0
for(some loop condition):
mysql_query("UPDATE `details` SET
`url_battlelog` = '".$stats[$out]['url_battlelog']."',
`url_bf3stats` = '".$stats[$out]['url_bf3stats']."',
`rank_img_medium` = '".$stats[$out]['rank_img_medium']."',
`country_name` = '".$stats[$out]['country_name']."',
`country` = '".$stats[$out]['country']."',
`country_flag` = '".$stats[$out]['country_flag']."',
`rank_number` = '".$stats[$out]['rank_number']."',
`score_total` = '".$stats[$out]['score_total']."',
`time_total` = '".$stats[$out]['time_total']."',
`dogtag_basic_img` = '".$stats[$out]['dogtag_basic_img']."',
`dogtag_basic` = '".$stats[$out]['dogtag_basic']."',
`dogtag_advance_img` = '".$stats[$out]['dogtag_advance_img']."',
`dogtag_advance` = '".$stats[$out]['dogtag_advance']."'
WHERE `name_player` = '".$stats[$out]['name_player']."'
")
or die(mysql_error());
for(2nd loop condition):
mysql_query("UPDATE `weapons` SET
`img` = '".$gun_img."',
`name` = '".$gun_name."',
`kit` = '".$gun_kit."',
`time` = '".$gun_time."',
`kills` = '".$gun_kills."',
`headshots` = '".$gun_hs."',
`shots` = '".$gun_shots."',
`hits` = '".$gun_hits."',
`star_total` = '".$gun_star_c."',
`star_img` = '".$gun_star_i."',
`star_need` = '".$gun_star_n."',
`rank_curr` = '".$gun_rank_c."',
`rank_all` = '".$gun_rank_w."',
`desc` = '".$gun_desc."',
`category` = '".$gun_cat."',
`range` = '".$gun_range."',
`fire_rate` = '".$gun_fire_rate."',
`ammo` = '".$gun_ammo."',
`auto_fire` = '".$gun_fire_auto."',
`burst_fire` = '".$gun_fire_burst."',
`single_fire` = '".$gun_fire_single."',
`unlock_total` = '".$unlock_total."',
`unlock_done` = '".$unlock_done."',
`unlock_p` = '".round($unlock_p)."'
WHERE `name_player` = '".$stats[$out]['name_player']."'
")
or die(mysql_error());
問題是隻有第2個表(武器)正在更新,第1個表(細節)沒有顯示任何更改。不顯示任何錯誤。MySQL更新2表在單個php腳本中使用2個mysql_query()
我有相同類型的腳本插入數據到這兩個表和它的工作正常。
我是MySQL和PHP的新手。對不起,英語......
你是什麼意思? –
回聲您的SQL更新細節,然後嘗試在您的查詢瀏覽器中執行您的查詢,你會看到你的查詢有什麼問題 –
抱歉..但如何做到這一點? –