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我想制作一个食谱书籍,用户可以制作新食谱,因此他可以从下拉字段中的预先输入的成分中进行选择。使用php下拉字段填充带有SQL数据的html表单
我的代码看起来像这样至今:
<html>
<body>
<form method="post" >
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "tagebuch1";
// LOGIN TO DATABASE SCRIPT WRITTEN FOR MYSQLI
$mysqli = new mysqli($servername, $username, $password, $dbname);
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
// END OF LOGIN TO DB SCRIPT
$query_ak='SELECT * FROM privnahrungsmittel';
$result = $mysqli->query($query_ak);
?>
<select name="exa" >
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo '<option value="'.$row['idprivNahrungsmittel'].'">'.$row['brennwert'].'</option>';
}
?>
</select>
<input type="submit" value="GO!" name="go"/>
</form>
</body>
</html>
但我只得到这样一个空的输出,虽然表中填充内容:
任何人都可以请指教错误在哪里?
谢谢
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