Question

我有2页。一个来自我的msql数据库的结果，其中有一个更新来自和sql查询。我做了一个按钮，以便在尝试更新时获取该字段的ID，但是当我输入信息并提交没有任何反应时，它似乎只是刷新页面。

<?php

$db_host = '';
$db_user = '';
$db_pass = '';
$db_name = '';


$con = new mysqli($db_host, $db_user, $db_pass, $db_name);
  ?>
<html>
<form method="post" >
<input type="hidden" name="id" value="<?php echo $row['ID']?>">
<tr>        
  <td>Title</td>
<td>
<input type="text" name="title" size="20" value="<?php echo $row[title]?>">
</td>
</tr>
<tr>
  <td>Description</td>
<td>
<input type="text" name="description" size="40" value="<?php echo $row[description]?>">
</td></tr>
<tr>
<td align="right">
<input type="submit" name="submit" value="submit">
</td>
</tr>
</form>
</html>
<?php

if(isset($_POST["submit"])) {
$id = $_POST['id'];
$sql = "SELECT * FROM diary";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
if (isset($_POST[title])){
$title = mysql_real_escape_string(trim($_POST['title']));
}else{
$title = NULL;
}

if (isset($_POST[description])){
$description = mysql_real_escape_string(trim($_POST['description']));
}else{
$description = NULL;
}




$query = "UPDATE diary SET title='$title', post='$description' WHERE id= ?";           
 mysql_query($query);
if ($conn->query($sql) === TRUE) {
header("location: results.php");
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
?>

任何帮助都会非常适合

PHP更新表单使用mysqli不会更新

0 个答案: