Using php with database in a javascript

Question

This code is when i hardcore the sentence "Have a nice day!", it will echo out the exact same line. My question is what if i want to retrieve sentence from the database, instead of hard-coding it.

<?php
    $php_var = "Have a nice day!";
?>
  <html>
  <head>
  <title>Hello</title>
  </head>
  <body>

  <script>
    var js_var = "<?php echo $php_var; ?>";
    //var js_var = "Try123";
    document.writeln(js_var);
   //document.writeln(js_var);
  </script>
  </body>
  </html>

I am suppose to do something like this is it? but it cant work. It printed out "SELECT * FROM sen WHERE id=1 ;" on the page.

<?php
   $con = mysql_connect(localhost,root,password,database_name);
   $php_var = "SELECT * FROM sen WHERE id=1 ;";
?>


    <script>
  var js_var = "<?php echo  $php_var ; ?>";
  //var js_var = "Try123";
 document.writeln(js_var);
 //document.writeln(js_var);
</script>

currently you only assign your query to a variable you're supposed to run it on your connection and then extract the result, im currently looking up how to do this again. — kpp
– kpp, Commented Jun 25, 2014 at 6:53

Wouter0100 · Accepted Answer · 2014-06-25 09:14:29Z

4

You're not executing the query and fetching the result. Something like this should work:

<?php
$con = mysqli_connect(localhost,root,password,database_name);
$php_var = mysqli_fetch_assoc(mysqli_query($con, "SELECT * FROM sen WHERE id=1 LIMIT 1")); 
?>

<script>
  var js_var = "<?php echo  $php_var['id']; ?>";
  //var js_var = "Try123";
 document.writeln(js_var);
 //document.writeln(js_var);
</script>

Please be aware of some things:

Don't forgot error handling on the right way. (Not or die)
Check if the MySQL connecton was successfully made.
Possibility of MySQL injection
I've updated mysql_* to mysqli_*, this because mysql_* is deprecated and will being removed in the future.

edited Jun 25, 2014 at 9:14

answered Jun 25, 2014 at 6:55

Wouter0100

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4 Comments

gbestard Over a year ago

change it to mysqli_ ... there's no point on teaching deprecated functions

Cerbrus Over a year ago

Instead of showing someone how not to do it, just show them how it should be done, and add a sidenote about the deprecated functions.

user3773928 Over a year ago

@Wouter0100 Its giving this two errors- Warning: mysqli_query() expects at least 2 parameters, 1 given in Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in

Wouter0100 Over a year ago

@user3773928 Excuus me, I always use the Object oriented one. I've edited my answer, please try again. :)

Vaionixx · Accepted Answer · 2014-06-25 06:54:32Z

0

My suggestion is that you create a Rest api with json response backend in PHP and then have a javascript frontend using like JQuery $get or something.

answered Jun 25, 2014 at 6:54

Vaionixx

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1 Comment

Cerbrus Over a year ago

This answer is totally out of the scope of the question.

Ankit · Accepted Answer · 2014-06-25 07:16:21Z

0

Remove ; from the query. Use $php_var = "SELECT * FROM sen WHERE id=1";

answered Jun 25, 2014 at 7:16

Ankit

1303 bronze badges

1 Comment

Jigar Over a year ago

-1. ; doesn't matter. And whole mysqli_query() and mysqli_fetch_assoc() are missing. Read Wouter0100's answer.

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