Re: [PHP-DB] Re: jpgraph and mysql passing array

2011-03-27 Thread Chris Stinemetz
On Sat, Mar 26, 2011 at 9:44 PM, Brian Smither  wrote:
> Would you try an experiment?

Thanks for for the suggestion but this didn't seem to work. I beleive
it is a header conflict based on what I have researched.
I just can't seem to find a solution.

>
> In this line:
> $sql = ("SELECT * FROM evdobanding WHERE Market like '%$term' and Cell_Sect = 
> '$term2' and Date = '$term3' ORDER BY distance asc");
>
> Change to:
> $sql = "SELECT * FROM evdobanding WHERE Market like '%".$term."' and 
> Cell_Sect = '".$term2."' and Date = '".$term3."' ORDER BY distance asc";
>
> That is, remove the parenthesis and concatenate the variables into the string.
>

The code in the PHP block:

  echo "TEST DATA PRINT";
  print_r($datay,false);

is passing all the values from the sql query because I can see them on
the page with the jpgraph error.

For example:

TEST DATA PRINTArray ( [0] => 665 [1] => 7093 [2] => 3 [3] =>
45816 [4] => 70848 [5] => 41365 [6] => 35676 [7] => 22434 [8] => 5450
[9] => 29131 [10] => 35244 [11] => 48614 [12] => 51748 [13] => 55497
[14] => 79042 [15] => 732 [16] => 1375 [17] => 1094 [18] => 897 [19]
=> 1122 [20] => 3059 [21] => 5350 [22] => 5080 [23] => 3082 [24] =>
2737 [25] => 326 [26] => 1334 [27] => 736 [28] => 469 [29] => 127 [30]
=> 105 [31] => 111 [32] => 197 [33] => 208 [34] => 950 [35] => 9 [36]
=> 9 [37] => 19 [38] => 8 [39] => 17 [40] => 90 [41] => 1917 [42] =>
1289 [43] => 2051 [44] => 1534 [45] => 1 [46] => 2 [47] => 4 [48] => 4
[49] => 7 [50] => 8 [51] => 9 [52] => 18 [53] => 22 [54] => 5 [55] =>
3 [56] => 1 [57] => 1 [58] => 2 )

JpGraph ERROR 25121



>From the research I have done so far using Jpgraph I have found this
link to a similar issue:

http://ubuntuincident.wordpress.com/tag/jpgraph/

but I am unable to get my arrays from the sql query to populate the graph.

Thank you in advance,

Chris

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[PHP-DB] Re: jpgraph and mysql passing array

2011-03-26 Thread Brian Smither
Would you try an experiment?

In this line:
$sql = ("SELECT * FROM evdobanding WHERE Market like '%$term' and Cell_Sect = 
'$term2' and Date = '$term3' ORDER BY distance asc");

Change to:
$sql = "SELECT * FROM evdobanding WHERE Market like '%".$term."' and Cell_Sect 
= '".$term2."' and Date = '".$term3."' ORDER BY distance asc";

That is, remove the parenthesis and concatenate the variables into the string.



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[PHP-DB] Re: jpgraph and mysql passing array

2011-03-23 Thread David Robley
Chris Stinemetz wrote:

> Hello,
> 
> 
> 
> I hope you can help me. The Jpgraph forum seems to be pretty
> uneventful with participation.
> 
> I am trying to pass arrays through the image tag to populate my
> variables for my graph.
> 
> I keep getting an error that there is no input. Do you have any
> suggestions?
> 
> I would greatly appreciate it! I have been trying to figure this out
> for several hours.
> 
> Below is my code:
> 
> I need some help with finding a solution. For some reason my graph is
> not showing up when it evident that my arrays are being passed.
> 
> When I use:
> 
> echo "TEST DATA PRINT";
> print_r($datay,false);
> 
> ---
> 
> I keep getting the error:
> 
> Empty input data array specified for plot.
> 
> ---
> 
> My php script is as follows:
> 
> ---
>  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
> http://www.w3.org/1999/xhtml";>
> 
> KCRF Dashboard
>  ini_set('display_errors', 1);
> error_reporting(E_ALL);
> 
> $term=$_POST['term'];
> $term2=$_POST['term2'];
> $term3=$_POST['term3'];
> 
> //include ('err_reporting.php');
> require ('PHP_Scripts/config.php');
> require_once ('jpgraph/jpgraph.php');
> require_once ('jpgraph/jpgraph_line.php');
> 
> $sql = ("SELECT * FROM evdobanding WHERE Market like '%$term' and
> Cell_Sect = '$term2' and Date = '$term3' ORDER BY distance asc");
> 
> $result = mysql_query($sql);
> 
> while($row = mysql_fetch_array($result))
> {
> $datay[] = $row["MBusage"];
> $datax[] = $row["Distance"];
> }
> 
> echo "TEST DATA PRINT";
> 
> print_r($datay,false);
> 
> ?>

> Thank you,
> 
> 
> 
> Chris

If $datay shows as empty, my first step would be to check that your SQL
query doesn't return an empty set.


Cheers
-- 
David Robley

"This ocean is calm," said the sailors specifically.
Today is Pungenday, the 10th day of Discord in the YOLD 3177. 


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