我想從我的mysql數據庫中進行多個下拉菜單。我想要4下拉精確。這是我現在所擁有的：來自mysql的多個下拉菜單
<?php
mysql_connect('#', '#', '#');
mysql_select_db('test');
$sql = "SELECT wie, waar, metwie, voeruig FROM data";
$result = mysql_query($sql);
echo "<select name='test'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['wie'] . "'>" . $row['wie'] . "</option>";
}
echo "</select>";
?>
舉例來說，這樣的：
mysql_connect('#', '#', '#');
mysql_select_db('test');
$sql = "SELECT wie FROM data";
$result = mysql_query($sql);
echo "<select name='test1'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['wie'] . "'>" . $row['wie'] . "</option>";
}
echo "</select>";
$sql = "SELECT waar FROM data";
$result = mysql_query($sql);
echo "<select name='test1'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['waar'] . "'>" . $row['waar'] . "</option>";
}
echo "</select>";
$sql = "SELECT metwie FROM data";
$result = mysql_query($sql);
echo "<select name='test2'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['metwie'] . "'>" . $row['metwie'] . "</option>";
}
echo "</select>";
$sql = "SELECT voeruig FROM data";
$result = mysql_query($sql);
echo "<select name='test3'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['voeruig'] . "'>" . $row['voeruig'] . "</option>";
}
echo "</select>";
?>
請避免使用_deprecated_ MySQL的功能，還寫升技爲什麼/這個答案如何幫助。 – user5173426
正如HawasKaPujaari所述，避免使用mysql。使用mysqli。您可以使用條件switch語句是這樣的：
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "test");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
/* Select queries return a resultset */
if ($result = $mysqli->query("SELECT wie, waar, metwie, voeruig FROM data")) {
printf("Select returned %d rows.\n", $result->num_rows);
while ($row = mysql_fetch_array($result)) {
switch ($row) {
case "wie":
echo "<select name='wie'>";
echo "<option value='" . $row['wie'] . "'>" . $row['wie'] . "</option>";
echo "</select>";
break;
case "waar":
echo "<select name='waar'>";
echo "<option value='" . $row['waar'] . "'>" . $row['waar'] . "</option>";
echo "</select>";
break;
case "metwie":
echo "<select name='metwie'>";
echo "<option value='" . $row['metwie'] . "'>" . $row['metwie'] . "</option>";
echo "</select>";
break;
}
case "voeruig":
echo "<select name='voeruig'>";
echo "<option value='" . $row['voeruig'] . "'>" . $row['voeruig'] . "</option>";
echo "</select>";
break;
}
/* free result set */
$result->close();
}
?>
它顯示我一個空白頁 –
嘗試更換每個 情況下，「文」： 與 情況下$行[「文本」]： – dotKn0ck
PHP和MySQL是不同的軟件。你在這裏做的是連接php與mysql使用mysql_*()函數 在你的情況下你得到的是一個數組在php中。你可以將這個數組用於任何你想要的。如果你想打印陣列是爲調試使用：
echo "<pre>";
print_r($row);
echo "</pre>";
從此您將得到陣列結構 那麼CA使用不同的下拉列表中的數組元素
注：mysql_*()也不安全。使用mysqli_*或PDO。
嘗試下面的代碼。希望它使用_deprecated_ MySQL的功能，幫助您
<?php
$wie=array();
$waar =array();
$metwie =array();
$voeruig =array();
while ($row = mysql_fetch_array($result)){
$wie[]=$row["wie"];
$waar[]=$row["waar"];
$metwie[]=$row["metwie"];
$voeruig[]=$row["voeruig"];
}
?>
<select name="wie">
<?php
foreach($wie as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?>
<php \t
}
?>
</option>
<select name="waar">
<?php
foreach($waar as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?> </option>
<php \t
}
?>
</select>
<select name="metwie">
<?php
foreach($metwie as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?></option>
<php \t
}
?>
</select>
<select name="voeruig">
<?php
foreach($voeruig as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?> </option>
<php \t
}
?>
</select>
避免。 PS。 _我想做多個下拉菜單_你的意思是相關的保管箱嗎？ – user5173426
那到底是什麼問題？ –
只有一個有多個選項的下拉有 –