以模式显示搜索栏查询结果
[英]Display search bar query results in modal
问题描述
我有一个搜索栏，供用户输入查询。 单击“搜索”后，应显示查询结果的模态。
我的index.php输出仍然没有显示在模式中。 当我单击“搜索”时，模态弹出并带有一个空的主体。 如何从index.php中获取输出以显示在模式主体中？
我的脚本有问题吗？ 我需要在模态车身上添加一些东西吗？
的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'); $('#mymodal:visible .modal-content .modal-body').html(e); } }); </script> <!-- The Modal --> <div class="modal" id="mymodal"> <div class="modal-dialog"> <div class="modal-content"> <!-- Modal Header --> <div class="modal-header"> <h4 class="modal-title">Modal Heading</h4> <button type="button" class="close" data-dismiss="modal">×</button> </div> <!-- Modal body --> <div class="modal-body"> </div> <!-- Modal footer --> <div class="modal-footer"> <button type="button" class="btn btn-danger" data-dismiss="modal">Close</button> </div> </div> </div> </div>
search.php中
<?php include_once('db.php'); //Connect to database if(isset($_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); ?>
1 个解决方案
解决方案1
2 2018-11-28 06:56:20
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);
?>
以模式显示来自搜索栏的查询
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