php通过表格更新表 [英] php update table by form

问题描述

我尝试从表单更新表。

我有3页。第一个用编辑链接查询我表格中的所有行。
当点击编辑（第2页）时，代码将获取$ id并将其放入url中。 $ id从网址中提取，用于填充表单。

我的问题是将更新后的表单信息传递到我的表中。基本上没有更新。

第二页

 < ;？php 
 
包含'../db/config.php'; 
包含'../db/opendb.php'; 
 
 $ id = $ _GET [id]; 
 
 $ order =SELECT * FROM tableName where id ='$ id'; 
 $ result = mysql_query（$ order）; 
 $ row = mysql_fetch_array（$ result）; 
 
？> 
 
< form method =postaction =edit_data.php> 
< input type =hiddenname =idvalue =<？php echo$ row [id]？>>> 
< tr> 
< td>标题< / td> 
< td> 
< input type =textname =titlesize =20value =<？php echo$ row [title]？>>> 
< / td> 
< / tr> 
< tr> 
< td>发布< / td> 
< td> 
< input type =textname =postsize =40value =<？php echo 
$ row [post]？>> 
< / td>< / tr> 
< tr> 
< td align =right> 
< input type =submitname =submit valuevalue =Edit> 
< / td> 
< / tr> 
< / form> 
  

* 固定第三页

  include'../db/config.php'; 
包含'../db/opendb.php'; 
 
 if（isset（$ _ POST [id]））{
 $ id = mysql_real_escape_string（trim（$ _ POST ['id']））; 
} else {
 $ id = NULL; 
} 
 
 if（isset（$ _ POST [title]））{
 $ title = mysql_real_escape_string（trim（$ _ POST ['title']））; 
} else {
 $ title = NULL; 
} 
 
 if（isset（$ _ POST [post]））{
 $ post = mysql_real_escape_string（trim（$ _ POST ['post']））; 
} else {
 $ post = NULL; 
} 
 
 $ query =UPDATE tableName SET title ='$ title'，post ='$ post'WHERE id ='$ id'; 
 mysql_query（$ query）; 
 
？> 
  

提前致谢：）

*编辑谢谢大家

解决方案

  if（isset（$ _ POST [title]））{
 $ title = mysql_real_escape_string（trim（$ _ POST ['title']））; 
} else {
 $ title = NULL; 
} 
 
 if（isset（$ _ POST [post]））{
 $ post = mysql_real_escape_string（trim（$ _ POST ['post']））; 
} else {
 $ post = NULL; 
 
 
 
 $ query =UPDATE tableName SET title ='$ title'，post ='$ post'WHERE id ='$ id'; 
 mysql_query（$ query）; 
  

我还会推荐mysqli函数而不是mysql，我可能不会调用变量和表列'post'以避免混淆。

I'm trying to update a table from a form.

I have 3 pages. The first one queries all of the rows from my table with an "edit" link. When edit is clicked (page 2) the code pulls the $id and puts it in the url. The $id is pulled from the url and is used in a query to fill a form.

My problem is passing the updated form info to my table. Basically the update isn't happening.

Second page

<?php

  include '../db/config.php';
  include '../db/opendb.php';

$id = $_GET["id"]; 

  $order = "SELECT * FROM tableName where id='$id'";
  $result = mysql_query($order);
  $row = mysql_fetch_array($result);

  ?>

  <form method="post" action="edit_data.php">
  <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>Post</td>
    <td>
<input type="text" name="post" size="40" value="<?php echo
  "$row[post]"?>">
</td></tr>
<tr>
<td align="right">
<input type="submit" name="submit value" value="Edit">
</td>
    </tr>
</form>

* fixed third page

                   include '../db/config.php';
      include '../db/opendb.php';

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

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

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

$query = "UPDATE tableName SET title='$title', post='$post' WHERE id='$id'";           
mysql_query($query); 

?>

thanks in advance : )

*edit thanks everyone

解决方案

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

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


$query = "UPDATE tableName SET title='$title', post='$post' WHERE id='$id'";           
mysql_query($query); 

I would also recommend mysqli functions instead of mysql and I probably wouldn't call a variable and table column 'post' to avoid confusion.

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