$str = "";
$str .=  "<img src=\'http://chart.apis.google.com/chart?chs=240x100";;
$str .= "&amp;chd=t:";

this code looks a lot neater/nicer.




________________________________
From: Asad Yousaf <[email protected]>
To: [email protected]
Sent: Tuesday, 3 March, 2009 7:18:35 PM
Subject: Re: A problem...

Niketheguy

try this code
it builds the string and echos it once thus we eliminate the chance that echo 
might place line breaks in your <img > tag

<html>
<head>
<title>
Nike's Second Attempt on Charts API with php
</title>
</head>
<body>

<?php

  include("db.php");
  $sql="SELECT * from test";
  $result = mysql_query($sql);
  // We will use a string variable $str to generate the <img tag>

  $str = "";
  $str =  "<img src=\'http://chart.apis.google.com/chart?chs=240x100";;
  $str = $str."&amp;chd=t:";
  $rows = mysql_num_rows($result);

  $i=1;
  while($row = mysql_fetch_array($result))
  {
    if($i < $rows)
    {
      $str = $str.$row['value'].",";
      $i=$i+1; 
    }
    else
    {
      $str = $str.$row['value'];
      $i=$i+1;
    }
  }
  $str = $str."&amp;cht=lc";
  // We will close the src attribute with \' and to print escape character ' we 
shall precede it with \

  $str = $str."&amp;chxt=x,y\'";
  $str = $str." alt=\'Sample chart\' />";

  //we shall echo the $str that will display the graph
  echo $str;

?>
</body>
</html>

Asad








On Tue, Mar 3, 2009 at 11:42 PM, niketheguy <[email protected]> wrote:


I am currently trying to incorporate charts api into my project by
interfacing it with my MYSQL DB.I am using PHP..this is the code i
have written..It seems right to me but the chart does not get
displayed..would appreciate help in this matter
Thanks in advance.

Here is the code...
<html>
<?php

include("db.php");
$sql="SELECT * from test";
$result = mysql_query($sql);
echo "<img src='http://chart.apis.google.com/chart?chs=240x100 ";
echo "&amp;chd=t:";
$rows = mysql_num_rows($result);

$i=1;
while($row = mysql_fetch_array($result))
{

if($i < $rows){
echo $row['value'].",";
$i=$i+1;}

else
{
echo $row['value'];
$i=$i+1;
}


}
echo " &amp;cht=lc";
echo " &amp;chxt=x,y";
echo " alt='Sample chart'/></img>";


?>
</html>


The table contains values of type INT whc im trying to display through
the chart.I double checked the generated source of this page and it is
perfectly alright with no anomalies observable to me..I hope someone
could help me..Thanks.




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