我正在嘗試構建一個賓果游戲，我想在選擇（下拉）中顯示我以前的記錄！如何在選擇下拉串列中從資料庫中獲取我的資料---下面是我的 php 檔案
<html>
<style>
#rec_mode{
background-image: url('register.png');
background-size: 100% 100%;
width: 100px;
height: 50px;
border: none;
outline: 0px;
-webkit-appearance: none;
}
</style>
<body>
<select name = "select_history" id="rec_mode">
<option selected="true" disabled="disabled">
<?php
require_once 'config.php';
// $hist = mysqli_query($mysqli, "SELECT name FROM `movie_names` ORDER BY movieID DESC");
$hist = mysqli_query($mysqli,"SELECT m.name FROM movie_names m INNER JOIN host_table ht WHERE m.movieID = ht.random_num ORDER BY ID DESC");
while ($row = $hist->fetch_assoc())
{
echo "<option value=\"select_history\">".$row['name']."</option>";
}
?>
</option>
</select>
</body>
</html>
uj5u.com熱心網友回復：
為什么在 php 代碼之后添加選項結束標記。不應該是這樣的嗎。
<select name = "select_history" id="rec_mode">
<option selected="true" disabled="disabled">Select Option</option>
<?php
require_once 'config.php';
$hist = mysqli_query($mysqli,"SELECT m.name FROM movie_names m INNER JOIN host_table ht WHERE m.movieID = ht.random_num ORDER BY ID DESC");
while ($row = $hist->fetch_assoc()){
echo "<option value=\"select_history\">".$row['name']."</option>";
}
?>
</select>
uj5u.com熱心網友回復：
首先從資料庫中選擇舊的選定名稱，例如“哈利波特”，然后使用 if 條件在 while 回圈中選擇舊名稱
$hist = mysqli_query($mysqli,"SELECT m.name FROM movie_names m INNER JOIN
host_table ht WHERE m.movieID = ht.random_num ORDER BY ID DESC");
$Bingo = "harry potter"; // previously selected value
while ($row = $hist->fetch_assoc())
{
$selected = "";
if($row["name"]==$Bingo) $selected = "selected"; // If you given Id use colunm name id
echo "<option value=\"select_history\" $selected >".$row['name']."</option>";
}
謝謝
uj5u.com熱心網友回復：
首先，您必須正確關閉第一個選項標簽。此外，您可能希望選項值是 db 中的 id 列。請使用表格中正確的列名。但它應該是這樣的：
<select name = "select_history" id="rec_mode">
<option selected="true" disabled="disabled"> -- SELECT -- </option> <!-- close it here -->
<?php
require_once 'config.php';
$hist = mysqli_query($mysqli,"SELECT m.movieID, m.name FROM movie_names m INNER JOIN host_table ht WHERE m.movieID = ht.random_num ORDER BY movieID DESC"); // make sure query is correct, test it in your mysql client
while ($row = $hist->fetch_assoc())
{
echo "<option value='{$row['movieID']}'>{$row['name']}</option>";
}
?>
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標籤：javascript php html jQuery 阿贾克斯