我正在进行显示图像缩略图的实时搜索。这很好用。问题是...当我单击缩略图时，它不会将数据从表单输入字段发送到 movie.php 页面。 movie.php 页面需要那些来加载适当的内容。如果有人能提供帮助，我们将不胜感激！
搜索页面
<script src="https://code.jquery.com/jquery-3.5.1.min.js"></script>
<script>
$(document).ready(function(){
$('.search-box input[type="text"]').on("keyup input", function(){
/* Get input value on change */
var inputVal = $(this).val();
var resultDropdown = $(this).siblings(".result");
if(inputVal.length){
$.get("search_backend.php", {term: inputVal}).done(function(data){
// Display the returned data in browser
resultDropdown.html(data);
});
} else{
resultDropdown.empty();
}
});
// Set search input value on click of result item
$(document).on("click", ".result", function(){
$(this).parents(".search-box").find('input[type="text"]').val($(this).text());
$(this).parent(".result").empty();
});
});
</script>
</head>
<body>
<div class="search-box">
<input type="text" autocomplete="off" placeholder="Search for a movie" />
<div class="result"></div>
</div>
搜索后端
<?php include "../includes/db_connector.php";
if(isset($_REQUEST["term"])){
$search_keyword = $_REQUEST["term"];
// Prepare a select statement
$sql = "SELECT title, img FROM jeffbox WHERE title LIKE '%" . $search_keyword . "%'";
if($stmt = mysqli_prepare($db_connect, $sql)){
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "s", $param_term);
// Set parameters
$param_term = $search_keyword . '%';
// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
$result = mysqli_stmt_get_result($stmt);
// Check number of rows in the result set
if(mysqli_num_rows($result) > 0){
// Fetch result rows as an associative array
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
$img = $row["img"];
$title = $row["title"];
?>
<form id="form-search" method="post" action="movie.php">
<input id="btn_top" style="display: none;" type="text" name="title" id="title" value="<?php echo $title ?>" />
<input id="btn_top" style="display: none;" type="text" name="latest" id="latest" value="<?php echo $title ?>" />
<input id="new_releases" type="image" name="image" src="images/<?php echo $img ?>">
</form>
<?php
} } else { echo "<p>No movies found</p>"; }
}
}
// Close statement
mysqli_stmt_close($stmt);
}
// close connection
mysqli_close($db_connect);
?>
我试过多次搜索都无济于事...
通常你会使用提交按钮来触发表单提交，比如
<input type=submit value="Submit">
但是，如果你想让图片触发表单提交，其中一种方法是使用下面的javascript：
<script>
function submitForm(originator)
{
var pN=originator.parentNode;
while (true)
{
if (pN&&pN.nodeName=='FORM')
{
pN.submit();
break;
}
pN=pN.parentNode;
}
}
</script>
然后换行：
<input id="new_releases" type="image" name="image" src="images/<?php echo $img ?>">
到
<img id="new_releases" src="images/<?php echo $img ?>" onclick="javascript:submitForm(this);">
另一方面，您提到当您单击图像时，表单中的所有数据字段都被“清除”，只有当您的脚本中某处清空表单数据时才会发生，请修复它，否则目标表单无法从提交的表单中接收数据。