25/10/2013, 10:57
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| | Fecha de Ingreso: julio-2005
Mensajes: 73
Antigüedad: 18 años, 10 meses Puntos: 0 | |
Respuesta: Porque no Funciona? error en consulta este es el codigo con los echo para mostrar lo que sale en cada consulta. Código PHP: error_reporting(E_ERROR | E_WARNING | E_PARSE); mysql_connect("localhost", "root", ""); mysql_select_db("dashboard"); $sql = mysql_query("SELECT site_id FROM granjas where granja_tipo=2"); while ($row = mysql_fetch_array($sql)) { $granja[] = $row['site_id']; $i++; } echo implode(",", $granja) . "<br>"; echo "Las Granjas son " . $i . "<br>"; $sql2 = mysql_query("SELECT DISTINCT (YEAR( evento )) as anio FROM partos_server"); while ($row = mysql_fetch_array($sql2)) { $anio[] = $row['anio']; $a++; } echo implode(",", $anio) . "<br>"; echo "Los Años son " . $a . "<br>"; echo "<table>"; echo "<tr>"; echo "<td>Sitio</td>"; echo "<td>prom Nacido</td>"; echo "<td>Fecha</td>"; echo "<td>Semana</td>"; echo "</tr>"; for ($j = 0; $j < $i; $j++) { for ($b = 0; $b < $a; $b++) { $s = 0; unset($semana); $sql3 = mysql_query("SELECT DISTINCT(semana) as semana FROM partos_server where YEAR(evento)=" . $anio[$b]); while ($row = mysql_fetch_array($sql3)) { $semana[] = $row['semana']; $s++; } echo implode(",", $semana) . " AÑO " . $anio[$b] . "<br>"; for ($t = 0; $t < $s; $t++) { $query = @mysql_query("SELECT AVG(vivos) as promedio,site_id,year(evento) as anio FROM `partos_server` WHERE site_id = " . $granja[$j] . " and YEAR(evento)=" . $anio[$b] . " and semana = " . $semana[$t]) or die("Error en Consulta.- " . mysql_error()); echo "SELECT AVG(vivos) as promedio,site_id,year(evento) as anio,semana FROM `partos_server` WHERE site_id =" . $granja[$j] . " and YEAR(evento)=" . $anio[$b] . " and semana =" . $semana[$t] . "<br>"; $reg = mysql_fetch_array($query, MYSQL_BOTH); echo "<tr>"; echo "<td>" . $site = $reg['site_id'] . "</td>"; echo "<td>" . $promedio = round($reg['promedio'], 2) . "</td>"; echo "<td>" . $anio = $reg['anio'] . "</td>"; echo "<td>" . $semana = $reg['semana'] . "</td>"; $reg = mysql_fetch_array($query, MYSQL_BOTH); echo "</tr>"; } } } echo "</table>";
y esto es lo que me arroja como resultado
Código:
20,21,22,23,24,25,31
Las Granjas son 7
2012,2010,2011,2013
Los Años son 4
39,34,36,35,38,33,44,43,37,41,42,40,45,46,13,14,17,1,21,3,15,5,26,19,16,2,22,18,20,4,24,23,11,31,12,32,25,6,7,27,30,8,29,28,10,9,48,47,50,51,49,52 AÑO 2012
SELECT AVG(vivos) as promedio,site_id,year(evento) as anio,semana FROM `partos_server` WHERE site_id =20 and YEAR(evento)=2012 and semana =39
Error en Consulta.- You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '/' at line 2
Sitio prom Nacido Fecha Semana
20 11.9 2012
el array $semana[$t] esta agarrando valores como una / , d , t ... etc. y no entiendo porque.. gracias por tu ayuda. |