Awesome! Now lets take a look at the javascript end of things. You said
before that you have a form and you are using jQuery to submit it, do you
have code for that already, or do you need to write new code?
On Monday, December 10, 2012 2:45:34 AM UTC-5, Chrystopher Medina wrote:
>
> yes u are right.... my new result is this :
>
> {"cols":[{"label":"Type","type":"string"},{"label":"Count","type":"number"}],"rows":[{"c":[{"v":"word
>
> of mouth recomendation"},{"v":2}]},{"c":[{"v":"magazine
> advertisement"},{"v":1}]},{"c":[{"v":"roadside
> advertisement"},{"v":1}]},{"c":[{"v":"google"},{"v":2}]}]}
>
> yes my friend i need 4 answers .is for that i add this :
>
> $query = "SELECT b.id_respuesta , COUNT(b.id_respuesta) AS cnt
> FROM huesped a, rompe_encuesta b
> WHERE
> b.id_huesped = a.id_huesped AND
> b.id_aspecto = 1 AND
> b.id_respuesta >= 8 AND
> b.id_respuesta <= 11 AND // this
> a.fecha BETWEEN '$var1' AND '$var2'
> GROUP BY b.id_respuesta;";
>
> I have modified the code like this:
> switch ((int) $r['id_respuesta']) {
>
> case 8:
> $type = 'word of mouth recomendation';
> break;
> case 9:
> $type = 'magazine advertisement';
> break;
> case 10:
> $type = 'roadside advertisement';
> break;
> case 11:
> $type = 'google';
> break;
> default:
> die('{"error":"Error in SQL query: unknown \'id_respuesta\'"}');
> }
>
>
> and this is all my file php. Check it please.
>
>
> <?php
> $var1="$_POST[fecha1]";
>
> $var2= "$_POST[fecha2]";
>
> $server="localhost";
> $username="root";
> $password="chrystopher";
> $databasename="encuestasavandaro";
>
> $con = mysql_connect($server,$username,$password) or die ('Error
> connecting to server');
> mysql_select_db($databasename,$con);
>
> $query = "SELECT b.id_respuesta , COUNT(b.id_respuesta) AS cnt
> FROM huesped a, rompe_encuesta b
> WHERE
> b.id_huesped = a.id_huesped AND
> b.id_aspecto = 1 AND
> b.id_respuesta >= 8 AND
> b.id_respuesta <= 11 AND
> a.fecha BETWEEN '$var1' AND '$var2'
> GROUP BY b.id_respuesta;";
>
>
> $table = array();
>
> $table['cols']=array(
> array('label' => 'Type' , 'type' => 'string'),
> array('label' => 'Count' , 'type' => 'number')
> );
>
> $rows = array();
> if (!mysql_query($query,$con))
> {
> die('Error: ' . mysql_error());
> }else{
>
> $sqlquery= mysql_query($query);
> while($r = mysql_fetch_assoc($sqlquery)){
> $temp = array();
> // using (int) $variable typecasts the variable as an integer, usefull
> when the SQL is returning numbers as strings
>
> switch ((int) $r['id_respuesta']) {
>
> case 8:
> $type = 'word of mouth recomendation';
> break;
> case 9:
> $type = 'magazine advertisement';
> break;
> case 10:
> $type = 'roadside advertisement';
> break;
> case 11:
> $type = 'google';
> break;
> default:
> die('{"error":"Error in SQL query: unknown \'id_respuesta\'"}');
> }
>
> $temp[] = array('v' => $type);
> $temp[] = array('v' => (int) $r['cnt']);
> $rows[] = array('c' => $temp);
> }
> }
> $table['rows'] = $rows;
>
> $jsonTable = json_encode($table);
> echo $jsonTable;
>
> ?>
>
>
>
>
>
>
>
>
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