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คือผมจะทำการเขียนคำสั่ง Update แบบแยกการทำงาน
ผลของการรัน
Code (PHP)
1. UPDATE sv_received SET Officer_ID = '1' ,chan_compl_ID = '2' ,way_advise_ID = '1' ,inform = '2' ,pname_ID = '1' ,name = '3' ,s_name = '4' ,home_num = '5' ,road = '6' ,tel = '7' ,email = 'meijicpipo@gmail.com' WHERE Query Error 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 1
ทำการประกาศเรียกใช้ Function Update ข้อมูล และส่ง พาราดิเตอร์ ไป 3 ชุด ข้อมูล
Code (PHP)
1. $updateReceived = fncUpdateData(
2. 'sv_received' ,
3. "Officer_ID = '" . $Officer_ID . "',chan_compl_ID = '" . $chan_compl_ID . "',way_advise_ID = '" . $way_advise_ID . "',inform = '" . $inform . "',pname_ID = '" . $pname_ID . "',name = '" . $name . "',s_name = '" . $s_name . "',home_num = '" . $home_num . "',road = '" . $road . "',tel = '" . $tel . "',email = '" . $email . "' " ,
4. "sv_received_ID = '" . $sv_received_ID . "'"
5. );
Function update ใน mysql
Code (PHP)
1. function fncUpdateData( $tblname , $Command , $Condition ){
2. echo "<br>" ;print_r( $tblname ); echo "<br>" ;print_r( $Command ); echo "<br>" ;print_r( $Condition ); echo "<br>" ;
3. $objSQL = "UPDATE $tblname SET $Command WHERE $condition " ;
4. print_r( $objSQL );
5. $objQuery = mysql_query( $objSQL ) or die ( "Query Error " .mysql_error());
6.
7. return $objQuery ;
8. }
Tag : PHP, MySQL, HTML/CSS
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Date :
2015-08-20 10:36:09 |
By :
meijicpipo |
View :
1194 |
Reply :
1 |
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