Javascript 如何根据MySQL中的自动填充？
我没有PHP、HTML或JavaScript方面的经验，我请求帮助 我有显示一个select的php代码，这个select的字符串来自MySQL数据库和两个输入字段。 MySQL表格g_成就格式：Javascript 如何根据MySQL中的自动填充？,javascript,php,html,mysql,Javascript,Php,Html,Mysql,我没有PHP、HTML或JavaScript方面的经验，我请求帮助 我有显示一个select的php代码，这个select的字符串来自MySQL数据库和两个输入字段。 MySQL表格g_成就格式： id; achievement; craft_1_points; craft_2_points; 0; Medal Gold; 5; 8; 1; Medal Silver; 10;
id; achievement; craft_1_points; craft_2_points;
0; Medal Gold; 5; 8;
1; Medal Silver; 10; 15;
echo "<label for=\"achievement_id\">Achievement</label><br/>";
$sql2 = "SELECT * FROM g_achievement";
$result_select2 = mysql_query($sql2);
/*drop-down list*/
echo "<select name = 'achievement'>";
while($object2 = mysql_fetch_object($result_select2)){
echo "<option value = '$object2->achievement'>$object2->achievement</option>";
}
echo "</select>";
//============================
echo "<label for=\"craft_1\">Craft 1</label><br/>";
echo "<input type=\"number\" name=\"craft_1\" size=\"30\"><br/>";
echo "<label for=\"craft_2\">Craft 2</label><br/>";
echo "<input type=\"number\" name=\"craft_2\" size=\"30\"><br/>";
请帮帮我。谢谢。为此，您必须以两种方式使用Javascript 您可以从表中提取所有这些数据到客户端，然后通过Javascript从下拉列表中选择任何选项，您可以在输入框中填充相关值。 在从下拉列表中选择任何选项时，可以对服务器端进行Ajax调用，从服务器端获取数据，并使用Javascript填充客户端的数据。 下面是我上面描述的第一种方式的一个例子
<?php
// Following is a sample data. You can replace this data by getting it from your DB
$dataSource = array(
0 => array(
'achievement' => 'Medal Gold',
'craft_1_points' => 5,
'craft_2_points' => 8,
),
1 => array(
'achievement' => 'Medal Silver',
'craft_1_points' => 10,
'craft_2_points' => 15,
)
);
$dataSourceInJson = json_encode($dataSource);
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
<script src="https://code.jquery.com/jquery-2.2.4.min.js" integrity="sha256-BbhdlvQf/xTY9gja0Dq3HiwQF8LaCRTXxZKRutelT44=" crossorigin="anonymous"></script>
<script type="text/javascript">
$( document ).ready(function() {
var dataSource = '<?php echo $dataSourceInJson; ?>';
$( "#achievements" ).click(function() {
var achievementSelected = $(this).val();
$.each(JSON.parse(dataSource), function( index, value ) {
var achievementType = value.achievement;
if(achievementSelected == achievementType){
// Extract related Craft values from JSON object
var craft_1_points = value.craft_1_points;
var craft_2_points = value.craft_2_points;
// Place craft value in respective INPUT box
$('#craft_1_points').val(craft_1_points);
$('#craft_2_points').val(craft_2_points);
}
});
});
});
</script>
</head>
<body>
<select id='achievements' name="achievements">
<option value="">Select Achivements</option>
<?php
foreach ($dataSource as $key => $value)
{
echo "<option value='$value[achievement]'>$value[achievement]</option>";
}
?>
</select>
<br>Craft 1<br>
<input type="text" id="craft_1_points" name="craft_1_points" value="">
<br>Craft 2<br>
<input type="text" id="craft_2_points" name="craft_2_points" value="">
</body>
</html>
搜索php ajax级联选择你很好，非常感谢！只有我在Wordpress中使用了这段代码，我需要用jQuery替换$。很高兴知道这对您有帮助。
set dbConn=Server.CreateObject("ADODB.Connection")
dbConn.Open strProvider
rowCount=0
sqlTable = "select id, name from [sysobjects] where type in ('U','V','P') and category<>2 Order By Name "
set rsSchema=dbConn.execute(sqlTable)
%>
<select name="getTable" class="selectpicker form-control show-tick" data-live-search="true">
<%
do until rsSchema.EOF %>
<option value='<%=rsSchema(0)%>'><%= rsSchema(1)%>
<% rsSchema.MoveNext
loop
rsSchema.Close
set rsSchema = Nothing %></option>
</select>