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php - 下拉搜索列表无法将变量设置为等于所选选项

转载 作者:行者123 更新时间:2023-11-29 16:00:10 26 4
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我有一个下拉列表,显示不同 session 的名称。我可以选择某个 session ,但当我选择 session 时,我想添加一个提交按钮,以便我可以获得所选 session 的变量。

我是数据库新手,但我尝试添加表单,但我似乎无法让它在 php 代码中工作。数据库连接并显示所有 session 都很好,我只是不知道如何获取等于所选选项的变量。

<!DOCTYPE HTML>  
<html>
<head>
<style>
.error {color: #FF0000;}
</style>
</head>
<body>

<?php
//Declare variables
$db_host = "";
$db_username = "";
$db_pass = "";
$db_name = "";
$db_table = "";
//Connect to phpMyAdmin
$con=mysqli_connect("$db_host","$db_username","$db_pass","$db_name");

// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

mysqli_select_db($con,"$db_name") or die ("No database");

$result=mysqli_query($con,"select * From conferenceList");

echo "<select id='searchddl'>";
echo "<option> -- Search Conference Name -- </option>";
while($row=mysqli_fetch_array($result))
{
echo "<option>$row[name]</option>";
}
echo "</select>";


//Close phpMyAdmin
mysqli_close($con);
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/chosen/1.8.7/chosen.jquery.min.js"></script>
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/chosen/1.8.7/chosen.min.css" />

<script>
$("#searchddl").chosen();
</script>

</body>
</html>

我期望一个变量等于所选的相同内容。所以我希望我的变量等于 echo "<option>$row[name]</option>"; enter image description here

最佳答案

<script>
$( "#searchddl" ).chosen().change(function() { // execute query every time select option changes
var val = $( "#searchddl" ).chosen().val(); // get value from selected option
$.ajax({
url: "myphpfile.php", // call external PHP file for query execution
type: "get",
data: {
myselectvalue: val // passes value as GET parameter (?=myselectvalue=val)
},
success: function(response) {
tmp = response // store variable from PHP output
// do Something

},
error: function(xhr) {
// do Something else
}
});
</script>

在单独的 PHP 文件中,您可以从 get 变量获取值:

if (isset($_GET['myselectvalue']) && strlen($_GET['myselectvalue']) { // make sure it's populated

$valueFromForm = $_GET['myselectvalue'];

// Execute queries, echo results, etc.

}

我在此解决方案中使用了 GET,但 POST 也可以工作。您只需在 ajax 调用中将 type: "get" 更改为 type: "post" 并将 $_GET[] 更改为 PHP 中的$_POST[]

关于php - 下拉搜索列表无法将变量设置为等于所选选项,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56247415/

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