Question

我做了这个博客，我需要一点帮助，这是打印帖子的index.php，与他们的图片相同......

<?php
if (!isset($_SESSION['name'])) {
}else{

echo "<a href=\"edit.php?id=".$image."\"><strong>EDIT</strong>";
}
?>

这里是edit.php

 <?php
 $_GET['id']
 $sql = mysql_query("SELECT * FROM BlogData id");
$row = mysql_fetch_array($sql){
$title = $row['title'];
$content = $row['content'];
$category = $row['category'];
$image = $row['id'];
$extension= $row['ext'];

 ?>

 <center>
 <form action='edit.php' method='post' enctype="multipart/form-data">
    Title: <input type='text' name='title' placeholder="$title" /><br>
    Date: <input type='text' name='category' placeholder="$category" /><br>
 <textarea rows="10" cols="50" name='content'>$content</textarea><br>
    <?php 
echo "<img width='228' height='60' src='uploads/".$image.".".$extension."'>";
  ?>
Select image to upload:
     <input type="file" name="fileToUpload" id="fileToUpload">
 <br><br><br>
     <input type="submit" value="edit!" name="submit">
 </form>
 <img src="images/tutorialpost2.jpg">
 </center>

 ?>

所以我有了这个EDIT按钮，我希望它可以工作。它应该能够编辑我的所有MySQL表使用当前ID。我该怎么做？

Answer 1

if (isset($_SESSION['name'])) {
  <button>
}

使用“编辑”按钮编辑/更新mysql表

1 个答案: