how to use a string variable from a php file to a javascript in html

Question

my final goal is to convert data from database to a array, then i can print a pretty nice graph with a jquery plugin "flot" in html file:

<script language="javascript" type="text/javascript" src="jquery.flot.js"></script>
<script type="text/javascript" >
$(function test() {   
var d3= [[0, 3], [4, 8], [8, 5], [9, 13]];   
$.plot($("#placeholder"), [d3]);
});
</script>

in php file

echo"<script language='javascript' >\n";  
$k = 0;  
    $agevi = mysql_result($result, $k, "Age"); 
    $snvi = mysql_result($result, $k, "StuNo"); 
settype($agevi, "integer"); 
settype($snvi, "integer"); 
$smallarray = array($snvi,$agevi);  
    $bigarray = array($smallarray);
    $i++;   
while ($i < $number) 
    { 
      $agev = mysql_result($result, $i, "Age"); 
  $snv = mysql_result($result, $i, "StuNo");  
  settype($agev, "integer"); 
  settype($snv, "integer");  
  $smallarray = array($snv,$agev);   
  array_replace($smallarray,$smallarray);
  array_push($bigarray,$smallarray); 
  echo"\n"; 
  $i++;    
  }  
  echo 'var stunoarr = '.json_encode($bigarray).';'; 
  }    
      echo"\n</script>";  
  echo  json_encode($bigarray) ; 
      echo "\ndocument.write(\"bigarray array is: <b>" . json_encode($bigarray) .  "</b>\")";    
      mysql_free_result($result); 
      mysql_close();

what i have in php output is work:

[[1,12],[2,5],[3,6],[4,2],[5,7],[6,2],[7,12],[8,3],[9,6],[10,8],[11,4]] document.write("bigarray array is: [[1,12],[2,5],[3,6],[4,2],[5,7],[6,2],[7,12],[8,3],[9,6],[10,8],[11,4]]")

so people suggesting use json, yes i did, and i also put

<script>
var myvar = <?= json_encode($bigarray); ?>; 
</script>

at the end of php file outside ?>

how exactly should get the variable from php to javascript in order to produce the graph then..? tried hundrd of ways now still not working...thanks in advance! :D

Making something work with functionality like flot means that you understand what flot expects, in addition to understanding how the language you're using to manipulate the output itself works. — Jared Farrish
– Jared Farrish, Commented Oct 10, 2011 at 3:21
flot can be easily used as long as i have integer array, and i made it work in the php file :[[1,12],[2,5],[3,6],[4,2],[5,7],[6,2],[7,12],[8,3],[9,6],[10,8],[11,4]] — user987013
– user987013, Commented Oct 10, 2011 at 3:25
Have you checked the web console for javascript issues, or is the script seemingly not executing? — JackWink
– JackWink, Commented Oct 10, 2011 at 3:31
Instead why would you want to make an ajax call and get the required data from DB using php ? — Dhiraj
– Dhiraj, Commented Oct 10, 2011 at 4:03

Dhiraj · Accepted Answer · 2011-10-10 04:31:35Z

1

in javascript file

<script language="javascript" type="text/javascript" src="jquery.flot.js"></script>
<script language="javascript" type="text/javascript" src="http://code.jquery.com/jquery-latest.js"></script>
    <script type="text/javascript" >
    $(function test() {
    getData();
    });
    function getData(){
      $.ajax({
                    url: "yourfile.php",
                    type: "GET",
                    data: '',
                    cache: false,
                    success: function (result) {
                        showGraph(result);
                    }
            });
    }
function showGraph(data){
 $(data).find('node').each(key,val){
  var age = $(val).find('agev').text();
  var stuno = $(val).find('snv').text();
  // populate your d3 array here
}
$.plot($("#placeholder"), [d3]);
}
    </script>

in yourfile.php

header('Content-Type: text/xml');

$k = 0;  
    $agevi = mysql_result($result, $k, "Age"); 
    $snvi = mysql_result($result, $k, "StuNo"); 
settype($agevi, "integer"); 
settype($snvi, "integer"); 
$smallarray = array($snvi,$agevi);  
    $bigarray = array($smallarray);
$resultXML = new SimpleXMLElement(stripslashes('<data></data>'));
    $i++;   
while ($i < $number) 
    { 
      $agev = mysql_result($result, $i, "Age"); 
  $snv = mysql_result($result, $i, "StuNo");  
  settype($agev, "integer"); 
  settype($snv, "integer");  
  $temp = $resultXML->addChild('node');
  $temp->addChild('snv',$snv);
  $temp->addChild('agev',$agev);
  $i++;    
  }  

  }    
     mysql_free_result($result); 
      mysql_close();  
echo $resultXML->asXML();

answered Oct 10, 2011 at 4:31

Dhiraj

33.7k10 gold badges65 silver badges78 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

user987013 Over a year ago

on html page i tried using array=[]; array.push(age,stuno); and d3.push(array); should be working right? then its not generating graph...so i delete the three line $(data).find('node').each(key,val){ var age = $(val).find('agev').text(); var stuno = $(val).find('snv').text();

user987013 Over a year ago

on html page i tried using array=[]; array.push(age,stuno); and d3.push(array); should be working right? then its not generating graph...so i delete the three line $(data).find('node').each(key,val){ var age = $(val).find('agev').text(); var stuno = $(val).find('snv').text(); } and define a array[[1,2],[2,3]]; then it works...is it the code having error or should add something else to make it work? thanks so much for helping!

user987013 Over a year ago

haha problem solved $(data2).find('node').each(function(){ var stuno = $(this).find('snv').text(); var age = $(this).find('agev').text();

Collectives™ on Stack Overflow

how to use a string variable from a php file to a javascript in html

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related