $_POST["chpno1"];
$z = $_POST["qnumber1"];
$con=mysqli_connect("localhost","compete7_bhunia","pr393ss","compete7_iitmock");
if (mysqli_connect_errno())
{ echo "Failed to connect to MySQL: " . mysqli_connect_error(); }
$result = mysqli_query($con,"SELECT question, optiona, optionb, optionc, optiond,
coroption FROM $x WHERE qnumber = $z AND chapno = $y");
while($row = mysqli_fetch_array($result))
{
$px = $row['question'];
}
mysqli_close($con);
?>
</body>
</html>
This $px is showing its desired data that has been retrieved from database. Now, I want that this variable $px should be passed to the same html-javascript file. How to get it at javascript. The relevant part is as below.
<script type="text/javascript">
var abc = "<?=echo $px ?>";
alert(abc);
</script>
This is not showing the value of $px. Pl. help.
Avoid using short open tags.May be in your php short open tags is not enabled. So try like this..
var abc = "<?php echo $px ?>";
short_open_tag directive) with PHP 5.3 or below (but since PHP 5.4.0,
Actually, in the php.ini-production file provided with PHP 5.3.0, they are disabled by default:
$ grep 'short_open' php.ini-production
; short_open_tag
short_open_tag = Off
So, using them in an application you want to distribute might not be a good idea: your application will not work if they are not enabled.
Use like this :
<?php echo $px ?>";
As others have stated, you have a syntax error, your output line should read like
var abc = "<?php echo $px ?>";
However, this is not safe (and don't use short open tags <?=, as they may be disabled in PHP < 5.4): consider that $px contains a double quote, then your syntax will be malformed; it is even possible to inject JavaScript code this way. Don't do this!
A safer method is to use json_encode:
var abc = <?php echo json_encode($px) ?>;
(Note that you don't need to surround this with quotes.)
On a side note, you are vulnerable to SQL injection. Please use prepared statements to prevent this.
I want to set the value of a JavaScript variable into a PHP variable. How can that be done?
Kindly give me some suggestions.
My code is :
document.getElementById("dummy").value= <?php echo ("document.getElementById('txtplan').value"); ?> //document.formn.txtplan.options[document.formn.txt'+fld+'.selectedIndex].value //eval(''); $_POST['txtplan']
alert(document.getElementById("dummy").value);
<?php
$planname = $_POST["dummy"];
$plandetails = mysql_query("SELECT * FROM membershipplan where planname='$planname'");
while($row1 = mysql_fetch_array($plandetails))
{?>
document.getElementById("txtduration").value=<?php echo $row1['planduration']; ?>
document.getElementById("txtplancontact").value=<?php echo $row1['plannoofcontacts']; ?>
document.getElementById("txtplanamount").value=<?php echo $row1['planamount']; ?>
<?php
}
?>
});
am not do with this means kindly give me the alternative way for the above process
JS -> PHP = impossible (only if you send that info to PHP using POST or GET)
PHP -> JS = possible var text = <?php echo( $text ); ?> ;
The only reason for that is that the PHP code is executed on your server.
You cant do that i.e you cant assign a javascript variable to a php variable
(unless you are using cookie or ajax) but u can assign a php variable to a javascript variable using
var a=<?php echo $a ?> ;
but be careful with that as the php variables will be executed in the server.
The answer is simple. Impossible. Because,
Java script is a client side scripting. Client machine can handle
this script, If it is possible to assign javascript value to PHP
variable that leads to some sort of vulnerability.
So this can't be done.
If you want to accomplish something by doing this kind of assignment, Definitly there would be a way of doing it. Post another question what do you want to do, you will be showered with answers.