我有3个页面,一个html,javascript和php文件。当在html页面中按下一个按钮时,php页面在数据库中搜索按钮的名称为键的行,并带回多行。我想在html页面中显示所有这些行,但是当我发回数据时,它不会显示。这是我的三页代码。
Html页面
<button name = 'science'> Science</button>
<button name = 'engineering'> Engineering</button>
<button name = 'math'> Math</button>
<button name = 'law'> Law</button>
<button name = 'arts'> Arts</button>
<button name = 'business'> Business</button>
<button name = 'fiction'> Fiction</button>
<button name = 'selfhelp'> Self Help</button>
<button name = 'labequipment'> Lab Equipment</button>
<p id = 'feedback'></p>
<script type ="text/javascript" src ="jquery.js"></script>
<script type ="text/javascript" src ="catselect.js"></script>
javascript网页
$(":button").click(function(){
var cat = $(this).attr("name");
$.get("trier.php",{input: cat}, function(data){
$("#feedback").html(data);
});
});
php页面
if(isset($_GET['input'])){
$catt = $_GET['input'];
$info = "SELECT * FROM books WHERE (Category = '$catt')";
while(mysqli_query($conn,$info)){
$information = mysqli_query($conn,$info);
$intel = mysqli_fetch_assoc($information);
echo $intel['Title'];
echo nl2br("\n");
echo $intel['Course'];
}
}
我想显示很多行,包括&#39;标题&#39;专栏和课程&#39;在php页面的标签中的columnn,id =&#34; display&#34;。关于如何修改我的代码来做任何想法?
答案 0 :(得分:1)
$(":button")
应为$("button")
你可以使用表格的php echo html,但我建议使用json或使用angularjs
ID不应经常更改,id = "display" better use class="display"
而不是
答案 1 :(得分:0)
你可以像@Yanjun Lin建议使用html表那样简单 在你的PHP上: 如果(isset($ _ GET [ '输入'])){ $ catt = $ _GET ['input'];
$info = "SELECT * FROM books WHERE (Category = '$catt')";
$html='';
while(mysqli_query($conn,$info)){
$information = mysqli_query($conn,$info);
$intel = mysqli_fetch_assoc($information);
$html=$html.'<tr>';
//echo $intel['Title'];
//echo nl2br("\n");
//echo $intel['Course'];
$html=$html.'<td>'.$intel['Title'].'</td>';
$html=$html.'<td>'.$intel['Course'].'</td>';
$html=$html.'</tr>';
}
echo $html;
在你的javascript上:
$("button").click(function(){
var cat = $(this).attr("name");
$.get("trier.php",{input: cat}, function(data){
$("#feedback").html(data);
});
在您的HTML上(将<p>
标记更改为<table>
代码):
<button name = 'science'> Science</button>
<button name = 'engineering'> Engineering</button>
<button name = 'math'> Math</button>
<button name = 'law'> Law</button>
<button name = 'arts'> Arts</button>
<button name = 'business'> Business</button>
<button name = 'fiction'> Fiction</button>
<button name = 'selfhelp'> Self Help</button>
<button name = 'labequipment'> Lab Equipment</button>
<table id = 'feedback'></table>
<script type ="text/javascript" src ="jquery.js"></script>
<script type ="text/javascript" src ="catselect.js"></script>
我没有测试你的php,但希望它有效......还JSFiddle它只会在javascript和html中看起来像这样。也可以考虑使用json作为正确的标准,如@Yanjun Lin建议