这是我在 Stack Overflow 上的第一篇文章.我知道这个问题之前已经被问过很多次了.我经历了很多答案,尝试了所有答案(显然正确的方法除外),但不知道该尝试什么了.
this is my first post here at Stack Overflow. I know the question has been asked many times before. I went through many answers, tried all of them (except the correct approach obviously) and don't know what to try anymore.
我有一个 SQL 表,其中每一行都有一个编辑"按钮.单击它时,我将所选行的 id 传递给 edit.php
.在那里,我得到它并根据用户从表单输入的 id 更新给定的行.第一列是 id,设置为 AUTO_INCREMENT
.
I have an SQL table where every row has an "edit" button. When clicking it, I pass over the id of the selected row to edit.php
. There, I get it and update the given row based on the id with the user input from the form. The first column is id which is set to AUTO_INCREMENT
.
顺便说一句,无论我使用 WHERE id=$id";
还是 WHERE id='$id'";
On a side note, I get the same error, no matter if I use WHERE id=$id";
or WHERE id='$id'";
我认为最接近正确方法的代码如下,并在代码下方生成错误消息:
The code which I think is closest to the correct approach is as follows and generates the error message below the code:
<html>
<title>
Video Archiv - New
</title>
<body>
<?php
include("connect.php");
$id=$_GET['id'];
echo "Details von Video #$id editieren:<br /><br />";
if(isset($_POST['update']))
{
$sql = "UPDATE VideoArchiv
SET ('".$_POST["titel"]."','".$_POST["schauspieler"]."')
WHERE id=$id";
$result = mysqli_query($connect,$sql);
if (mysqli_query($connect,$sql) === TRUE)
{
echo "Record updated successfully";
}
else
{
echo "Error updating record: " . $connect->error;
}
}
?>
<form action="edit.php" method="post">
<label> Titel:</label><br/>
<input type="text" name="titel" required><br/>
<label>Schauspieler</label><br/>
<input type="text" name="schauspieler" required><br/>
<br />
<button type="submit" name="update">Speichern</button>
</form>
<?php
include("back.php");
?>
</body>
</html>
错误信息:
错误更新记录:您的 SQL 语法有错误;检查与您的 MySQL 服务器版本相对应的手册,了解在第 2 行的 '('a','d') WHERE id=9' 附近使用的正确语法
非常感谢您的帮助,对于重复的问题很抱歉,但我真的找不到解决方案并且非常绝望.
Thanks a lot for your help and sorry for the duplicate question, but I really can't find the solution and am pretty desperate.
以下代码给出了这个错误:
The following code gives this error:
致命错误:未捕获的错误:在/homepages/25/d72758610/htdocs/multimedia/edit.php:30 中的 bool 上调用成员函数 bind_param() 堆栈跟踪:#0 {main} throw in/homepages/25/d72758610/htdocs/multimedia/edit.php 第 30 行
Fatal error: Uncaught Error: Call to a member function bind_param() on bool in /homepages/25/d72758610/htdocs/multimedia/edit.php:30 Stack trace: #0 {main} thrown in /homepages/25/d72758610/htdocs/multimedia/edit.php on line 30
<html>
<title>
Video Archiv - New
</title>
<body>
<?php
include("connect.php");
$id=$_GET['id'];
$title = $_POST["titel"];
$schauspieler = $_POST["schauspieler"];
if(empty($title))
{
echo "error";
}
elseif(empty($schauspieler))
{
echo "error";
}
else
{
$sql = "UPDATE users SET title=?, schauspieler=? WHERE id=?";
$stmt= $connect->prepare($sql);
$stmt->bind_param("ssi", $title, $schauspieler, $id);
if($stmt->execute())
{
echo "Succes";
}
else
{
echo "something went wromg";
}
}
?>
<form action="edit.php" method="post">
<label> Titel:</label><br/>
<input type="text" name="titel" required><br/>
<label>Schauspieler</label><br/>
<input type="text" name="schauspieler" required><br/>
<br />
<button type="submit" name="update">Speichern</button>
</form>
<?php
include("back.php");
?>
</body>
</html>
非常简单,可以避免 sql 注入并使用最新代码,并且您的 SQL 语法有错误.
Very simple to avoid sql injections and use up to date codes and You have an error in your SQL syntax.
这是一个例子:
include("connect.php");
$id=$_GET['id'];
$title = $_POST["titel"];
$schauspieler = $_POST["schauspieler"];
if(empty($title)){
echo "error";
}elseif(empty($schauspieler)){
echo "error";
}else{
$sql = "UPDATE VideoArchiv SET title=?, schauspieler=? WHERE id=?";
$stmt= $connect->prepare($sql);
$stmt->bind_param("ssi", $title, $schauspieler, $id);
if($stmt->execute()){
echo "Succes";
}else{
echo "something went wromg";
}
}
查看更多:https://phpdelusions.net/mysqli_examples/update
更新:第一个代码对你有用,但如果你仍然想使用程序化的方式,那么我们这个:
UPDATE : First code will work for you, but if you still want to use procedural way then us this :
include("connect.php");
if ($_SERVER["REQUEST_METHOD"] == "POST") {
//Check if we get id
$Testid = $_GET['id'];
if(empty($Testid)){
echo "id is empty";
}else{
$id = $_GET['id'];
}
$title = $_POST["titel"];
$schauspieler = $_POST["schauspieler"];
if(empty($title )){
echo "error". $title;
}elseif(empty($schauspieler)){
echo "error". $schauspieler;
}else{
$sql = "UPDATE VideoArchiv SET title=?, schauspieler=? WHERE id=?";
$stmt = mysqli_prepare($connect, $sql);
mysqli_stmt_bind_param($stmt, 'ssi', $title, $schauspieler, $id);
mysqli_stmt_execute($stmt);
}
}
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<label> Titel:</label><br/>
<input type="text" name="titel" required><br/>
<label>Schauspieler</label><br/>
<input type="text" name="schauspieler" required><br/>
<br />
<button type="submit" name="update">Speichern</button>
</form>
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