我的搜索结果没有显示在同一个窗口上,我希望结果显示在同一个窗口中。我找到了同样的问题,但代码与我正在使用的不同,所以我无法与之相关:Search wont show on same page
方案1:
如果我放入action="search_result2.php"
- 它会将结果重定向到另一页
方案2:
如果我在下面的代码中使用action=""
,它没有做任何事情
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script>
$(document).ready(function(){
$("#results").show();
});
</script>
<script type="text/javascript">
$(document).ready(function(){
$("#search").on('click',function() {
var find = $('#find').val();
var field = $('#field').val();
$.post('search_result2.php',{find:find, field:field}, function(data){
$("#results").html(data);
});
return false;
});
});
</script>
</head>
<body>
<div id="container" style="width:auto">
<div id="mainContent">
<h2>Search</h2>
<form name="search" method="post" action="">
Seach for: <input type="text" name="find" id="find" /> in
<Select NAME="field" id="field">
<Option VALUE="testA">A</option>
<Option VALUE="testB">B</option>
<Option VALUE="testC">C</option>
<Option VALUE="testD">D</option>
</Select>
<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" id="search" value="Search" />
</form>
<div id="results">
</div>
</div>
</div>
</body>
</html>
这是我的search_result2.php:
<?php
//This is only displayed if they have submitted the form
if (isset($_POST['searching']) && $_POST['searching'] == "yes")
{
echo "<h2>Results</h2><p>";
//If they did not enter a search term we give them an error
if (empty($_POST['find']))
{
echo "<p>You forgot to enter a search term";
exit;
}
// Otherwise we connect to our Database
mysql_connect("host", "username", "passw") or die(mysql_error());
mysql_select_db("testdb") or die(mysql_error());
// We preform a bit of filtering
$find = strtoupper($_POST['find']);
$find = strip_tags($_POST['find']);
$find = trim ($_POST['find']);
$field = trim ($_POST['field']);
//Now we search for our search term, in the field the user specified
$data = mysql_query("SELECT * FROM testtable WHERE upper($field) LIKE'%$find%'");
//And we display the results
while($result = mysql_fetch_array( $data ))
{
echo $result['testA'];
echo " ";
echo $result['testB'];
echo "<br>";
echo $result['testC'];
echo "<br>";
echo $result['testD'];
echo "<br>";
echo "<br>";
}
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($data);
if ($anymatches == 0)
{
echo "Sorry, but we can not find an entry to match your query<br><br>";
}
//And we remind them what they searched for
echo "<b>Searched For:</b> " .$find;
}
?>
答案 0 :(得分:1)
如果您想要在不刷新页面的情况下加载同一页面,则需要发出ajax请求。
如果您可以重新加载页面,则php部分必须与原始链接位于相同的“位置”。 例如,如果您将该代码放在带有表单的同一文件的顶部(并使用.php扩展名重命名),它应该可以工作(如果php可以在该文件夹中解释)。