PHP Ajax drop down box

本文关键字：box down drop Ajax PHP | 更新日期: 2023-09-27

<?php
require ("init.php");
?>
<head>
<script src="ajax.js"></script>
<script src="common.js"></script>
</head>
<body>
<?php
$query = "SELECT * FROM User1";
$result = mysqli_query($connection, $query);
echo '<form> ';
echo "Select a Users:"; 
echo '<select name="users" onchange="showUser(this.value)">';
while ($row=mysqli_fetch_assoc($result)){
    echo $row=['Username'];
    echo '<option value="'.$row=['Username'].'">'.$row=['Username'].'</option>';
}
echo '</select></form>';
?>
<div id="txtHint"><b>User info will be listed here.</b></div>
</body>

我遇到的问题是下拉框仅多次显示数组，并且不显示我的数据库中的用户名，但是与我的数据库的连接正在工作，就像尝试调试时一样，它只回显出一个用户名

IT应该是这样的：

while ($row=mysqli_fetch_assoc($result)){
    //echo $row=['Username']; //Invalied here
    echo '<option value="'.$row['Username'].'">'.$row['Username'].'</option>';
}

不需要那个=签名。

正确：

echo $row=['Username'];// This is unnecessary.
echo '<option value="'.$row=['Username'].'">'.$row=['Username'].'</option>';// what is = doing here?

自：

echo '<option value="'.$row['Username'].'">'.$row['Username'].'</option>';