PHP 形式與 while 循環
[英]PHP form with while loop
問題描述
我在生成可以打印基於 SQL 查詢的表單的頁面時遇到了一些麻煩。 在這種情況下，我需要執行查詢以列出名稱為 Porcac1x 的表的所有行。 內容需要是顯示當前變量值並且可以更新的輸入字段。 這正是我卡住的地方。 如何創建基於 php while 循環的可變表單？ 使用附加的代碼，我可以列出內容並顯示所有變量，但是在創建表單操作以更新值時遇到了麻煩。 我想明確一點，我不關心安全性，因為代碼將在本地環境中運行，我是唯一可以訪問的人。
這是當前 output 但是保存按鈕當然不起作用
<html> <head> <meta charset="utf-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"> <meta name="description" content="$1"> <meta name="viewport" content="width=device-width, initial-scale=1"> <link rel="stylesheet" type="text/css" href="style.css"> <title>test page</title> </head> <body><form action="" method="post"> <?php $servername = "localhost"; $username = "root"; $password = "root"; $dbname = "root"; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: ". $conn->connect_error); } $sql = "SELECT `id`, `title`, `amount` FROM `expenses` WHERE name='Porcac1x';"; $result = $conn->query($sql); if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "ID: <input type='text' name='".$row["id"]."' value='".$row["id"]."'> Title: <input type='text' name='".$row["title"]."' value='".$row["title"]."'> Amount: <input type='text' name='".$row["amount"]."' value='".$row["amount"]."'> <button type='submit' name='save'>Save</button><br>"; } } else { echo "0 results"; } if(isset($_POST['save'])){ $myID = $_POST["id"];//??? < Issue $myTitle = $_POST["title"];//??? < Issue $myAmount = $_POST["amount"]; //??? < Issue echo $myID; echo $myTitle; echo $myAmount; $sqlUpdate = "UPDATE expenses SET title='$myTitle', amount ='$myAmount' WHERE id='$myID';"; echo $sqlUpdate; if ($conn->multi_query($sqlUpdate) === TRUE) { echo "Record updated successfully"; $risposta= "Record updated successfully"; } else { echo "Error updating record: ". $conn->error; $risposta= "Error updating record: ". $conn->error; } $conn->close(); }?> </form> </body> </html>
1 個解決方案
解決方案1
1 已采納 2020-06-21 16:56:59
就像我在評論中所說的那樣，你需要數組表示法。 （或每行 1 個表格）
這是每行 1 個表單元素的解決方案：
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="description" content="$1">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" type="text/css" href="style.css">
<title>test page</title>
</head>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "root";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT `id`, `title`, `amount` FROM `expenses` WHERE name='Porcac1x';";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<form method="post">';
echo "ID: <input type='text' name='id' value='".$row["id"]."'> Title: <input type='text' name='title' value='".$row["title"]."'> Amount: <input type='text' name='amount' value='".$row["amount"]."'> <button type='submit' name='save'>Save</button><br>";
echo '</form>';
}
} else {
echo "0 results";
}
if(isset($_POST['save'])){
$myID = $_POST["id"];//??? < Issue
$myTitle = $_POST["title"];//??? < Issue
$myAmount = $_POST["amount"]; //??? < Issue
echo $myID;
echo $myTitle;
echo $myAmount;
$sqlUpdate = "UPDATE expenses SET title='$myTitle', amount ='$myAmount' WHERE id='$myID';";
echo $sqlUpdate;
if ($conn->multi_query($sqlUpdate) === TRUE) {
echo "Record updated successfully";
$risposta= "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
$risposta= "Error updating record: " . $conn->error;
}
$conn->close();
}
?>
</body>
</html>
PHP中的While循環形式
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