Php 我有一个显示用户名的下拉列表。如何使用Ajax显示他们的信息?
我的下拉列表显示我从数据库获得的用户名。我想实现一个Php 我有一个显示用户名的下拉列表。如何使用Ajax显示他们的信息?,php,jquery,mysql,ajax,Php,Jquery,Mysql,Ajax,我的下拉列表显示我从数据库获得的用户名。我想实现一个displayInfo功能,这样当有人选择一个用户时,它会自动在下面显示他/她的信息 当某人选择其姓名时,我如何显示用户的信息 这是我的下拉代码: <?php //connect $conn = mysqli_connect("localhost","user","123abc"); mysqli_select_db($conn, "use
displayInfo
功能,这样当有人选择一个用户时,它会自动在下面显示他/她的信息
当某人选择其姓名时,我如何显示用户的信息
这是我的下拉代码:
<?php
//connect
$conn = mysqli_connect("localhost","user","123abc");
mysqli_select_db($conn, "users");
//query
$sql= mysqli_query($conn, "SELECT person_id,first_name FROM users");
echo "<select name='dropdown' onchange='displayInfo' id='dropdown'>";
while ($row = mysqli_fetch_array($sql))
{
//display friends' first names on dropdown
if($row['person_id'] == $row['first_name']) {
echo "<option value='" . $row['person_id'] . "' selected>" . $row['list_name'] . "</option>";
} else {
echo "<option value='" . $row['person_id'] . "'>" . $row['first_name'] . "</option>";
}
}
echo "</select>";
在选项select上调用Javascript函数的最佳方法如下
<?php
//connect
$conn = mysqli_connect("localhost","user","123abc");
mysqli_select_db($conn, "users");
//query
$sql= mysqli_query($conn, "SELECT person_id,first_name FROM users");
echo "<select name='dropdown' onchange='displayInfo' id='dropdown'>";
while ($row = mysqli_fetch_array($sql))
{
//display friends' first names on dropdown
if($row['person_id'] == $row['first_name']) {
**echo "<option onchange='myFunction(this);' value='" . $row['person_id'] . "' selected>" . $row['list_name'] . "</option>";**
} else {
echo "<option onchange='myFunction(this);' value='" . $row['person_id'] . "'>" . $row['first_name'] . "</option>";
}
}
echo "</select>";
<script>
function myFunction(control){
$(control).val() //to access value or any other functionality
}
</script>
描述重写,代码formatted@yan777dro请接受ans,这将很有帮助:)