javascript not working with php variable when retrieving information from linux command

Question

Below image show my code.

$usernam = sudo grep -w "navneet" "/var/www/mcloud/modules/demo_cmp/.cmpinfo.txt" | cut -d ":" -f7;

function check()

{
    var username = "<?php echo $usernam; ?>";
    alert(usernam);
    return false;
}

I am retrieving value of $usernam variable by linux command and I want to use this variable value in javascript. But unfortunately javascript stop working after adding value of php variable. Please help me to short out my problem.

enter image description here

Sorry for that but have no idea to add all kind of code means javascript,html,and php code simulteniously — Lokesh Bhandari, Jul 16 '15 at 06:35
@LokeshBhandari You just need to give us the code you provided in the screenshot. Not sure why that's harder than giving a screenshot. Anyways it's because you are missing a semi-colon. An easy way to notice that is to [turn error reporting on in PHP](http://stackoverflow.com/questions/1053424/how-do-i-get-php-errors-to-display). — Spencer Wieczorek, Jul 16 '15 at 06:40
Problem is with retriving value from linux command please any one check it — Lokesh Bhandari, Jul 16 '15 at 06:43

score 1 · Answer 1 · answered Jul 16 '15 at 09:20

1

I got the solution.

I have add trim in my linux command which used to retrieve my record.

$usernam = trim(sudo grep -w "navneet" "/var/www/mcloud/modules/demo_cmp/.cmpinfo.txt" | cut -d ":" -f7);

answered Jul 16 '15 at 09:20

Lokesh Bhandari

65
6

score 0 · Answer 2 · edited Jul 17 '15 at 11:21

0

<?php
    $sim = 'hello world';

?>
<script type="text/javascript">
    var sim = '<?php echo $sim; ?>';

</script>

Other than that, if you really want to "interact" between PHP and JavaScript you should use Ajax.

edited Jul 17 '15 at 11:21

Nikos Paraskevopoulos

36,975
10
83
85

answered Jul 16 '15 at 06:35

Ashish Kumar Saxena

3,570
8
24
39

still this code not work – Lokesh Bhandari Jul 16 '15 at 06:39
Problem is with linux command use in php variable. – Lokesh Bhandari Jul 16 '15 at 06:50

score 0 · Answer 3 · answered Jul 16 '15 at 06:46

0

<script>
function check(){
var username = "<?php echo $usernam; ?>";
alert(username);
}
</script>

What you are doing wrong is placing quotes with JS variable username which should not be there.

answered Jul 16 '15 at 06:46

Deepak jha

278
5
18

problem is with retrieving variable from linux command – Lokesh Bhandari Jul 16 '15 at 06:47

score 0 · Answer 4 · answered Jul 16 '15 at 06:59

Open Google Chrome or other browser with developer/debugging capabilities. Navigate to the page with your code. View source code or open Developer tools and check the lines referent to var username in Element or Source panel. Also check Console panel.

Also, you need validate/clean/transform PHP variable $usernam in PHP to ensure it a valid string that suit your needs (an error or warning message is not an username).

score 0 · Answer 5 · answered Jul 16 '15 at 08:22

I give you a sample code. Think that you have to execute the php with a system or exec function. If you only write the code inside a variable, the variable will only have the text nor the result.

<?php
  $usernam = system('sudo grep -w "navneet" "/var/www/mcloud/modules/demo_cmp/.cmpinfo.txt" | cut -d ":" -f7');
?>

<script>
  function check(){
    var username = "<?php echo $usernam; ?>";
    alert(username);
  }
</script>

Check if that is useful for you. I wish it

Kavita · Answer 6 · 2015-07-16T08:15:47.847

-1

Remove single quote from alert i.e.

  alert(username);

okay sorry for this answer. I only refer your screen shot.

Try this one

{
var username = "<?php echo $usernam; ?>";

alert(usernam); replace with alert(username);

 return false;
}

edited Jul 16 '15 at 08:15

answered Jul 16 '15 at 08:03

Kavita

77
7

javascript not working with php variable when retrieving information from linux command

6 Answers6