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PHP error


Go to solution Solved by gmc1103,

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Hi

 

Can someone help me with this

<table id="dg" class="easyui-datagrid" url="get_atividadesByUserId.php?=<?php echo $userid ?>"
            style="width:1800px;height:500px; border:1px solid #ccc;" title="Gestão das Atividades"
            toolbar="#toolbar" pagination="true" idField="id"
            rownumbers="true" fitColumns="true" resizable="true">
            <thead>
               <tr>
                  <th field="idativade" width="20">Nº</th>
                  <th field="cargo" width="200">Cargo</th>
                  <th field="atividade" width="200">Atividade</th>
                  <th field="data" width="50">Data</th>
                  <th field="hora" width="50">Hora</th>
                  <th field="local" width="100">Local</th>
                  <th field="inter" width="200">Intervenientes</th>
                  <th field="notas" width="150">Notas</th>
               </tr>
            </thead>
         </table>

and my php

<?php
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$iduser= $_GET['userid'];
include 'conn.php';
mysql_query("SET NAMES 'utf8'");
$rs = mysql_query('SELECT  `pae_atividades`.`idativade` ,  `pae_cargo`.`cargo` ,  `pae_atividades`.`atividade` ,  `pae_atividades`.`data` ,  `pae_atividades`.`hora` ,  `pae_atividades`.`local` ,  `pae_atividades`.`inter` , `pae_atividades`.`notas` ,  `utilizador`.`nome` 
FROM  `ebspma_paad_ebspma`.`pae_atividades` 
INNER JOIN  `ebspma_paad_ebspma`.`utilizador` ON (  `pae_atividades`.`idutilizador` =  `utilizador`.`idutilizador` ) 
INNER JOIN  `ebspma_paad_ebspma`.`pae_cargo` ON (  `pae_atividades`.`idcargo` =  `pae_cargo`.`idcargo` ) 
WHERE  `pae_atividades`.`data` >= CURDATE( ) 
AND  `pae_atividades`.`idutilizador`= $iduser ORDER BY  `pae_atividades`.`data` ASC ');
$result = array();
while($row = mysql_fetch_object($rs)){
	array_push($result, $row);
}
echo json_encode($result);
?>

I'm getting this error

 

<br />
<b>Notice</b>:  Undefined index:  userid in <b>/home/ebspma/public_html/atividades/get_atividadesByUserId.php</b> on line <b>4</b><br />
<br />
<b>Warning</b>:  mysql_fetch_object(): supplied argument is not a valid MySQL result resource in <b>/home/ebspma/public_html/atividades/get_atividadesByUserId.php</b> on line <b>14</b><br />
[]
 
The url passed is right
 
get_atividadesByUserId.php?=1261
 
i have the id correct
 
so whats wrong?
 
Thanks
Link to comment
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Thank you

 

The first error has gone but i still have the second

<?php
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$iduser= $_GET['userid'];
include 'conn.php';
mysql_query("SET NAMES 'utf8'");
$rs = mysql_query('SELECT  `pae_atividades`.`idativade` ,  `pae_cargo`.`cargo` ,  `pae_atividades`.`atividade` ,  `pae_atividades`.`data` ,  `pae_atividades`.`hora` ,  `pae_atividades`.`local` ,  `pae_atividades`.`inter` , `pae_atividades`.`notas` ,  `utilizador`.`nome` 
FROM  `ebspma_paad_ebspma`.`pae_atividades` 
INNER JOIN  `ebspma_paad_ebspma`.`utilizador` ON (  `pae_atividades`.`idutilizador` =  `utilizador`.`idutilizador` ) 
INNER JOIN  `ebspma_paad_ebspma`.`pae_cargo` ON (  `pae_atividades`.`idcargo` =  `pae_cargo`.`idcargo` ) 
WHERE  `pae_atividades`.`data` >= CURDATE( ) 
AND  `pae_atividades`.`idutilizador`= $iduser ORDER BY  `pae_atividades`.`data` ASC ');
$result = array();
while($row = mysql_fetch_object($rs)){
	array_push($result, $row);
}
echo json_encode($result);
?>

Line 14

 

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/ebspma/public_html/atividades/get_atividadesByUserId.php on line 14
[]

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https://forums.phpfreaks.com/topic/294476-php-error/#findComment-1505251
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Thank you

 

The first error has gone but i still have the second

<?php
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$iduser= $_GET['userid'];
include 'conn.php';
mysql_query("SET NAMES 'utf8'");
$rs = mysql_query('SELECT  `pae_atividades`.`idativade` ,  `pae_cargo`.`cargo` ,  `pae_atividades`.`atividade` ,  `pae_atividades`.`data` ,  `pae_atividades`.`hora` ,  `pae_atividades`.`local` ,  `pae_atividades`.`inter` , `pae_atividades`.`notas` ,  `utilizador`.`nome` 
FROM  `ebspma_paad_ebspma`.`pae_atividades` 
INNER JOIN  `ebspma_paad_ebspma`.`utilizador` ON (  `pae_atividades`.`idutilizador` =  `utilizador`.`idutilizador` ) 
INNER JOIN  `ebspma_paad_ebspma`.`pae_cargo` ON (  `pae_atividades`.`idcargo` =  `pae_cargo`.`idcargo` ) 
WHERE  `pae_atividades`.`data` >= CURDATE( ) 
AND  `pae_atividades`.`idutilizador`= $iduser ORDER BY  `pae_atividades`.`data` ASC ');
$result = array();
while($row = mysql_fetch_object($rs)){
	array_push($result, $row);
}
echo json_encode($result);
?>

Line 14

 

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/ebspma/public_html/atividades/get_atividadesByUserId.php on line 14

[]

mysql is deprecated you should use mysqli, have you tried 

while($row = mysql_fetch_object($result)){
	array_push($result, $row);
} 

Your trying to get the result from the query without actually getting the results

 

Also, 

$result = $rs->fetch();
Edited by Tom10
Link to comment
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mysql is deprecated you should use mysqli...

 

...or PDO. More information about the different APIs can be found here:

http://php.net/manual/en/mysqlinfo.api.choosing.php

 

 

...have you tried 

while($row = mysql_fetch_object($result)){
	array_push($result, $row);
} 

 

Based on the code originally posted, the resultset is being saved to the $rs variable.

$rs = mysql_query('SELECT  `pae_atividades`.`idativade` ... ORDER BY  `pae_atividades`.`data` ASC ');

$result contains something else.

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