Re: [請益] Checkbox判斷寫入資料庫問題
※ 引述《arianda (火鍋)》之銘言：
: ※ 引述《kiiro (阿鬼)》之銘言：
: 因為你checkbox選幾個, php那邊就只接到幾個, 所以你試試看你的checkbox的value
: 用你goods_no的key值
: 然後用foreach
: foreach ($_POST['goodselect'] as $goodskey) {
: $insertSQL = sprintf("INSERT INTO opengoods (open_id, goods_no, goods_name,
: goods_price) VALUES (%s, %s, %s, %s)",
: GetSQLValueString($_POST['open_id'], "text"),
: GetSQLValueString($_POST['goods_no'][$goodskey], "int"),
: GetSQLValueString($_POST['goods_name'][$goodskey], "text"),
: GetSQLValueString($_POST['goods_price'][$goodskey], "text"));
: mysql_select_db($database_Herewego, $Herewego);
: $Result2 = mysql_query($insertSQL, $Herewego) or die(mysql_error());
: }
感謝a大指教，問題已解決(Y)
後來我稍微將程式碼改了一下：
1.先將checkbox一一設值 $i
<?php $i=0; do { ?>
<input name="goodselect[]" type="checkbox" id="goodselect[]"
value="<?php echo $i; ?>" />
<?php $i+=1;
} while ($row_goods = mysql_fetch_assoc($goods)); ?>
2.再利用foreach迴圈抓checkbox已核取的value值 $i，將資料寫入資料庫
foreach ($_POST['goodselect'] as $goodskey) {
$insertSQL = sprintf("INSERT INTO opengoods (open_id, goods_no, goods_name,
goods_price) VALUES (%s, %s, %s, %s)",
GetSQLValueString($_POST['open_id'], "text"),
GetSQLValueString($_POST['goods_no'][$goodskey], "int"),
GetSQLValueString($_POST['goods_name'][$goodskey], "text"),
GetSQLValueString($_POST['goods_price'][$goodskey], "text"));
mysql_select_db($database_Herewego, $Herewego);
$Result2 = mysql_query($insertSQL, $Herewego) or die(mysql_error());
}
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