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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in / เกิดจาก config.php รึป่าวคับ |
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มันฟ้องว่า พารามิเตอร์ที่ใส่ให้มันไม่ใช่ result resource เบื้องต้นคือ sql statement คุณผิดพลาดครับ
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| Date :
2011-08-29 15:12:15 |
By :
ikikkok |
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ตรงไหนคับระหว่าง
$result = mysql_query($sql);
while($row= mysql_fetch_array($result)) {
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| Date :
2011-08-29 15:47:06 |
By :
seya |
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ดูในเงื่อนไข $sql คับ
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| Date :
2011-08-29 15:56:17 |
By :
siammbk |
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ช่วยดูให้หน่อยคับ
<? include("include/config.php");?>
<?
$sql="SELECT
tb_scheme.scheme_id
, tb_scheme.scheme_name
, tb_scheme.scheme_detail
, tb_scheme.scheme_date
, tb_personnel.Personnel_name
FROM
nk_school.tb_scheme
INNER JOIN nk_school.tb_personnel
ON (tb_scheme.Personnel_id = tb_personnel.Personnel_id);";
$result = mysql_query($sql);
while($row= mysql_fetch_array($result)) {
$id=$row[scheme_id];
$name=$row[scheme_name];
$detail=$row[scheme_detail];
$date=$row[scheme_date];
$Personnel_name=$row[Personnel_name];
?>
<tr>
<td align="center"><? echo"$id"?><br />
-----</td>
<td align="center"><? echo"$name"?><br />
----------------------</td>
<td align="center"><? echo"$detail"?><br />
-------------------</td>
<td align="center"><? echo"$date"?><br />
------</td>
<td align="center"><? echo"$Personnel_name"?><br />
-----------</td>
</tr>
<? }?>
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| Date :
2011-08-29 16:07:10 |
By :
seya |
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 Load balance : Server 04
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