Question

我有这个下拉列表，显示了不同会议的名称。我可以选择某个会议，但是在选择会议时，我想添加一个提交按钮，以便获取所选会议的变量。

我是数据库的新手，但是我尝试添加一个表单，但是我似乎无法使其在php代码中正常工作。数据库连接并显示所有会议都很好，我只是想不出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>

我希望变量等于选择的相同对象。所以我希望我的变量等于echo "<option>$row[name]</option>"; enter image description here

Answer 1

<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变量中获取值：

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GET，但POST也可以正常工作。您只需要在ajax调用中将type: "get"更改为type: "post"，在PHP中将$_GET[]更改为$_POST[]。

下拉搜索列表无法使变量等于所选选项

1 个答案: