在非对象上调用成员函数 bind_param().mysqli

这是我的 php 代码:

include ("dbinfo.php");if(isset($_POST['editsave'])){$edittitle=$_POST["edittitle"];$editurl=$_POST["editurl"];$editdesc=$_POST["editdesc"];$editid=$_POST["editid"];$mysqli = $GLOBALS['dbc'];$stmt = $mysqli->prepare("更新链接 SET title = ?,描述 = ?哪里 id = ?");$stmt->bind_param('ssd',$_POST['edittitle'],$_POST['editdesc'],$_POST['editid']);$stmt->执行();$stmt->close();}

这是我的表格:

 <form role="form" action="edit.php" method="post"><div class="form-group"><label for="title">标题</label><input type="hidden" name="editid" value="<?echo $id; ?>"><button type="submit" name="editsave" class="btn btn-primary">保存更改</button></div></表格>

当我按下提交时，我得到这个:

致命错误:在第 27 行的/storage/content/x/xxx/中的非对象上调用成员函数 bind_param().

第 27 行是:

$stmt->bind_param('ssd',

我不熟悉mysqli.我已经尝试解决这个问题几天了，但我快疯了.

解决方案

这表示里面的查询 mysqli::prepare() 导致错误.

根据文档:

<块引用>

mysqli_prepare() 返回一个语句对象，如果发生错误则返回 FALSE.

尝试正确转义 desc，这是 MySQL 中的保留关键字( 不是正确转义):

$stmt = $mysqli->prepare('UPDATE links SET title = ?, `desc` = ? WHERE id = ?');

This is my php code:

include ("dbinfo.php");

if(isset($_POST['editsave'])){
$edittitle=$_POST["edittitle"];
$editurl=$_POST["editurl"];
$editdesc=$_POST["editdesc"];
$editid=$_POST["editid"];

$mysqli = $GLOBALS['dbc'];
$stmt = $mysqli->prepare("UPDATE links SET title = ?, 
  desc = ? 
  WHERE id = ?");
$stmt->bind_param('ssd',
  $_POST['edittitle'],
  $_POST['editdesc'],
  $_POST['editid']);
$stmt->execute();
$stmt->close();
}

This is my form:

    <form role="form" action="edit.php" method="post">
    <div class="form-group">
      <label for="title">Title</label>
      <input type="text" name="edittitle" class="form-control" id="title" placeholder="Enter title" maxlength="70" value="<?php echo ($title); ?>">
    </div>
    <div class="form-group">
      <label for="url">URL</label>
      <input type="text" name="editurl" class="form-control" id="url" value="<?php echo ($url); ?>" disabled>
    </div>
    <div class="form-group">
      <label for="desc">Description</label><small> (max 500 characters)</small>
      <textarea class="form-control" name="editdesc" id="desc" rows="5" maxlength="500"><?php echo ($desc); ?></textarea>
    </div>
    <div>
      <input type="hidden" name="editid" value="<? echo $id; ?>">
      <button type="submit" name="editsave" class="btn btn-primary">Save changes</button>
    </div>
  </form>

When i press submit i get this:

Fatal error: Call to a member function bind_param() on a non-object in /storage/content/x/xxx/ on line 27.

Line 27 is:

$stmt->bind_param('ssd',

I'm not familiar with mysqli. I have tried to fix the problem for a few days now and I'm getting crazy.

解决方案

This means that the query inside mysqli::prepare() resulted in an error.

According to the doc:

mysqli_prepare() returns a statement object or FALSE if an error occurred.

Try to properly escape desc, which is a reserved keyword in MySQL ( is not proper escaping):

$stmt = $mysqli->prepare('UPDATE links SET title = ?, `desc` = ? WHERE id = ?');