<?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>';