UPDATE查詢不能正常運行php
[英]UPDATE query is not working php
問題描述
要盡量保持簡短。 我在grid.php文件中有一個while循環來填充表格......
<?php while($product = $products->fetch_assoc()) { ?>
<tr>
<td><?php echo $product['cd_id']?></td>
<td><?php echo $product['cd_title']?></td>
<td><?php echo $product['cd_musician_fname']?></td>
<td><?php echo $product['cd_musician_lname']?></td>
<td><?php echo $product['cd_price']?></td>
<td><a href="edit.php?id=<?echo $product['cd_id'];?>" type="button" class="btn btn-primary">Edit</a></td>
<td><a href="delete.php?id=<?echo $product['cd_id'];?>" type="button" class="btn btn-danger">Delete</a></td>
</tr>
<?php } ?>
如果我單擊第一個錨標記將我帶到edit.php文件，這里是該文件的頭代碼。
<?php include '_includes/db.php';
$cd_id = trim($_GET['id']);
$message = '';
include '_includes/connection.php';
if($db->connect_error){
$message = $db->connect_error;
}else{
$sql = "SELECT * FROM CD WHERE cd_id = $cd_id";
$result = $db->query($sql);
$row = $result->fetch_assoc();
if($db->error){
$message = $db->error;
}
}
?>
現在我將展示edit.php的html
<!-- Product Musician last name-->
<fieldset class="form-group">
<label for="cd_musician_lname">Musician's lirst name</label>
<input type="text" class="form-control" id="cd_musician_lname" name="cd_musician_lname" value="<?php echo $row['cd_musician_lname'];?>">
</fieldset>
<!-- End of Musician last name-->
<!-- Product price-->
<fieldset class="form-group">
<label for="cd_price">Product price</label>
<input type="text" class="form-control" id="cd_price" name="cd_price" value="<?php echo $row['cd_price'];?>">
</fieldset>
<!-- End of Product price-->
<!-- Form submit button-->
<a href="edit_confirm.php?id=<?echo $row['cd_id'];?>" type="submit" class="btn btn-primary">Update Record</a>
<a class="btn btn-primary" href="index.php" role="button">Go Back Home</a>
我有edit.php頁面工作正常，但如果我在字段中進行更改並單擊提交錨標記，我會將該行的所有字段都清空但PK。 這是最終的edit_confirm.php文件的代碼
<?php
include '_includes/db.php';
$cd_id = trim($_GET['id']);
$cd_title = $_POST['cd_title'];
$cd_musician_fname = $_POST['cd_musician_fname'];
$cd_musician_lname = $_POST['cd_musician_lname'];
$cd_price = $_POST['cd_price'];
$message = '';
include '_includes/connection.php';
if($db->connect_error){
die("Connection failed: ".$db->connect_error);
} else {
$sql = "UPDATE CD SET cd_title='".$cd_title."', cd_musician_fname='".
$cd_musician_fname."', cd_musician_lname='".
$cd_musician_lname."', cd_price='".$cd_price."' WHERE cd_id = $cd_id ";
$db->query($sql);
var_dump($sql);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<?php include '_includes/main-head.php';?>
</head>
<body>
<?php include '_includes/main-navbar.php';?>
<div class="container">
<hr>
<?php
if($db->query($sql) === TRUE){ ?>
<h1>Record updated successfully.</h1>
<?php echo $cd_title; ?>
<?php echo $record->affected_rows ?>
<p> record was updated in the database.</p></br>
<?php } else { ?>
<p>Error updating the record: </p> <?php $db->error; ?>
<?php }; ?>
<hr>
<a class="btn btn-primary" href="index.php" role="button">Go Back Home</a>
</div>
<?php include '_includes/main-script.php';?>
</body>
</html>
如果你在edit_confirm.php中注意到我做了一個var_dump來查看變量中的值是什么，它顯示為空。
我需要幫助。 先感謝您。
2 個解決方案
解決方案1
2 已采納 2015-11-01 23:43:58
人們更好的方法是讓測試記錄是否正在更新變得簡單
formsample.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
include("connection.php");
$id = $_GET['id'];
$query= "select * from clients where id = '$id'";
$sql = mysqli_query($connect, $query);
$result = mysqli_fetch_assoc($sql);
?>
<form action="process.php" method="post">
<input type="text" name="name" id="name" value="<?php echo $result['name'] ?>" />
<input type="text" name="email" id="email" value="<?php echo $result['email'] ?>" />
<input type="hidden" name="id" id="id" value="<?php echo $id?>" />
<input type="submit" />
</form>
</body>
</html>
process.php
<?php
include("connection.php");
$id = $_POST['id'];
$name = $_POST['name'];
$email= $_POST['email'];
$query = "UPDATE clients set name= '$name', email ='$email' where id = '$id'";
$sql = mysqli_query($connect, $query);
?>
解決方案2
1 2015-11-01 23:47:21
<a href="edit_confirm.php?id=<?echo $row['cd_id'];?>" type="submit" class="btn btn-primary">Update Record</a>
這不是提交表格的正確方式 - 它根本不起作用。
你需要有一個表單開始和結束標記，目標地址在表單元素的action屬性中，方法也在那里，應該發布此表單（ method="POST" ）。 在您的代碼中，您有一個鏈接，您應該有一個提交輸入，因此它不會提交數據，它只會將您重定向到該URL。 你應該有這樣的東西：
<input type="submit" value="Update Record" />
PHP更新查詢不起作用
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