我是 PHP HTML 的新手，在此站点上找到了一些帮助，但无法使代码作为我的脚本的一部分工作
单击“提交”按钮后，一条记录将发布到数据库表(交易)，但字段(代码)为空，它应该包含一个 3 个字母的大写字符串，例如美国广播公司
ALERT 显示该值存储在变量 $code 中，但将一个空字符串发布到表中
新手非常感谢您的帮助
<?php
// Include config file
require_once 'config.php';
// Define variables and initialize with empty values
$code = "";
$code_err = "";
// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST")
{
// Validate Code
$code = strtoupper($code);
// Check input errors before inserting in database
if(empty($code_err)){
// Prepare an insert statement
$sql = "INSERT INTO Transactions (Code)
VALUES (?)";
if($stmt = mysqli_prepare($link, $sql)){
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "s", $param_code);
// Set parameters
$param_code = $code;
// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
// Records created successfully. Redirect to landing page
$url = 'http://localhost:8888/portfolio/index.php';
echo '<META HTTP-EQUIV=Refresh CONTENT="0; URL='.$url.'">';
exit();
} else{
echo "Something went wrong. Please try again later.";
}
}
// Close statement
mysqli_stmt_close($stmt);
}
}
// Close connection
mysqli_close($link);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Create Record</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.css">
<link href="style1.css" rel="stylesheet" type="text/css"/>
<style type="text/css">
.wrapper{
width: 450px;
margin: 0 auto;
}
</style>
<script>
function getValue(obj){
alert(obj.value);
$code=(obj.value);
**alert($code);**
}
</script>
<!--Display heading at top center of screen-->
<div>
<center><h3>Peter's Portfolio - Shares</h3></center>
</div> <!-- end of Div -->
</head>
<body style="background-color:#fcf8d9">
<div class="wrapper">
<div class="container-fluid">
<div class="row">
<div class="col-md-6">
<div class="page-header">
<h2>Create Transaction Record</h2>
</div>
<form class="form-horizontal" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<!-- ASX CODE -->
<div class="form-group">
<label for="name" class="control-label col-xs-6">ASX Code:</label>
<div class="col-xs-6">
<?php
$conn = new mysqli('localhost', 'root', 'root', 'Portfolio') or die ('Cannot connect to db');
$result = $conn->query("SELECT Code, Coy_Nm from Companies ORDER BY Code");
echo "<html>";
echo "<body>";
echo "<select Name='Code' ID='Code'onchange='getValue(this)'>";
while ($row = $result->fetch_assoc()) {
unset($Code, $Coy_Nm);
$Code = $row['Code'];
$Coy_Nm = $row['Coy_Nm'];
echo '<option value="'.$Code.'">'.$Coy_Nm.'</option>';
}
echo "</select>";
echo "<input type='hidden' value='submit'>";
//$code=$_POST['Code'];
//echo $code;
?>
</div>
</div>
<!--Submit and Cancel Buttons-->
<input type="submit" class="btn btn-primary" value="Submit">
<a href="index.php" class="btn btn-default">Cancel</a>
</form>
</div>
</div>
</div>
</div>
</body>
</html>
最佳答案
Javascript 和 PHP 不能简单地相互实时通信。
试试这个
function getValue(obj){
alert(obj);// check if obj has something in it before proceeding to "value"
alert(obj.value);
}
关于javascript - 从选择中获取选定值并发布到 MySQL 表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/48416513/