jqplot not evaluating string

Question

I am using jqplot and have constructed an array with the following PHP code

if($fiveRes){
    $lastNum = mysql_num_rows($fiveRes);

    $testarray = array();

    while($row = mysql_fetch_array($fiveRes, MYSQL_ASSOC)) {
        $testarray[] = array($row['GameDate'],  floatval($row['WKP']));
    }

    echo json_encode($testarray);

}

This code outputs the correct code that I need to insert into the jqplot function. Here is the array generated by the code above:

[["2011-12-24",0],["2011-12-19",14],["2011-12-08",22],["2011-12-04",14],["2011-11-27",12]]

So currently i'm printing this array to the screen, and then using jQuery .text() to capture the string and place it inside a variable. I can echo out the var that I assigned the array string to and it works correctly, however when I pass it into the jqplot function it does nothing.

var p1array = $('.col.first .parray').text();

alert(p1array); //Alerts the correct array formatted like above.

    var plot1 = $.jqplot('jqplot0', [p1array], {
          title:'Last 5 Fantasy Points',
          axes:{
            xaxis:{
              renderer:$.jqplot.DateAxisRenderer,
              tickOptions:{
                formatString:'%b&nbsp;%#d'
              } 
            },
            yaxis:{
              tickOptions:{
                formatString:''
                }
            }
          },
          highlighter: {
            show: true,
            sizeAdjust: 7.5
          },
          cursor: {
            show: false
          }
      });

To make it even more complicated, if I copy the string that the PHP generates and I hardcode it into a variable in JS it works. Any ideas why the jqplot plugin won't evaluate the string I get from using $(this).text();.

Here is the jQplot example i'm modeling after: http://www.jqplot.com/tests/cursor-highlighter.php

I don't know why the text() is not working but Have you tried/or is there any problem using p1array = ; — TigerTiger, Aug 08 '12 at 16:55
what happens if you use p1array without the brackets? ie var plot1 = $.jqplot('jqplot0', p1array, {... — Jim, Aug 08 '12 at 16:56
@HappyApe Does that mean I would have to move my JS inside of the PHP file? — jsheffers, Aug 08 '12 at 16:59
hmm good question - yes, I tested it with js in php - but I recon its something to do with type-casting when you use text() method — TigerTiger, Aug 08 '12 at 17:01
Test this -> craete then var p1array = $('#arraystring').val(); — TigerTiger, Aug 08 '12 at 17:04
Not resolved, but here is a jsfiddle to depict the problem : http://jsfiddle.net/rDMea/ — sdespont, Aug 08 '12 at 17:26
@HappyApe I tried adding the hidden input, but it's not working. — jsheffers, Aug 08 '12 at 18:09

score 0 · Accepted Answer · edited May 23 '17 at 12:27

0

Using this method helped me solve this problem. Instead of using jQuery to access the printed variable I inserted a little bit of javascript at the bottom of my PHP page to capture the local variable in PHP and then used that to pass into the jQplot function.

Pass a PHP string to a JavaScript variable (and escape newlines)

edited May 23 '17 at 12:27

Community

1
1

answered Aug 09 '12 at 16:18

jsheffers

1,602
19
41

score 0 · Answer 2 · answered Aug 09 '12 at 16:43

0

Use [p1array] directly with out braces and try... it should work just like p1array

answered Aug 09 '12 at 16:43

Anand thakkar

479
3
8

jQplot uses that syntax, it does not matter if the square brackets are around it I tried. – jsheffers Aug 09 '12 at 17:05
can you try var p1array = eval('{' + $('.col.first .parray').text() + '}'); this might help you out ....... hope so – Anand thakkar Aug 10 '12 at 05:27

jqplot not evaluating string

2 Answers2