2011-09-23 125 views
0

我有PHP代碼,它爲每個列表項創建一個帶有單選按鈕的列表。兩個提交表單

$result = mysql_query("SELECT * FROM attitudes WHERE x_axis = '$famID'",$db); 
$rowcheck = mysql_num_rows($result); 
while ($row_user = mysql_fetch_assoc($result)) 

foreach ($row_user as $col=>$val) { 
    if ($col != $famID && $col != 'x_axis') { 
     list($famname) = mysql_fetch_row(mysql_query("SELECT familyname FROM families WHERE families_ID='$col'",$db)); 
     echo "col $col famname $famname is val $val."; 
     echo "<input type = \"radio\" name = \"whichfam\" value = \"$col\" />"; 
     echo "</br>"; 

    } 
} 

然後,我有一個在底部提交按鈕(和表單標籤整件事)

我想有兩種可能的提交。此代碼旨在讓玩家提升或降低價值。他們點擊其中一個單選按鈕,然後選擇「上升」或「下降」。然後它應該發佈到後端並執行代碼來提高或降低該值。我知道如何在jQuery中做到這一點,但我不知道如何在PHP中有兩個SUBMIT按鈕。

我該怎麼做?

(具體地講,我怎麼做兩個按鈕提交工作,後端代碼應該比較簡單,$ _ POST或其他)

回答

5

這是你在找什麼?

<button type="submit" name="submit1" value="submit1">submit1</button> 
<button type="submit" name="submit2" value="submit2">submit2</button> 

然後

if(isset($_POST["submit1"])) { 

} else if(isset($_POST["submit2"])) { 

}