[英]How to make the result of the query to be selected on a dropdown menu
我有這個數據庫:
**SBU **SBU_Users
*id | *SBU_Name *users_id | *sbu_id
1 | SBU1 1 | 3
:
4 | SBU4
這是我的代碼:
$query = "SELECT sbu.SBU_Name, sbu_users.sbu_id FROM sbu, sbu_users WHERE sbu_users.sbu_id=sbu.id AND sbu_users.users_id=".$users_id."";
$result = mysql_query($query, $db) or die (mysql_error($db));
while ($row = mysql_fetch_array($result))
{
extract($row);
}
echo "<select name=sbuuser>";
echo "<option value='1' ";if ($sbu_id==1){echo "selected";} else {echo "";}echo ">SBU1</option>";
echo "<option value='2' ";if ($sbu_id==2){echo "selected";} else {echo "";}echo ">SBU2</option>";
echo "<option value='3' ";if ($sbu_id==3){echo "selected";} else {echo "";}echo ">SBU3</option>";
echo "<option value='4' ";if ($sbu_id==4){echo "selected";} else {echo "";}echo ">SBU4</option>";
我在選擇正確的SBU_Name時遇到問題,因為我的代碼一直在選擇最后一個SBU_Name。 請幫我。
也許是這樣的嗎?
$query = "SELECT sbu.SBU_Name, sbu_users.sbu_id FROM sbu, sbu_users WHERE sbu_users.sbu_id=sbu.id AND sbu_users.users_id=".$users_id."";
$result = mysql_query($query, $db) or die (mysql_error($db));
echo "<select name=sbuuser>";
while ($row = mysql_fetch_array($result))
{
extract($row);
echo "<option value='1' ";if ($sbu_id==1){echo "selected";} else {echo "";}echo ">SBU1</option>";
echo "<option value='2' ";if ($sbu_id==2){echo "selected";} else {echo "";}echo ">SBU2</option>";
echo "<option value='3' ";if ($sbu_id==3){echo "selected";} else {echo "";}echo ">SBU3</option>";
echo "<option value='4' ";if ($sbu_id==4){echo "selected";} else {echo "";}echo ">SBU4</option>";
}
echo "</select>";
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