Question

目标：将名称（输入文本）和信息（文本区域包含多行）插入数据库，并在提交表单后，在同一页面上，两列用于显示相同列，名称和信息中的数据，但信息列下的数据。我已经为名称前面的每一行做了一个按钮，用作slideToggle，用于显示/隐藏其中包含从“信息”列中检索到的数据

问题-当我单击第一行的按钮时，它会滑动并显示所有与所有条目相关的所有信息，而不是仅显示与第一条目相关的信息

*其他-表单中已添加了一个输入，但已隐藏，用于id（自动递增）

----index.php-----

<?php  include('php_code.php'); ?>
<?php 
	if (isset($_GET['edit'])) {
		$id = $_GET['edit'];
		$update = true;
		$record = mysqli_query($db, "SELECT * FROM records WHERE id=$id");

		if (count($record) == 1 ) {
			$n = mysqli_fetch_array($record);
			$name = $n['name'];
			$acc = $n['acc_no'];
			$info = $n['info'];
		}
	}
?>
<html>
<head>
<title>JSK</title>
<link rel="stylesheet" href="style.css" type="text/css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
	$('form').hide();
	$('p').hide();
	$('#sp').hide();
    $("#inf").click(function(){
    	$("p").slideToggle();
    	$('#sp').slideToggle();
    });
$("#fo").click(function(){
        $("form").slideToggle();
 	    });
});
</script>
</head>
<body>
	<div class="container">
		<div class="left">
		<?php if (isset($_SESSION['message'])): ?>
		<div class="msg">
		<?php 
			echo $_SESSION['message']; 
			unset($_SESSION['message']);
		?>
		</div>
		<?php endif ?>

		<?php $results = mysqli_query($db, "SELECT * FROM records"); ?>
<table>
	<thead>
		<tr>
			<th>Name</th>
			<th>Account No</th>
			<th>Info</th>
			<th colspan="2">Action</th>
		</tr>
	</thead>
	<?php while ($row = mysqli_fetch_array($results)) { ?>
		<tr>
			<td><?php echo $row['name']; ?></td>
			<td><?php echo $row['acc_no']; ?></td>
			<td><button id="inf" onclick="myFunction()">show</button></td>
			<td>
				<a href="index.php?edit=<?php echo $row['id']; ?>" class="edit_btn" >Edit</a>
			</td>
			<td>
				<a href="php_code.php?del=<?php echo $row['id']; ?>" class="del_btn">Delete</a>
			</td>
		</tr>
		<tr  id="sp"> <td colspan="4"><p> <?php echo $row['info']; ?> </p></td>
		</tr>
		<?php } ?> 
</table>
<div id="fotable" align="center">
<button id="fo">Add New/Edit</button> 
</div>

	<form method="post" action="php_code.php" >
		<input type="hidden" name="id" value="<?php echo $id; ?>">

		<div class="input-group">
			<label>Name</label>
			<input type="text" autocomplete="off" name="name" value="<?php echo $name; ?>">
		</div>
		<div class="input-group">
			<label>Account No</label>
			<input type="text" name="acc" value="<?php echo $acc; ?>">
		</div>
		<div class="input-group">
      <label for="info">Info</label>

      <textarea class="form-control" rows="8" name="info" id="info"><?php echo $row['info']; ?></textarea>

    </div>
		<div class="input-group">
			<?php if ($update == true): ?>
	<button class="btn" type="submit" name="update" style="background: #556B2F;" >update</button>
<?php else: ?>
	<button class="btn" type="submit" name="save" >Add Account</button>
<?php endif ?>
		</div>
	</form>
</div><!-- left closed -->
<div class="right">

hello



</div>	<!-- right closed -->
</div> <!-- container closed -->
</body>
</html>

---php_code.php-----

<?php 
	session_start();
	$db = mysqli_connect('localhost', 'root', '', 'jskrecords');
	

	// initialize variables
	$name = "";
	$acc = "";
	$info = "";
	$id = 0;
	$update = false;

	if (isset($_POST['save'])) {
		$name = $_POST['name'];
		$acc = $_POST['acc'];
		$info = $_POST['info'];

		mysqli_query($db, "INSERT INTO records (name, acc_no, info) VALUES ('$name', '$acc', '$info')"); 
		$_SESSION['message'] = "Account saved"; 
		header('location: index.php');
	}
	if (isset($_POST['update'])) {
	$id = $_POST['id'];
	$name = $_POST['name'];
	$acc = $_POST['acc'];
	$info = $_POST['info'];

	mysqli_query($db, "UPDATE records SET name='$name',acc_no='$acc',info='$info' WHERE id=$id");
	$_SESSION['message'] = "Account updated!"; 
	header('location: index.php');
}
if (isset($_GET['del'])) {
	$id = $_GET['del'];
	mysqli_query($db, "DELETE FROM records WHERE id=$id");
	$_SESSION['message'] = "ACC deleted!"; 
	header('location: index.php');
}

?>

Answer 1

概念：

如果要使用php脚本从同一mysql表创建多个表单，则需要为每个表单赋予唯一的ID。例如

<form method="post" action="php_code.php" id="form<?= $id ?>">

然后将data-target =“＃form”添加到类为'inf'的按钮。它将存储表单的ID。

<button class="inf" data-target="#form<?= $id ?>">show</button>

当单击按钮时，我们知道要从数据目标中打开哪个表单。

<script>
$('.container').on('click','button.inf',function(e){
 e.preventDefault();
 var formid=$(this).attr('data-target');  //get value of data-target attribute 
       //...proceed to play toggle with this form 'formid'

在按钮上使用slideToggle显示/隐藏信息

1 个答案: