php completes execution first, then javascript runs

Question

According to the main acceptance of the title, how do you explain the following:

       ...
       $temp1=$_POST['expert_id']; ?>
       <script type="text/javascript">
          var jstemp1 =<?php echo json_encode($temp1); ?>;
       </script>
       <?php         

       $temp1=$_POST['answers_id']; ?>
       <script type="text/javascript">
          var jstemp2 =<?php echo json_encode($temp1); ?>;
       </script>

Suppose $_POST['expert_id']=1 and $_POST['answers_id']=2. My thought is that $temp1 will equal to 2 when Javascript code begins to execute since PHP code executes first. Therefore, jstemp1 would equal to 2 and jstemp2 would equal to 2. However, to my big surprise, jstemp1=1 and jstemp2=2. Can you explain that to me?

Because you've reassigned the value of the PHP variable $temp1 from $_POST['expert_id'] (presumably 1) to $_POST['answers_id'] (presumably 2). — CD001
– CD001, Commented Jul 10, 2015 at 10:34
You did not see my point. When Javascript begins to run, $temp1=2, isn't it? — Unknown developer
– Unknown developer, Commented Jul 10, 2015 at 10:39
Nope - imagine you were writing all your JavaScript from PHP with echo statements; echo <script type... blah blah blah> ... </script> you're still pushing the code to the output buffer (screen) but the JS won't be executed as it's output (same as with just loading an normal HTML page) — CD001
– CD001, Commented Jul 10, 2015 at 10:54

CD001 · Accepted Answer · 2015-07-10 10:43:24Z

2

Strip out all the JavaScript and you should see what's going on server side; you are literally doing:

$temp1=$_POST['expert_id'];
echo json_encode($temp1);

$temp1=$_POST['answers_id'];
echo json_encode($temp1);

So if $_POST['expert_id'] = 1 and $_POST['answers_id'] = 2 ...

$temp1=1;
echo json_encode($temp1);
// outputs 1

$temp1=2;
echo json_encode($temp1);
// outputs 2

So what you'll have client-side is:

<script type="text/javascript">
    var jstemp1 =1;
</script>

<script type="text/javascript">
    var jstemp2 =2;
</script>

All the server-side code (PHP) executes before the client-side code (JavaScript)

answered Jul 10, 2015 at 10:43

CD001

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Comments

Alex Andrei · Accepted Answer · 2015-07-10 10:41:39Z

1

Your PHP code is inline with HTML/Javascript it will be executed first on the server, true,
but the evaluation is be done in the sequence you have wrote it.

You assign a value to a variable, then print it out.
After that you assign a different value to the same variable, over-writing it, then print it out.

This is perfectly normal.

answered Jul 10, 2015 at 10:41

Alex Andrei

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Comments

JM Mayo · Accepted Answer · 2015-07-10 10:40:23Z

0

PHP executes the code in server side before the page is sended to the browser.

The browser receives:

<script type="text/javascript">
  var jstemp1 =1;
</script>
<script type="text/javascript">
  var jstemp2 =2;
</script>

And when the page is loaded javascript code is executed.

answered Jul 10, 2015 at 10:40

JM Mayo

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