[英]Select values from SQL db and display in dropdown, submit form
來自db的值顯示在下拉列表中,我已經將其與未從db中拉出的文本值一起使用,但是當我嘗試使用來自db的值並單擊“提交”時,頁面將重新加載並且什么也沒有發生。
我的代碼:
<form action="test.php" method="post">
<br />
<label for="name">Choose a name</label><br />
<select name="selected_value">
$query = "SELECT * FROM items WHERE user_id = $user_id";
$select_items = mysqli_query($connection, $query);
confirmQuery($select_items);
while($row = mysqli_fetch_assoc($select_items)) {
$item_id = $row['item_id'];
$item_name = $row['item_name'];
echo "<option value='$item_name'>{$item_name}</option>";
</select>
<button type="submit" name="submit_form">Submit</button>
</form>
if(isset($_POST['submit_form'])){
$selected_value = $_POST['selected_value'];
echo $selected_value;
}
您是否有錯誤的解決方法?
$ _POST ['selected_value'] = $ selected_value;
應該
$ selected_value = $ _POST ['selected_value'];
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