我有一个搜索栏供用户输入查询。单击“搜索”后，应该会出现一个带有查询结果的模式。
我的 index.php 输出仍然没有显示在模式中。当我单击“搜索”时，模态弹出一个空体。 如何让 index.php 的输出显示在模态框的正文中？
我的脚本有问题吗？我需要向 modal-body 添加一些东西吗？
index.php
<head>
<title>Search</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="stylesheet" type="text/css" href="style.css"/>
</head>
<body>
<form method="POST" action="#">
<input type="text" name="q" placeholder="Enter query"/>
<input type="button" name="search" value="Search" data-toggle="modal" data-target="#mymodal">
</form>
</body>
<script>
$.ajax({ type: "GET",
url: 'search.php',
success: function(data){ debugger $('#mymodal').modal('show');
$('#mymodal:visible .modal-content .modal-body').html(e); } });
</script>
<!-- The Modal -->
<div class="modal" id="mymodal">
<div class="modal-dialog">
<div class="modal-content">
<!-- Modal Header -->
<div class="modal-header">
<h4 class="modal-title">Modal Heading</h4>
<button type="button" class="close" data-dismiss="modal">×</button>
</div>
<!-- Modal body -->
<div class="modal-body">
</div>
<!-- Modal footer -->
<div class="modal-footer">
<button type="button" class="btn btn-danger" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>search.php
<?php
include_once('db.php'); //Connect to database
if(isset($_REQUEST['q'])){
$q = $_REQUEST['q'];
//get required columns
$query = mysqli_query($conn, "SELECT * FROM `words` WHERE `englishWord` LIKE '%".$q."%' OR `yupikWord` LIKE '%".$q."%') or die(mysqli_error($conn)); //check for query error
$count = mysqli_num_rows($query);
if($count == 0){
$output = '<h2>No result found</h2>';
}else{
while($row = mysqli_fetch_assoc($query)){
$output .= '<h2>'.$row['yupikWord'].'</h2><br>';
$output .= '<h2>'.$row['englishWord'].'</h2><br>';
$output .= '<h2>'.$row['audio'].'</h2><br>';
$audio_name = $row['audio'];
$output .= '<td><audio src="audio/'.$audio_name.'" controls="control">'.$audio_name.'</audio></td>';
}
}
echo $output;
}else{
"Please add search parameter";
}
mysqli_close($conn);
?>最佳答案
使用search.php的这些代码
<?php
include_once('db.php'); //Connect to database
if(isset($_REQUEST['q']))
{
$q = $_REQUEST['q'];
$query = mysqli_query($conn, "SELECT * FROM `words` WHERE `englishWord` LIKE '%".$q."%' OR `yupikWord` LIKE '%".$q."%'") or die(mysqli_error($conn));
$count = mysqli_num_rows($query);
if($count == 0){
$output = '<h2>No result found</h2>';
}else{
while($row = mysqli_fetch_assoc($query)){
$output .= '<h2>'.$row['yupikWord'].'</h2><br>';
$output .= '<h2>'.$row['englishWord'].'</h2><br>';
$output .= '<h2>'.$row['audio'].'</h2><br>';
$audio_name = $row['audio'];
$output .= '<td><audio src="audio/'.$audio_name.'" controls="control">'.$audio_name.'</audio></td>';
}
}
echo $output;
}else{
"Please add search parameter";
}
mysqli_close($conn);
?>
关于javascript - 以模式显示搜索栏查询结果，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53511400/