มันแจ้ง
Warning: odbc_result() [function.odbc-result]: Field CUSTOM_INV.CUSTOM_INV not found in D:\AppServ\www\official\Export\Search_Export_Result.php on line 81
$sSql = "SELECT * FROM EXPORT INNER JOIN CUSTOM_INV ON (EXPORT.ID=CUSTOM_INV.ID_EXPORT)
INNER JOIN COMMERCIAL_INV ON (EXPORT.ID=COMMERCIAL_INV.ID_EXPORT)
WHERE 1=1";
ผิดป่าวคะ
Warning: odbc_exec() [function.odbc-exec]: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax error (missing operator) in query expression '(EXPORT.ID=CUSTOM_INV.ID_EXPORT) INNER JOIN COMMERCIAL_INV ON (EXPORT.ID=COMMERCIAL_INV.ID_EXPORT)'., SQL state 37000 in SQLExecDirect in D:\AppServ\www\official\Export\Search_Export_Result.php on line 47
Warning: odbc_fetch_array(): supplied argument is not a valid ODBC result resource in D:\AppServ\www\official\Export\Search_Export_Result.php on line 74
ไม่ค่อยแม่นเรื่องการ join เท่าไหร่คะ
Date :
2010-07-29 09:26:01
By :
ปังคุง
No. 8
Guest
สงสัยต้องหันกลับมาใช้ = แล้วมั้ง
ไม่ วันนิ่งคะ
$sSql = "SELECT * FROM EXPORT,CUSTOM_INV,COMMERCIAL_INV WHERE EXPORT.ID=CUSTOM_INV.ID_EXPORT AND
EXPORT.ID=COMMERCIAL_INV.ID_EXPORT AND 1=1";