php - mysql 的多个下拉菜单

标签 php mysql

我想从我的 mysql 数据库中的数据中创建多个下拉列表。准确地说,我想要 4 个下拉菜单。这就是我现在所拥有的:

<?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>";
?>

最佳答案

例如,这个:

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>";
?>

关于php - mysql 的多个下拉菜单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32800127/

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